A-PDF Merger DEMO : Purchase from www.A-PDF.com to remove the watermark
ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT MATHEMATICS Higher 2
9740 / 01
Paper 1
19 August 2010
JC 2 PRELIMINARY EXAMINATION Time allowed: 3 hours Additional Materials: List of Formulae (MF15) READ THESE INSTRUCTIONS FIRST Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in. Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in the question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages.
[Turn Over
ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010 MATHEMATICS 9740 Higher 2 Paper 1
/ 100
Index No: Form Class: ___________ Name: _________________________ Calculator model: _____________________
Arrange your answers in the same numerical order. Place this cover sheet on top of them and tie them together with the string provided.
Question no.
Marks
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1 Page 2 of 6
1
n +1
The nth term of a sequence is given by un = ( −1) n 2 , for n ≥ 1 . The sum of the first n terms is denoted by Sn. Use the method of mathematical induction to show that n +1 n ( n + 1) for all positive integers n. [4] Sn = ( −1) 2
2
The 3 flavours of puddings produced by a dessert shop are mango, durian and strawberry. A mango pudding requires 5g of sugar and 36ml of water. A durian pudding requires 6g of sugar and 38ml of water. A strawberry pudding requires 4g of sugar and 40ml of water. The puddings are sold in pairs of the same type at $1.60, $2.20 and $1.80 for mango, durian and strawberry respectively. On a particular day, 754g of sugar and 5972ml of water were used to make the puddings and all the puddings made were sold except for a pair of strawberry puddings. The collection from the sale of puddings was $142.40. Formulate the equations required to determine the number of each type of pudding made on that day. [4]
3
The diagram below shows the graph of y = f(x) . The curve passes through the origin and has a maximum point at A ( 4 , 4 ) and asymptotes x = −2 and y = 2 . y
( 4 , 4)
y = f(x)
A
2
x
−2
Sketch on separate diagrams, the graphs of [3] (i) y = 1 f(x) (ii) y = f '(x) [3] showing clearly asymptotes, intercepts and coordinates of turning points where possible.
4
2π . Find the modulus and argument of 3 −i −i , where w* is the complex conjugate of w. Hence express in the form a + ib , where w* w* a and b are real, giving the exact values of a and b in non-trigonometrical form. [4]
The complex number w has modulus 3 and argument
n
−i Find the possible values of n such that is purely imaginary. w*
[2] [Turn Over
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1 Page 3 of 6
Solve the equation z 4 − i =0 , giving the roots in the form reiα , where r > 0 and [3] −π < α ≤ π . The roots represented by z1 and z2 are such that arg ( z1 ) > arg ( z2 ) > 0 . Show z1 , z2 and
5
z1 + z2 on an Argand diagram. Deduce the exact value of arg ( z1 + z2 ) .
6
[3]
An economist is studying how the annual economic growth of 2 countries varies with time. The annual economic growth of a country is measured in percentage and is denoted by G and the time in years after 1980 is denoted by t. Both G and t are taken to be continuous variables. (i) Country A is a developing country and the economist found that G and t are can be modeled dG G + 1 by the differential equation . Given that, when t = 0 , G = 0 , find G in terms of t. = dt 2 [4] (ii) Comment on the suitability of the above differential equation model to forecast the future economic growth of Country A. [1] (iii) Country B is a developed country and the economist found that G and t can be modeled by dG G +1 the differential equation = − . dt 2 Given that Country B has been experiencing decreasing economic growth during the period of study, sketch a member of the family of solution curves of the differential equation model for Country B. Hence, comment on the economic growth of Country B in the long term. [2]
7 (i) Given that f ( x ) = ecos
−1
x
, where −1 ≤ x ≤ 1 , find f ( 0 ) , f ′( 0 ) and f ′′( 0 ) . Hence write down
the first three non-zero terms in the Maclaurin series for f ( x ) . Give the coefficients in terms [4] of e kπ , where k ∈ . 3 (ii) Given that g= ( x ) tan x + sec x , where x is sufficiently small for x and higher powers of x to be neglected. Deduce the first three non-zero terms in the series expansion of g ( x ) . π
π
π
Hence, show that f ( x ) + e 2 g ( x ) ≈ 2e 2 + x 2 e 2 . (iii) Explain clearly why it is inappropriate to state that where a ∈ .
8 (i)
[3]
∫
a
−a
π
π
π
f ( x ) + e 2 g ( x ) dx ≈ ∫ 2e 2 + x 2 e 2 dx , a
−a
[1]
Ar 2 + Br + C , where A, B and C are constants to be Show that 1 − 2 + 1 = r ! ( r + 1) ! ( r + 2 ) ! ( r + 2 )! found. [2] n
(ii) Hence find
3r 2 + 3r − 3 . ∑ ( r + 2 )! r =1
[3] ∞
(iii) Give a reason why the series
3r 2 + 3r − 3 converges, and write down its value. ∑ r + 2 )! r =0 (
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1 Page 4 of 6
[2]
Relative to the origin O, two points A and B have position vectors given by a = 3i + j + 3k and b = 5i − 4 j + 3k respectively. (i) Find the length of the projection of OA on OB . [2] (ii) Hence, or otherwise, find the position vector of the point C on OB such that AC is perpendicular to OB. [2] (iii) Find a vector equation of the reflection of the line AB in the line AC. [3]
9
10(a) The first 2 terms of a geometric progression are a and b ( b < a ). If the sum of the first n terms is equal to twice the sum to infinity of the remaining terms, prove that a n = 3b n . [3] (b) The terms u1 , u2 , u3 ,... form an arithmetic sequence with first term a and having non-zero common difference d. (i) Given that the sum of the first 10 terms of the sequence is 105 more than 10u5 , find the common difference. [3] (ii) If u26 is the first term in the sequence which is greater than 542, find the range of values of a. [3]
11
The region R in the first quadrant is bounded by the curve = 2 y 2 a ( 2a − x ) , where a > 0 , and the line joining ( 2a, 0 ) and ( 0, a ) . The region S, lying in the first quadrant, is bounded 2 y 2 a ( 2a − x ) and the lines x = 2a and y = a . by the curve = (i) Draw a sketch showing the regions R and S. [1] (ii) Find, in terms of a, the volume of the solid formed when S is rotated completely about the x-axis. [4] (iii) By using a suitable translation, find, in terms of a, the volume of the solid formed when R is [4] rotated completely about the line x = 2a .
12
13
2 The curve C has the equation y = 3x + ax + 2 where a is a constant. x+a dy (i) Find and the set of values of a if the curve has 2 stationary points. [4] dx (ii) Sketch the curve C for a = 1, stating clearly the exact coordinates of any points of intersection with the axes and the equations of any asymptotes. [3] 2 Hence, find the range of values of k such that the equation 3x + x + 2= k(x + 1) has exactly 2 real roots. [2]
The curve has the parametric equations x = 5 2 , y = tan −1 t 1+ t (i) Sketch the curve for −2 ≤ t ≤ 2 . [1] (ii) Find the cartesian equations of the tangent and the normal to the curve at the point where t = 1 . [5] (iii) Find the area enclosed by the x-axis, the tangent and the normal at the point where t = 1. [3] [Turn Over Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1 Page 5 of 6
14
The functions f and g are defined as follows: f : x sin x ,
x∈ . ( x + 1)( 3 − x ) , 8 Sketch the graph of the function g, labeling clearly the exact values of the coordinates of turning point(s) and intersections with the axes, if any. [1] State the range of the function g in exact values. [1]
g:x
(i)
π
π π x ∈ − , , 2 2
(ii) Given that gf exists as a function. By considering the graphs of f and g, explain why gf (α ) ≠ gf ( β ) if −
π
≤α < β ≤
π
. 2 2 Hence what can be said about the function gf ?
[2] [1]
Without sketching the graph of gf , find the range of gf in the form [ a, b ] , giving the exact values of a and b. [1]
(iii) (a) Give a reason why fg does not exist as a function. [1] (b) Find the greatest exact value of k for which fg is a function if the domain of g is restricted to the interval [1, k ] . - End of Paper -
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1 Page 6 of 6
[2]
ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT MATHEMATICS Higher 2
9740 / 02
Paper 2
23 August 2010
JC 2 PRELIMINARY EXAMINATION Time allowed: 3 hours Additional Materials: List of Formulae (MF15) READ THESE INSTRUCTIONS FIRST Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in. Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in the question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 5 printed pages.
[Turn Over
ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010 MATHEMATICS 9740 Higher 2 Paper 2
/ 100
Index No: Form Class: ___________ Name: _________________________ Calculator model: _____________________
Arrange your answers in the same numerical order. Place this cover sheet on top of them and tie them together with the string provided. Question no.
Marks
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2 Page 2 of 5
Section A: Pure Mathematics [40 marks] 1
1
Find the exact value of
∫
−1
2
e2 x −
1 e
2 ( x −1)
[4]
dx.
The variable complex numbers z and w are such that z − 2 − i = 3 and arg ( w − 5 + 3i ) = π. (i) Illustrate both of these relations on a single Argand diagram. [2] (ii) State the least value of z − w . [1] (iii) Find the greatest and least possible values of arg ( z + 3) , giving your answers in radians correct to 3 decimal places. [4]
3
a Find in terms of a, the range of values of x that satisfy the inequality ln 2 x − ≥ 0 , x [4] where a > 1 .
4 (a) State the derivative of cos x3 . Hence, find 2 5
(b) Find the exact value of
∫ (x
2
− 5)
∫x −
3 2
5
sin x3 dx.
[4]
dx , in the form a 2 + b 3 , using the
10
substitution x = 5 sec θ . 5
[6]
x +1 z −1 . = y= 2 4 (i) Show that the line l1 is parallel to, but not contained in the plane p1. [2] (ii) Find the cartesian equation of the plane p2 which contains the line l1 and is perpendicular to the plane p1. [3] (iii) Find, in scalar product form, the vector equation of the plane p3 which contains the point [2] ( 4,1, −1) and is perpendicular to both p1 and p2.
The plane p1 has equation x + 2 y − z = 3 . The line l1 has equation
2 Another line l2 which is parallel to the vector 0 intersects the line l1 at the −3 point A ( −1, 0,1) . (iv) Given that the line l2 meets the plane p1 at the point B, find the coordinates of B. [4] (v) Find the sine of the acute angle between the line l2 and the plane p1, and hence, find the length of the projection of the line segment AB on the plane p1, giving your answer in surd form. [4] Section B: Statistics [60 marks] 6
Mr Raju, who owns a supermarket wishes to find out what customers think about the goods that he sells. He has been advised that he should take a random sample of his customers for this purpose. State, with reasons, which of the following sampling procedures is preferable. A. Select every 10th customer on each day in a typical week. B. Select the first 20 customers on each day in a typical week [2] [Turn Over Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2 Page 3 of 5
7
One day, Pinocchio went shopping and bought a pair of size 30 Wood Shoes. The right shoes have lengths which are normally distributed with mean 20 cm and standard deviation 0.14 cm. The left shoes have lengths which are normally distributed with mean 20.1 cm and standard deviation 0.11 cm. The length of the right shoe is independent of the length of the left shoe. When wearing the pair of shoes, Pinocchio takes six steps, heel to toe as shown in the diagram. Calculate the probability that the distance AB, from the back of the first step to the front of the sixth step, exceeds 120 cm. A
B
[3]
8
On Ulu Island the weights of adult men and women may both be taken to be independent normal random variables with means 75kg and 65 kg and standard deviations 4 kg and 3 kg respectively. Find the probability that the weight of a randomly chosen man and the weight of a randomly chosen woman differ by more than 1 kg. [3] Explain if this is equal to the probability that the difference in weight between a randomly chosen married woman and her husband is more than 1kg. [1]
9
Research has shown that before using an Internet service, the mean monthly family telephone costs is $72. A random sample of families which had started to use an Internet service was taken and their monthly telephone costs were : $70, $84, $89, $96, $74 Stating a necessary assumption about the population, carry out a test at the 5% significance level, whether there is an increase in the mean monthly telephone costs. [5] If the assumption stated above still holds, and if the standard deviation of the monthly telephone costs is $9.89, find the range of values of the mean monthly family telephone costs µ 0 that would lead to a reverse in the decision to the above test. [3]
10
Six overweight men registered at a slimming centre for a slimming programme. The following table records x, the height (to the nearest cm) and y, the weight (to the nearest 0.1 kg) of these six men. Man x (height in cm) y (weight in kg)
A 150 65.1
B 157 73.2
C 160 85
D 162 k
E 167 80.9
F 170 89.9
Given that the least square regression line of x on y line = is x 103.6 + 0.726 y , show that the value of k to the nearest 0.1 kg is 80.3. Hence or otherwise, find the least square regression line of y on x in the form = y ax + b , giving the values of a and b to the nearest 3 decimal places. [5] (ii) Based on the data given, use an appropriate regression line to predict the weight of an overweight man who is 165 cm tall. [2] (iii) Find the value of the product moment correlation coefficient between x and y and sketch the scatter diagram of y against x. A particular man among the 6 men who registered for the slimming programme is unusually overweight. Indicate who this man is. [3]
(i)
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2 Page 4 of 5
11
In a hotel, large number of cups and saucers are washed each day. The number of cups that are broken each day while washing averages 2.1. State in context, a condition under which a Poisson distribution would be a suitable probability model. [1] Assume that the number of broken cups and saucers follow a Poisson distribution. (i) Show that on any randomly chosen day, the probability that at least 3 cups are broken is 0.350 correct to 3 significant figures. [1] The probability that there will be at least two days in n days with at least 3 broken cups is more than 0.999. Find the least value of n. [3] (ii) The number of saucers broken each day averages 1.6, independently of the number of cups broken. The total number of cups broken and saucers broken during a week of 7 days is denoted by T. State a possible model for the distribution of T. [2] A random sample of 100 weeks is chosen. Using a suitable approximation, find the probability that the average weekly total number of broken cups and saucers does not exceed 26. [3]
12
Fish are bred in large batches and allowed to grow until they are caught at random for sale. When caught, only 20% of the fish measure less than 8 cm long. (i) What is the probability that the 10th fish caught is the sixth fish that is less than 8 cm long? [2] (ii) A large number, n, of fish are caught and the probability of there being 10 or fewer fish in the catch which measures less than 8 cm long is at most 0.0227 . Using a suitable approximation, derive the approximate inequality 10.5 − 0.2n ≤ −0.8 n . [4] Hence find the least possible number of fish to be caught. [2]
13
An automated blood pressure machine is being tested. Members of the public, p % of whom have high blood pressure (hypertension), try it out and are then seen by a doctor. She finds that 80% of those with hypertension and 10% of those with normal blood pressure have been diagnosed as hypertensive by the machine. The probability that a randomly chosen patient who was diagnosed as hypertensive by the machine actually has hypertension is 2 . 3
(i) Find the value of p [3] (ii) Hence, find the probability that a randomly chosen patient does not have hypertension, given that the machine diagnosed him as having normal blood pressure. [2] Comment on the usefulness of the machine. [1] 14
Ten balls are identical in size and shape of which 2 are red, 3 are blue and 5 are green. The two red balls are labeled ‘1’ and ‘2’, the three blue balls are labeled ‘1’, ‘2’ and ‘3’, and the five green balls are labeled ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’. (i) Find the number of ways of choosing 2 balls of identical colour. [2] (ii) Find the number of ways of choosing 6 balls if it includes at least one ball of each colour. [4] (iii) A person arranged 3 balls in a row with the numbered sides facing him forming a 3-digit number. Among these 3 balls, none of them are green. Find the number of possible 3-digit numbers facing that person. [The number formed is independent of the colours of the balls used. i.e. the number 112 is counted as one number whether the colour of the ball labeled ‘2’ is red or blue.] [3] - End of Paper Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2 Page 5 of 5
Anglo-Chinese Junior College H2 Mathematics 9740 2010 JC 2 PRELIM Marking Scheme Paper 1: 1
Let Pn denote the statement Sn = ( −1)
n +1
n ( n + 1) 2
LHS = S1 = u1 = ( −1) 12 = 1 RHS = ( −1) 2
For n = 1, LHS = RHS ∴ P1 is true.
y
y = f '(x)
1( 2 ) =1 2
k ( k + 1) 2 k + 2 ( k + 1)( k + 2 ) i.e. S k +1 = ( −1) 2
Assume Pk true for some k ∈ ℤ + , Prove that Pk+1 is true,
2
3 (ii)
i.e. Sk = ( −1)
k +1
(4,0) −2
x
A’
LHS = S k +1 = S k + uk +1 k ( k + 1) k +2 2 + ( −1) ( k + 1) 2 k +1 ( k + 1) = ( −1) k − 2 ( k + 1) 2 k +1 ( k + 1) = ( −1) [ −k − 2] 2 k + 2 ( k + 1)( k + 2 ) = ( −1) = RHS 2 Since P1 is true, and Pk is true ⇒ Pk+1 is true, by the principle of mathematical induction, Pn is true ∀ n ∈ ℤ + . Let x be the no of mango puddings produce. Let y be the no of durian puddings produce. Let z be the no of strawberry puddings produce. 5 x + 6 y + 4 z = 754 ……………………………(1) 36 x + 38 y + 40 z = 5972 ………………………(2) 0.8 x + 1.1y + 0.9( z − 2) = 142.4 ……………….(3) Solving (1), (2) & (3) using GC, x = 46, y = 42 and z = 68. y y= 1 f(x) = ( −1)
2
3 (i)
k +1
4
−i 1 = w* 3 π 2π −i arg = arg(−i ) − arg( w*) = − − − 2 3 w*
−i 1 π π 1 3 1 3 1 = cos + i sin = + i = + i w* 3 6 6 3 2 2 6 6 n
n
nπ nπ −i 1 + i sin = cos 6 6 w* 3 n
nπ −i =0 is purely imaginary, cos 6 w*
π
π
= ( 2k + 1) , k ∈ ℤ , 6 2 ∴ n = 3 ( 2k + 1) , k ∈ ℤ . n
5
z 4 − i = 0 ⇒ z 4 = i ⇒ z 4 = 1e π i + 2 kπ , e2
z4 =
z=e z=
y = 0.5
1 π i + 2 kπ 4 2
i
π 2
k = −2, −1, 0,1
, k = −2, −1, 0,1
7π 3π π 5π i − i − i i e 8 ,e 8 ,e 8 ,e 8
y z2 + z2
A’ (4, 0.25)
z1
x
−2
π = 6
5π 8 O
x=0 1
π
z2
x
8
2
z1 =
5π i e 8
z2 =
π i e8
−1 −1 f ′′( x ) = 2 2 1− x 1− x =
arg ( z1 + z2 ) =
π 8
1 4π + 2 8
6
(
dG G + 1 = dt 2 1 1 dG ∫ G + 1 = ∫ 2 dt ln G + 1 = 0.5t + C
(
)
−1 1 cos−1 x − xecos x (1 − x e 1 − x2
3π = 8 (ii)
−3 cos−1 x −1 −1 + −ecos x (1 − x 2 ) 2 ( −2 x ) e 2
π
)
π
−3 2 2
)
⇒ f ′′ ( 0 ) = e 2
π
π
∴f ( x) = e 2 − e 2 x +
e2 2 x + ... 2
g ( x ) = tan x + sec x −1
= tan x +
x2 x2 1 ≈ x + 1 − = x + 1 + ( −1) − + ... cos x 2 2
≈ 1+ x +
x2 2
G + 1 = ±e 0.5t +C G = −1 + Ae0.5t , where A = ±eC When t = 0 , G = 0 , A =1 ∴ G = −1 + e0.5t
(iii)
π
Examples of possible comments: The model is not suitable because … The economist is assuming that that there are no fluctuations in the economic growth in the future. The economist is assuming that the country will enjoy perpetual economic growth in the long term. The economist is assuming Country A is always experiencing positive and increasing economic growth in the future. Factors affecting economic growth remains unchanged.
π
= 2e 2 + e 2 x 2 (Shown) π
a
a
π
π
The statement ∫ f ( x ) + e 2 g ( x ) dx ≈ ∫ 2e 2 + x 2 e 2 dx is inappropriate as −a
π
−a
π
π
f ( x ) + e 2 g ( x ) ≈ 2e 2 + x 2 e 2 only when a is sufficiently small.
8 (i)
G
1− 2 + 1 r ! ( r + 1) ! ( r + 2 ) ! =
t
0
π π π π e 2 2 π2 x2 f ( x) + e 2 g ( x) ≈ e 2 − e 2 x + x + e 1 + x + 2 2
G = −1 + Be −0.5t , B = 1
( r + 1)( r + 2 ) − 2 ( r + 2 ) + 1 ( r + 2 )!
2 = r + r −1 ( r + 2 )! A = 1, B = 1, C = −1.
−1
(ii)
n
2
∑ 3r( r++32r)−! 3 r =1
n
2 = 3∑ r + r − 1 r + 2 )! ( r =1
In the long term, Country B is expected to be still in recession with an economic growth decreasing towards -1%.
7(i)
f ( x ) = ecos f ′( x ) =
−1
n = 3∑ 1 − 2 + 1 r ! ( r + 1) ! ( r + 2 ) ! r =1
π
⇒ f (0) = e 2
x
−1 1− x
2
ecos
−1
π
x
⇒ f ′ ( 0 ) = −e 2
3
4
(iii)
1 2 1 1! − 2! + 3! 1 2 1 + − + 2! 3! 4! 1 2 1 + − + 3! 4! 5! 1 2 1 + − + 4! 5! 6! = 3 ⋮ 1 2 1 − + + ( n − 2 ) ! ( n − 1) ! n ! + 1 − 2 + 1 ( n − 1) ! n ! ( n + 1) ! 1 2 1 + + − n ! ( n + 1) ! ( n + 2 ) ! n
5λ − 3 5 − 4λ − 1 . − 4 = 0 3λ − 3 −3 2 λ= 5
5 2 OC = −4 5 3 (iii)
= 3 1 − 1 + 1 2 ( n + 1) ! ( n + 2 ) !
(
∑ 3r( r++32r)−! 3
( ) = ( − 3 ) + 3 1 − 1 + 1 2 2 ( n + 1) ! ( n + 2 ) ! n
2 = − 3 + ∑ 3r + 3r − 3 2! r =1 ( r + 2 ) !
Length of the projection of OA on OB 3 5 4 1 20 = 1 ⋅ −4 = 50 = 2 2 or 2 3 50 3
Method 1:
O
C
10 (a)
From (i), OC = 2 2
A
B
5 1 OC = 2 2 −4 = 50 3
)
5 3 20 1 1 AB ' = OB ' − OA = − −4 − 1 = − 1 5 5 3 3 18
1 → 0 and 1 → 0. ( n + 1)! ( n+2 )! ∴ the series converges to 0.
As n → ∞,
(ii)
5 1 1 OB ' = − OB = − −4 5 5 3
Or use midpoint theorem, 1 OC = OB ' + OB 2 5 5 5 1 2 OB ' = 2OC − OB = 2 −4 − −4 = − −4 5 5 3 3 3
2
r =0
9(i)
2 Since OC = OB , OC : CB = 2 : 3 5
5 2 −4 5 3
3 20 Vector equation of line AB ' is r = 1 + α 1 , α ∈ ℝ ɶ 18 3 GP : a = a, r = b a n The sum to infinity of the remaining terms, S∞ = ar 1− r Sn = 2S∞
(
)
n a 1− rn = 2ar 1− r 1− r 1 − r n = 2r n
Method 2:
0 5 Line OB: r = 0 + λ −4 0 3 5 3 5λ − 3 AC = OC − OA = λ −4 − 1 = −4λ − 1 3 3 3λ − 3 Since AC ⊥ OB ,
3r n = 1 rn = 1 3
( ba ) = 13 n
3b n = a n (b) (i)
5
6
S10 = 105 + 10u5
(ii)
(i)
10 2a + 9d = 105 + 10 a + 4d ] [ ] 2[ 10a + 45d = 105 + 10a + 40d d = 21
2 y = 3x + ax + 2 x+a
=
u25 ≤ 542
u26 > 542
a + 24(21) ≤ 542 a ≤ 38
a + 25(21) > 542 a > 17
and
(
( 6a ) 2 − 4 ( 3 ) ( a 2 − 2 ) > 0
S
R
24a 2 > −24
x O
a 2 > −1 a 2 is always positive. ∴ a ∈ ℝ.
2a
When S is rotated completely about the x-axis, 2a a Required volume = π a 2 ( 2a ) − π ∫ ( 2a − x ) dx 0 2
= 2π a 3 −
)
For 2 stationary points,
a
= 2π a 3 −
)
( x + a )2
3x 2 + 6ax + a 2 − 2 = 0
y
2 y 2 = a ( 2a − x )
(ii)
(
3x 2 + 6ax + a 2 − 2
)
For stationary points, dy =0 dx
∴17 < a ≤ 38
11 (i)
(
2 dy ( x + a )( 6x + a ) − 3x + ax + 2 = 2 dx ( x + a)
2 π a ( 2a − x )
2
πa
−2
(ii)
2 y = 3x + x + 2 = 3x − 2 + 4 x +1 x +1
2a
0
( )
2a 2 2 = π a 3 cu. units
Points of intersection with the axes : (0,2)
(iii)
Asymptotes : y = 3x − 2
After a translation of 2a units in the negative x-direction, 2 y2 New equation is 2 y 2 = a ( 2a − ( x + 2a ) ) ⇒ x = − a When R is rotated completely about the line x = 2a ,
x = −1 .
The range of values of k such that the equation 3x 2 + x + 2 = k(x + 1) has exactly 2 real roots : k > 1.93 or k < −11.9
2
2 a 2y 1 2 Required volume = π ( 2a ) ( a ) − π ∫ − dy 0 3 a a
13 (i)
4 y5 2 5a 0 4 3 4 3 8 = π a − π a = π a 3 cu. units 3 5 15 4 = π a3 − π 3
and
12 7
8
and g : x ֏
π 8
( x + 1)(3 − x), x ∈ ℝ .
(i)
5
y
π (ii)
dy = 1 dt 1 + t 2
3π
dx = -5 ( 2t ) 2 dt 1+ t2
and
(
)
dy dy dt = . dx dt dx =
(
2
)
1 . 1+t 1 + t 2 −10t
-1
2
(1, π ) 2
2 8
y = g ( x)
1
0
3
x
(1 + t ) 2
=−
10t
π
When t = 1, dy = −1 dx 5
(iii)
(ii)
Rg = (−∞, ] 2
Equation of tangent : y−π 4 = −1 5 x− 5 2 y = −1 x+ 1 + π 5 2 4
y
)
(
Gradient of normal is 5
−
Equation of normal: y−π 4 =5 x− 5 2 y = 5x − 25 − π 2 4
(
y = f ( x)
1 f (α )
π
α
β
0
2
π 2
f (β ) -1
x
From the graph of y = f ( x) ,
π
y
)
(( ) (
π
≤α < β ≤ . 2 2 In fact, −1 ≤ f (α ) < f ( β ) ≤ 1 . f (α ) ≠ f ( β ) if −
y = g ( x)
π 2 3π 8
))
Area = 1 × 5 1 + π − 1 25 − π × π 2 2 4 5 2 4 4 2
= 13π = 1.60 (3s.f ) 80 -1
14
Functions
From the graph of y = g ( x) , gf (α ) < gf ( β )
π π Given f : x ֏ sin x, x ∈ − , 2 2
or gf (α ) ≠ gf ( β ) 9
0
1
x
10
if −1 ≤ f (α ) < f ( β ) ≤ 1 .
gf is one-one (or increasing).
π
Rgf = [0, ] . 2
(iii) (a)
π
π π
Rg = (−∞, ] ⊄ [− . ] = D f . 2 2 2 Alternative:
5π 5π π π ] is undefined because − ∉ [− , ] = D f . 8 8 2 2 Hence, fg does not exist as a function. fg (4) = f [ g (4)] = f [−
y
(iii) (b)
(1, π ) 2 y = g ( x)
-1
0
1
3
x y=−
g ( x) =
π
π 2
π 2
( x + 1)( 3 − x ) = −
π
8 2 ( x + 1)( 3 − x ) = −4
− x 2 + 2 x + 3 = −4 x2 − 2 x − 7 = 0 2 ± 32 x= 2 x = 1± 8 Since x ≥ 1, x = 1 + 8 . 1 ≤ k ≤ 1+ 8 Greatest value of k is 1 + 8 .
11
12
Anglo-Chinese Junior College H2 Mathematics 9740 2010 JC 2 PRELIM Marking Scheme
∫ x sin x dx = ∫ x ( x sin x ) dx 5
3
Paper 2: 1
1
3
∫
e2 x −
−1 1
=−
∫
1
2
−1
e2 x −
1 e
2 ( x −1)
1
dx +
∫
1
e2 x − 2
1 dx e 2 ( x −1)
1
1
1 1 1 2 1 = − e 2 x + e−2 ( x −1) + e 2 x + e −2 ( x −1) 2 2 2 −1 2 12 =
4 (b)
1 4 2 ( e + e − 4e + e−2 + 1) 2
2 (i)
3
1 1 = x 3 − cos x3 − ∫ − cos x3 3 3 x3 = − cos x3 + ∫ x 2 cos x3 dx 3 1 x3 = − cos x3 + sin x3 + c 3 3
dx
e2 ( x −1)
2
y
when x = 10, θ =
dx = 5 sec θ tan θ dθ
when x = 2 5, θ =
O
2
π
=
1 5
∫π π
=
2 (ii)
2 (iii)
3
4 (a)
2
θ − 5)
π 3
dx −
3 2
5 sec θ tan θ dθ
4
(5,−3) o
w
3 2
3
∫π ( 5sec
=
x
3
−
− 5)
2
4
10
π
i (2,1)
3 −3
∫ (x
z
26
π
x = 5 sec θ
2 5
3
2 3 x dx
Least value of z − w = 1
1 5
∫π
3
sec θ dθ tan 2 θ
4
3
π
cos θ dθ sin 2 θ
or
1 5
∫π
4
3
cot θ cos ec θ dθ
4
π
−1 1
1 1 3 = − 5 sin θ π 4
−1
Greatest arg ( z + 3) = tan + sin 5 −1 1 Least arg ( z + 3) = tan − sin −1 5
3 = 0.826 (3 dp) 26 3 = −0.432 (3 dp) 26
=
a ln 2 x − ≥ 0 where a > 1 . x a 2x − ≥ 1 x 2 x2 − x − a ≥0 x 1 + 1 + 8a ) 1 − 1 + 8a ) 2 x − x − 4 4 ≥0 x
5
(i)
1 2 2− 3 5 15
i.e. a =
1 p1 : r ⋅ 2 = 3 ɶ −1 2 1 1 . 2 = 2 + 2 − 4 = 0 4 −1
⇒ l1
π 1 [ − cos ec θ ]π 34 5
=
is parallel to p1.
1 5
b=−
2 15
−1 2 l1 : r = 0 + λ 1 , λ ∈ ℝ ɶ 1 4 −1 1 0 . 2 = −1 + 0 − 1 ≠ 3 1 −1
⇒ l1
is not contained in p1.
Alternative method: −1 + 2λ 1 λ i 2 = −2 ≠ 3 1 + 4λ −1
Since no solution for λ , ∴ l1 is parallel and not contained to p1
1 − 1 + 8a 1 + 1 + 8a ≤ x < 0 or x ≥ 4 4 d cos x3 ) = − 3 x 2 ( sin x 3 ) ( dx
(ii)
1
1 2 −3 2 × 1 = −3 2 −1 4 1 2
(iii)
(iv)
(v)
−3 −1 −3 p2 : r ⋅ 2 = 0 i 2 = 3 + 0 + 1 = 4 ɶ 1 11
p2 : −3x + 2 y + z = 4
2 4 2 p3 : r ⋅ 1 = 1 ⋅ 1 = 8 + 1 − 4 = 5 ɶ 4 −1 4
2 p3 : r ⋅ 1 = 5 ɶ 4
9
1 −1 2 l2 : r = 0 + µ 0 , µ ∈ ℝ p1 : r ⋅ 2 = 3 ɶ ɶ −3 1 −1 −1 + 2 µ 1 0 . 2 = 3 ⇒ − 2 + 5µ = 3 ⇒ µ = 1 ∴ point B is (1, 0, −2 ) 1 − 3µ −1
sin θ =
No , (i) The weight of a husband and wife may not be independent Or (ii) Randomness is not there ( a randomly chosen women but spouse is not randomly chosen) Or (iii) Distribution of weight of married woman is different from distribution of adult woman. Etc Telephone costs are assumed to be normally distributed.
To test H0 : µ = 72 against H1: µ > 72 at 5% level of significance _
Under H0 , T=
2 1 1 1 5 ⋅ 0 2 = 13 6 78 −3 −1
x− µ 0 s
n
t(5-1)
_
Test statistics : T=
x− µ 0 s
n
=
82.6 − 72 = 2.2219 10.6677 / 5
p value = P(T > 2.2219) = 0.0452 < 0.05
Reject H0 at the 5% level of significance. We conclude that there is sufficient evidence at the 5% level of significance that there is evidence of an increase in mean monthly costs.
Length of the projection of AB on p1 = AB cos θ
2 53 25 53 1 = 0 1 − 318 = 13 = = 78 6 6 78 −3
6
7
8
To test H0 : µ = µ 0 against H1: µ > µ 0 at 5% level of significance
Procedure A is preferable as it is unbiased. Early customers may not be typical customers in general.
_
Under H0 , Z =
Let R be the r.v for the length of a right shoe and L for the left shoe R N(20, 0.142) and L N(20.1 , 0.112) Method 1 X = R + L N(40.1, 0.0317) 3X N(40.1 x 3, 0.0317 x 32) P(X > 120) = 0.713 Method 2 R + L N(40.1, 0.0317) P(R + L > 120 3 ) = 0.713 Let M and W be the rv for the weight of an adult man and woman respectively. M N(75,42) and W N(65,32) W M ~ N(-10, 52) P( W − M > 1 ) = P(W – M > 1) + P(W – M< -1) = 0.978
x− µ0 9.89
n
N(0,1)
_
Test statistics : Z =
x − µ 0 82.6 − µ0 5 = = (82.6 - µ 0 ) 9.89 σ n 9.89 / 5
Do not reject H0 if P(Z > (82.6 - µ 0 ) (82.6 - µ 0 )
5 ) > 0.05 9.89
5 < 1.64485...........(1) 9.89
µ0 > 75.3
10 (i)
Or P( W − M > 1 ) = 1 - P( W − M < 1 ) = 1 – P(-1
x = 161 (from calculator or computation) 4
P(X ≥ 2) > 0.999 1 – P(X ≤ 1) > 0.999............(1) P(X ≤ 1) < 0.001 Using GC, n P(X ≤ 1) 21 0.00145 22 0.000984 least n is 22
when x = 161 , x = 103.6 + 0.726 y y = (161 − 103.6) / 0.726 = 79.06336088 using y = ∑ y n
1 79.06336088 = (65.1 + 73.2 + 85 + k + 80.9 + 89.9) 6 k = 80.3 Use G.C. to find regression line of y on x:
(ii)
y = −97.593 + 1.097 x
(ii)
Use y on x line to predict weight. When x = 165 , y = −97.593 + 1.097(165) y = 83.4 (1 d.p.) – using 3 d.p. of a and b to compute. or y = 83.5 - using full accuracy of a and b to compute.
(iii)
T Po(2.1x7 + 1.6 x 7) T Po(25.9) Method 1: Since n is large, _ 25.9 T (25.9, ) approx by central limit theorem 100 _
P( T ≤ 26) = 0.578 (to 3 sig fig) Method 2: 100 weeks, Y Po(2590) λ = 2590 > 10 . Normal approx to Poisson Y N(2590,2590) approx P(Y ≤ 2600) = P( Y < 2600.5) (With cc) = 0.578 (to 3 sig fig)
Using G.C., r = 0.893 y
12 89.9
(i)
Let X be the r.v for the number of fish which measures less than 8 cm long. 1 4 1 9 C5 ( )5 ( )4 ( ) = 0.00330 ( to 3 sf) 5 5 5
85.0 80.9 80.3
(ii)
73.2 x
65.1 150
157 160 162
167 170
C is unusually overweight.
11
Breakages occur randomly or Breakages occur independently or Mean number of breakages is a constant
(i)
Let A be the r.v for the number of broken cups per day A Po(2.1) P(A ≥ 3) = 1 – P(A ≤ 2) = 0.350369 = 0.350 (3 sig figs)
X B(n,0.2) Since n large and p= 0.2 , X N(0.2n,(0.2)(0.8)n) approx P(X ≤ 10) ≤ 0.0227 P(X < 10.5) ≤ 0.0227 10.5 − 0.2n ) ≤ 0.0227 P(Z < 0.4 n 10.5 − 0.2n ≤ -2.000929..........(1) 0.4 n 10.5 – 0.2n ≤ -0.800372 n Hence 10.5 – 0.2n ≤ -0.8 n approx Method 1 Using GC Y1 = 10.5 -0.2x + 0.8 x → Table Ans : 91
Method 2 : Use GC and graph
Let X be the r.v for the number of days with a least 3 broken cups out of n cups. X
5
6
13
7 balls: 6
Method 3: Hence (10.5 – 0.2n)2 ≥ (-0.8 n )2 Hence 4n2 - 484n + 11025 ≥ 0 approx……(2) From GC : n ≥ 90.6 or n ≤ 30.4 n ≥ 91 or n ≤ 30 (NA because does not satisfy (1) ) Least n = 91 Let H be the event that the member of public has hypertension Let D be the event that the machine diagnosed hypertension
number of red and blue balls is only 5.] 10 8 7 No. of ways = − + = 210 − (28 + 7) = 175 6 6 6
Alternative Method:
(i) D
Case Green Blue Red No. of ways 5 3 2 1 4 1 1 × × = 30 4 1 1
0.8 k
H 0.2 0.1
1-k
D’ D
H’ 0.9
(i)
2 P ( H ∩ D) P(H/D) = = 3 P( D) 2 0.8k = 3 0.8k + (1 − k )(0.1)
D ’
(1)
k = 0.2 p% = 20 % p = 20
(ii)
P(H’/D’) =
(1 − k )(0.9) P ( H ′ ∩ D′) = = 0.947 0.2k + (1 − k )(0.9) P ( D′)
(iii)
Examples of possible comments: (1) If it does not find you hypertensive then you can be reasonably confident that your blood pressure is normal. (2) If it diagnoses hypertension, then you should consult your doctor for further tests. (3) Any other logical comments with reference to the context of the question
14 (i)
2 Case 1: 2 red balls - = 1 2 3 Case 2: 2 blue balls - = 3 2 5 Case 3: 2 green balls - = 10 2 No. of ways = 1 + 3 + 10 = 14
(ii)
[Note: We can’t exclude green balls because total
2
3
2
1
3
3
1
2
4
2
3
1
5
2
2
2
6
1
3
5 3 2 3 × 2 × 1 = 60 5 3 2 × × = 30 3 1 2 5 3 2 × × = 60 3 2 1 5 3 2 2 × 2 × 2 = 20
5 3 2 1 × 3 × 2 = 5 Total 175 2
Excluding the green balls, we only have 1, 1, 2, 2, 3. Since we are ignoring the colours of the balls, we are forming 3-digit numbers from the 5 digits 1, 1, 2, 2, 3. Case 1: All 3 digits are distinct. The 3 digits are 1, 2, 3 and the number of ways of arranging them are 3! Case 2: 2 digits are identical. Step 1: Choose 2 digits that are identical (1, 1 or 2, 2): 2 Step 2: Choose a digit from the remaining digits (1, 3 or 2, 3): 2 Step 3: Arrange the 3 chosen digits in a row: 3! =3 2! 3! No. of ways = 3!+ ( 2 )( 2 ) = 6 + 12 = 18 2!
Case 1: No red ball. Choose 6 balls from a total of 8 (blue and green) 8 balls: . 6 Case 2: No blue ball. Choose 6 balls rom a total of 7 (red and green) 7
8
Anderson Junior College Preliminary Examination 2010 H2 Mathematics Paper 1 (9740/01)
1
Without using a graphic calculator, find the exact solution of the inequality 12 x + 29 ≤ 4, x ≠ ± 5 . 5 − x2 Hence solve the inequality
2
3
It is given that f(x) =
x(12 + 29 x) ≤4 . 5x2 −1
x tan 2 2x π
for 0 < x ≤ for
π 2
π 2
[7]
,
< x ≤ π.
(i)
Sketch the graph of y = f(x) for 0 < x ≤ π .
[2]
(ii)
Find the exact volume of revolution when the region bounded by the curve in (i), the line y = 2 and the y - axis is rotated completely about the x-axis.
[5]
Given that ln y = sin −1 2 x , show that
dy 1 − 4x 2 = 2 y . dx
[1]
By repeated differentiation of this result, (i)
Find the series expansion of y in ascending powers of x, up to and including the term in x 3 .
[4]
π
(ii) Hence deduce the approximate value of e 3 , giving your answer in the form of 1 a + b 3 , a, b ∈ Ζ where a and b are to be determined. 8
(
4
)
The curve C is given by the equation y = x
1 x
[2]
where x > 0. The variable point P
moves along the x−axis from x = 10 to x = 1 at a constant rate of 1 unit per second. Another variable point Q moves along the curve C so that the x−coordinates of both P and Q are the same at any point in time. (i) (ii)
dy in terms of x. dx The area of triangle OPQ is represented by A. Show that the rate of change of
Find
A is
1 3 ln 2 − unit2/s when t = 6. 2 2
Page 1 of 4 AJC / 2010 Preliminary Examination / 9740 / P1
[3]
[4]
5
6
7
8
d 1 x tan = . dx 2 1 + cos x x + sin x Hence, or otherwise, show that ∫ = dx x f ( x) + C , where f (x) is a single 1 + cos x trigonometric function to be determined and C is an arbitrary constant. Show that
f : x → 4 + sec x ,
[5]
π
< x ≤π . 2 (i) Show that the inverse function f -1 exists and find f -1 in similar form. (ii) On a single, clearly labelled diagram, sketch the graphs of f and f −1 . 1 Another function g is defined by g : x → ln , x>2 . x−2 Find the maximal domain of the function f for the composite function gf to be The function f is defined by
[2]
[3] [2]
defined. Find the corresponding range of gf .
[3]
A sequence of real numbers u1 , u2 , u3 , ….. satisfies the recurrence relation 4u= a un − 2 , a ∈ for all positive integers n and u1 = 1 . Express un in terms n +1 of a and n .
[4]
Find the range of values of a for which the sequence converges.
[1]
If the sequence converges, find the limit in terms of a .
[1]
The curve G has equation a y = x + b +
c where x ≠ 1 and a, b and c are x −1
constants with a > 0. Given that the curve G has two stationary points P and Q, (i)
Find the range of all possible values of c.
[2]
(ii)
Find the coordinates of P and Q in terms of a, b and c.
[2]
The point (0,1) lies on one of the asymptotes of G and the lines y = −1 and y = 5 are tangents to G. Find the values of a, b and c.
9
[4]
The non-zero complex numbers z1 and z2 satisfy the equation z2 − z1 z2 + z1 = 0 z Find the complex number 2 , given that its imaginary part is positive. z1 (i) The points P and Q represent z1 and z2 respectively in an Argand diagram. Describe, with reasons, the geometrical properties of triangle OPQ where O is the origin. 2
(ii)
2
Find the values of n for which z1 + z2 = 0. n
n
Page 2 of 4 AJC / 2010 Preliminary Examination / 9740 / P1
[3]
[3] [2]
10 (a)
The diagram below shows the graph of y = f(x). The curve cuts the x-axis at x = 0 , x = 3 and x = 5 . It has asymptotes x = 4 and y = −1 . There is a 5 minimum at the point A(-2, -3) and a maximum at the point C ,1 . 2 y
y = f(x) C
( 52 ,1) ×
x B (0, 0)
3
4
5
-1 × A (-2, -3)
(b)
11
Sketch the graph of = y 2 − f ( x) , indicating the asymptotes and corresponding points for A, B and C clearly.
[4]
Describe a series of linear transformations to show how the graph of 1 y = ln 2 , 0 0.
[4]
Part of an isosceles triangle, formed by numbers following a particular pattern is shown below:
1 5 7 21 23 25 73 75 77 79
1 3 9 27 81
5 11 13 29 31 33 83 85 87 89
…….
(i) (ii)
State the middle term in the nth row. Find the first and last term of the nth row.
[1] [2]
(iii) Show that the sum of the terms in the nth row is (2n − 1)3n −1 .
[2]
(iv) Find the minimum number of rows in the triangle for the number of terms in the triangle to exceed 500.
[3]
Page 3 of 4 AJC / 2010 Preliminary Examination / 9740 / P1
1 12(a) By using the substitution u = , or otherwise, show that t 5 1 ∞ t u2 = dt ∫1 (1 + t 3 )3 ∫0 (1 + u 3 )3 du . Hence evaluate the exact value of this integral.
[4]
(b) The diagram below shows the part of the curve C with parametric equations t t2 , for t ≥ 1 . = x = , y y 1+ t 3 1+ t 3
R
0
1 2
x
(i) C approaches to a point when t → ∞. Find the coordinates of this point. [1] 1 (ii) The region bounded by the curve, the x-axis and the line x = is denoted by R. 2 Using the result found in (a), find the exact area of R. [4]
13
The equations of a plane π1 and a line l are shown below:
2 π1 : r ⋅ 1 = 6 −2 x −1 y + 4 l: = = z −5 3 2 The point A has position vector 3i − j + 4k . (i)
Find the distance between point A and the plane π1 .
(ii)
B is another point such that AB = −5 j − 2k . Find the length of projection of
[2]
→
→
AB onto the plane π1 .
[2]
(iii) Using your answers in (i) and (ii), find the area of triangle ABC, where C is the reflection of A in the plane π1 .
[2]
(iv) Find the equation of the plane π2 which contains the line l and the origin. Hence, find the line of intersection between the planes, π1 and π2 .
[4]
END OF PAPER Page 4 of 4 AJC / 2010 Preliminary Examination / 9740 / P1
Anderson Junior College Preliminary Examination 2010 H2 Mathematics Paper 2 (9740/02)
Section A: Pure Mathematics (40 marks)
1
Find the values of a and b if the expansion of
9 + ax in ascending powers of x up 1 + bx 2
35 2 x . With these values of a and b, state 6 the range of values of x for which expansion is valid.
to and including the term in x 2 is 3 + x +
2
Express f ( x) =
x3 + 3x 2 + 2 x + 1 in partial fractions. x 2 + 3x + 2
[6]
[2]
Hence, or otherwise, show that N 1 2 1 . f (= x) N + N + 1) − ( ∑ 2 N +2 x =1
[3]
Using the sketch of y f ( x) where x > 0 shown below, or otherwise, show that = N
∑ f ( x) < ∫ x =1
N +1
1
f ( x) dx .
y
0
3
[3] y= f(x)
x
A complex number z satisfies z − a = a, a ∈ + . (i)
1 The point P represents the complex number w, where w = , in an Argand z diagram. Show that the locus of P is a straight line.
Sketch both loci on the same diagram and show that the two loci do not 1 intersect if 0 < a < . 2 1 1 (iii) For a = , find the range of values of arg z − , give your answer correct 2 a ° to 0.1 . 1 State the limit of arg z − when a approaches zero. a
[2]
(ii)
Page 1 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
[4]
[3] [1]
4
cos t Relative to the origin O , the points A and B have position vectors a = sin t and −1 cos 2t b = − sin 2t respectively, where t is a real parameter such that 0 ≤ t < π . 12 (i)
Show that a ⋅ b = m + cos(nt ) where m and n are constants to be determined.
(ii) Hence, find the exact value of t for which ∠AOB is a maximum.
[2] [3]
r (iii) Another point C has position vector c = s where r and s are real constants. 0 Given that A, B and C are collinear when t =
5
π
2
, find the values of r and s.
[3]
An underground storm canal has a fixed capacity of 6000 m3 and is able to discharge rainwater at a rate proportional to V , the volume of rainwater in the storm canal. On a particular stormy day, rainwater is flowing into the canal at a constant rate of 300 m3 per minute. The storm canal is initially empty. Let t be the time in minutes for which the rainwater had been flowing into the storm canal, (i)
show that V =
300(1 − e − kt ) , where k is a positive constant. k
[4]
A first alarm will be sounded at the control room when the volume of rainwater in the storm canal reaches 4500 m3 and a second alarm will be sounded when the storm canal is completely filled. Given that the first alarm was sounded 20 minutes after the rainwater started flowing into the storm canal. (ii)
Find the time interval between the first and second alarm. (Assuming the weather condition remains unchanged).
(iii) Briefly discuss the validity of the model for large values of t .
Page 2 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
[3] [1]
Section B: Statistics (60 marks) 6
7
Mary Lim has 7 cousins. In how many ways can she invite some or all of them to her birthday party?
[2]
At her birthday party, Mary sets up a round table of 8 seats with a different welcome gift at each seat. If all her cousins turn up for the party and given that 4 of them are from the Lee family, 2 are from the Yeo family and 1 is from the Tan family, find the number of ways they can be seated with Mary for a meal at the table if family members of the same surname are seated together but members of the Lee and Yeo families are not adjacent to each other.
[3]
After the meal, Mary and her cousins start to play a game. The game requires a formation of 2 facilitators and 2 teams of 3 members each. Find the number of possible formations if not all the members in each team have the same surname.
[3]
Chemical X will react with chemical Y to form compound Z. A scientist at a chemical plant wants to study the relation between chemicals X and Y by varying the amount of chemical Y (in milligrams) placed in a reaction flask containing 50mg of chemical X. When chemical Y is completely used up, the amount of chemical X (in milligram) left is recorded as shown in the table below. Chemical Y used (y mg) Chemical X left (x mg)
(a)
(b)
20 29.1
40 16.2
60 8.9
80 5.1
100 3.8
(i)
Find the equations of the least square regression lines of y on x and x on y.
[2]
(ii)
Using the appropriate regression line found in (i), estimate the least amount of chemical Y used given that chemical X has completely used up. Comment on the validity of your estimation.
[2]
(i)
A group of scientists propose alternative models of the form w = a + by , where w is a function of x, to describe the relation between chemicals X and Y. Model A: w = x 2 1 Model B: w = x Model C: w = ln x
(ii)
State, with a reason, which model is the most appropriate and find the corresponding least square regression line by performing a suitable transformation.
[2]
Hence, estimate the change in w when y increases by 5.
[1]
(iii) In the same experiment conducted earlier, the amount of compound Z formed, z (in miligrams), is measured. From the data collected, the linear product moment correlation coefficient between z and x is found to be − 0.9 and the least square regression line of z on x is given by z = −1.5 x + 12 . Find the least square regression line of x on z.
Page 3 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
[4]
8
In year 2009, the average length of time for cars parked at the Integrated Resort Haven (IRH) was 12 hours. The facilities are upgraded in 2010 and the management of IRH wants to find out if there is difference in the mean length of time for cars parked at IRH. Assuming that the resort and the car park are open at all times. The length of time, x hours, for cars parked at IRH were recorded for 200 randomly selected cars on a particular day and the following results were obtained: − 12) ∑ ( x=
(i)
9
80 ,
− 12) ∑ (x = 2
1425 .
Denoting the population mean and variance of the parking times by µ and σ 2 respectively, find unbiased estimates of µ and σ 2 .
[2]
(ii) Given that an appropriate hypothesis test carried out could not provide sufficient evidence to indicate a difference in the mean length of parking time, find the range of values of the significance level of this test.
[3]
(iii) State, with a reason, the range of the significance level of another test (without carrying out the test) such that the sample could not provide sufficient evidence that there is an increase in the mean length of parking time.
[1]
(iv) The sample of 200 cars could also be obtained using systematic sampling. Describe how this can be done.
[2]
A multiple-choice question (MCQ) consists of 5 suggested answers, only one of which is correct. For each of the questions set for a particular topic, there is a probability of p that a student, Alice, knows the correct answer, and whenever she knows the correct answer she selects it. If she does not know the correct answer, she randomly selects one of the 5 suggested answers. The events K and C are defined as follows: K: Alice knows the correct answer. C : Alice selects the correct answer. (i) (ii)
Find the probability, in terms of p, that Alice selects the correct answer. Describe what the event K’∩ C represents in the context of this question. 1 (iii) Given that P(K’|C) = , find the value of p. 16
[2] [1] [2]
Taking p to be 0.3 and Alice answers 1 MCQ daily from Monday to Friday. (iv) Given that she answers 3 MCQ correctly, find the probability that this happens in 3 consecutive days. (v)
Alice scores 3 marks for each correct answer, but loses 1 mark for each incorrect answer. Find the probability that Alice obtains a negative score for the 5 MCQ she attempted.
Page 4 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
[2]
[2]
10
At a newly opened shop, the number of orders for herbal chicken soup received in a randomly chosen 30-minute interval follows a Poisson distribution with mean 2.3. The shop is opened for 8 hours daily, from 11 am to 7 pm. (Assume that the orders received are independent.) (i) (ii)
11
Find the probability that there are at least 6, but less than 10 orders received in a randomly chosen one-hour period.
[2]
Find the probability that in 100 randomly chosen one-hour periods, the shop receives an average of more than 5 orders in a one-hour period.
[2]
(iii) The shop owner incurs a fixed operating cost of $250 per day. The cost price and selling price of a bowl of herbal chicken soup are $8 and $20 respectively. By using a suitable approximation, find the probability that he makes a profit of at least 40% of his total cost incurred per day.
[4]
(iv) After operating the shop for half a year, the shop owner wishes to assess if he should continue with the business. He decides to observe the lunch time crowd from 12 pm to 2 pm for 25 days selected at random. If there are less than 14 days with more than 10 orders during the lunch period, he will close down the shop. Comment on whether he should close down the shop.
[3]
Explain whether the Poisson distribution is a good model for the number of orders for herbal chicken soup in a day.
[1]
Peter bought an ice-cream machine. The amount of time taken by the machine to produce a large tub of ice-cream follows a normal distribution with mean µ minutes and standard deviation σ minutes. It is found that there is a 88% chance that the machine will take less than 60 minutes and a 70% chance that it will be more than 50 minutes. The amount of time taken by the machine to produce a small tub of ice-cream also follows a normal distribution with mean 20 minutes and standard deviation 2 minutes. The amount of time taken by the machine to produce a large tub of ice-cream and a small tub of ice-cream are independent of each other. (i)
Find µ and σ .
[3]
(ii)
Find the probability that the difference between the amount of time taken by the machine to produce 5 large tubs of ice-cream and thrice the amount of time taken to produce 3 small tubs of ice cream is more than 1 hour.
[4]
(iii) After using the machine for a year, Peter decides to test the functionality of the machine by using it to produce n large tubs of ice-cream (where n is large) and observing the time taken for each production. He will consider buying a new machine if there are more than 20 times that it takes at least 60 minutes to produce a large tub of ice-cream. Using a suitable approximation, find the greatest value of n such that the probability that he needs to consider buying a new machine is less than 0.2.
END OF PAPER Page 5 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
[5]
AJC H2 Maths _Prelim 2010_P1 (Solutions) 1 12 x + 29 (12 x + 29) − 4(5 − x 2 ) ≤4 ⇒ ≤0 5 − x2 5 − x2 2 ⇒ ( 5 − x)( 5 + x)(2 x + 3) ≤ 0 ⇒ x= -3/2, x ≤ − 5 or x ≥ 5
∴ y = 1 + 2x + π
Let e 3 = e sin
π
x= -3/2, x < − 5 or x > 5 ( x ≠ ± 5 )
3 x(12 + 29 x) ≤4 5x 2 − 1 Replace x by 1/x, 1 5 1 ⇒ x= -2/3, − 5
⇒ x= -2/3, −
2
1 1 < − 5 or > 5 x x 1 < x < 0 or 0 < x < 5 1
= sin −1 2 x ⇒ x =
(i)
0
π
3
(
)
⇒
5
Volume required = π ( 2)2 (π ) − π ∫ 2 tan 2 0
π
2 x dx − π ∫π x dx 2 2 π
π
⇒
1 1 = . x 1 + cos x 2 cos 2 2
x
x
x x x = ( x + sin x) tan - ∫ tan (2 cos 2 ) dx 2 2 2 x x = x tan + sin x tan − ∫ sin x dx 2 2 x x x x = = x tan + 2sin cos tan + cos x + A 2 2 2 2 x 2 x 2 x = x tan + 2sin + 1 − 2sin + A 2 2 2 x = x tan + C , where C = 1 + A 2
dy 1 − 4 x 2 = 2 y (shown) ------- (1) dx
d2y dy dy =2 ⇒ 1 − 4 x 2 − 4 x = 4 y ---- (2) dx dx 2 dx 3 2 2 d y d y dy d y dy 1 − 4 x 2 + 2 (− 8 x ) − 4 − 4 x 2 = 4 dx dx dx 3 dx dx
(
1 xy 2
d x 1 x tan = sec2 = dx 2 2 2
x + sin x
ln y = sin 2 x
d2y dy − 4 x 1 − 4 x 2 + dx 1 − 4 x 2 dx 2
A =
∫ 1 + cos x dx = ( x + sin x) tan 2 -∫ tan 2 (1 + cos x) dx
−1
⇒
x1/ x ln x dy y ln x = 1 − = 3/ 2 1 − x 2 dx x3/ 2 2
Using integration by parts,
7 = π 2 − 2π units3 4
π x 2 = 2π − π 2 tan − x − x 2 π 2 0 2 2
1 ln x x 1 dy 1 1 = − 3/ 2 ln x + 3/ 2 y dx x 2x
ln y =
dA 1 1 dy = y+ x dx 2 2 dx dy 1 = (1 − ln 2 ) When t = 6, x = 4 so y = 2 and dx 4 3 1 dA 1 1 1 so = ( 2 ) + ( 4 ) (1 − ln 2 ) = − ln 2 2 2 2 dx 2 4 dx = −1 , Given that dt dA dA dx 1 3 3 1 we have = × = − ln 2 ( −1) = ln 2 − units/sec dt dx dt 2 2 2 2
x
π π
)
)
d3y d2y dy 1 − 4 x 2 + 2 (− 12 x ) − 8 = 0 ---------------- (3) dx 3 dx dx dy d2y d3y When x = 0, y = 1 , = 2, = 4, = 16 2 dx dx dx 3
(
x
⇒
2
(
y = x1/
⇒
(ii)
1 dy 2 y dx = 1 − 4x 2
2
π 3 3 8 3 1 + 2 e 3 ≈ 1 + 2 4 + 3 4 = 8 11 + 5 3 4
3 4
x= -2/3,
2
3
2x
∴ a = 11, b = 5 4
(i)
(ii)
−1
4 x 2 16 x 3 8x 3 + + ... ∴ y ≈ 1 + 2 x + 2 x 2 + 2! 3! 3
)
Page 1 of 6
Page 2 of 6
6
Function f is a one-one function from the sketch of y = f ( x) , hence f-1 exist. Let y = 4 + sec x 1 ⇒ sec x = = y−4 cos x 1 -1 −1 ) , x ≤ 3. Hence f : x → cos ( x−4
g : x → ln
1 = − ln( x − 2) , x > 2 x−2
For Rf to be a subset of Dg , range of f need to be restricted to (2,3] , therefore the maximal domain of f is (
7
(ii)
y = f −1(x)
(iii)
y = f (x)
0
π 2
|
|
3 π
y=x
2π ,π ] . 3
9
2
1 a a ⇒ un +1 = un −1 − 1 + 2 4 4 2
1 1 a a a ⇒ un +1 = un − 2 − − 1 + 2 2 4 4 4 3 2 1 a a a ⇒ un +1 = un − 2 − 1 + + 2 4 4 4
c ⇒
y=
2
z 1 ± 1 − 4 1 ± 3i z z 2 2 = z2 − z1 z2 + z1 = 0 ⇒ 2 − 2 + 1 = 0 ⇒ 2 = 2 2 z z z1 1 1 z2 1 + 3i Since imaginary part is positive, ⇒ = 2 z1 z and arg 2 = tan −1 3 = π / 3 z1 |z2| = | z1| and arg (z2) = arg(z1) + π/3 Triangle OPQ form an equilateral triangle as OP = OQ and ∠POQ = 60o
(i)
( n −1) +1
2 n −1 1 a a a un − ( n −1) − 1 + + + ... + 2 4 4 4 n a n 1(1 − 1 a 4 ⇒ un +1 = u1 − 2 1− a 4 4 n
1+ b − 2 c a 1+ b + 2 c When x = 1 + c , ay = 1 + c + b + c ⇒ y = a x+b The equations of the asymptotes of C are x = 1 and y = . a x+b Given (0, 1) is a point on y = , we get a – b = 0 - - - (1) a Given that the line y = k does not pass through any point on C only for values of k in the interval (0, 2), the min turning pt corresponds to y = 0 and the max turning pt corresponds to y = 2: 1+ b − 2 c y= = −1 ⇒ a + b – 2 c = –1 a 1+ b + 2 c y= = 5 ⇒ 5a – b – 2 c = 1 a 3 9 Hence a = 1, b = 1 and c = , i.e. c = . 2 4 When x = 1 − c , ay = 1 − c + b −
Corresponding range of gf is [0, ∞) a 1 4un +1 = aun − 2 ⇒ un +1 = un − 4 2 aa 1 1 ⇒ un +1 = un −1 − − 44 2 2
a ⇒ un +1 = 4
From (i), the stationary points of C are at x = 1 ± c .
z2 = (1/ 2) 2 + ( 3 / 2) 2 = 1 z1
n
z n n (ii) z1 + z2 = 0 ⇒ 2 = −1 z1 n
i π3 nπ i (π + 2 k π ) ⇒ = π (1 + 2k ) ⇒ n = 3(1 + 2k ) for any integer k e = e 3 10 a
n −1
6−a a 2 6−a a 2 ∴ un = − − 4−a 4 4−a 4−a 4 4−a 2 a . If the sequence converges then < 1 ⇒ a < 4 . The sequence converges to − 4−a 4 c (i) ay = x + b + x −1 dy 1 c 2 ⇒ = 1 − = 0 when (x – 1) = c. dx a ( x − 1) 2 This equation has two real and distinct solutions only when c > 0. ⇒ un +1 =
8
10 b Page 3 of 6
1 −2 y = ln 2 = ln ( x − 3) = −2 ln(3 − x) x − 6x + 9 Page 4 of 6
11
The graph can be obtained from y = ln x by 1. translate of 3 unit in the negative x direction y = ln( x + 3) 2. reflect in the y-axis y = ln(− x + 3) 3. scale by factor 2, parallel to the y-axis 4. reflect in the x-axis ) i) For the nth row, middle term = 3n−1 ii) For the nth row, first term = 3n −1 − 2(n − 1) , last term = 3n −1 + 2(n − 1) iii) Sum of terms in nth row 2n − 1 n −1 ( 3 − 2(n − 1) ) + ( 3n −1 + 2(n − 1) ) = 2 2n − 1 n −1 n −1 = (2)(3 ) = (2n − 1)3 2
3 2 −1 ⋅ 1 4 −2 6 6 −1− 8 = − = − 2 = 3 unit 3 22 + 12 + 22 22 + 12 + 2 2
n
iv) No. of terms in each nth row row = (2n-1) Therefore, total no. of terms in n rows = ∑ (2r − 1) r =1
n
∑ (2r − 1) > 500
0 ⌢ 1 ii) AB = −5 , n = 2 2 −2 ɶ 2 + 1 + (−2) 2 length of projection 0 2 1 = AB × nˆ = −5 × 1 ɶ 3 −2 −2
r =1
n
2∑ r − n > 500
=
r =1
n(n + 1) 2 − n > 500 ⇒ n 2 − 500 > 0 2 Min no. of rows = 23 12
(i)
∞
∫1
t5 dt = 0 (1 + t 3 )3 ∫1 =
(ii) (iii)
1 1 = u5 − du 1 2 (1 + 3 )3 u u
1
u2
∫0 (u 3 + 1)3 du
3 −2 1 1 1 2 1 3u (1 + u 3 ) −3 du = (1 + u ) 3 0 3 −2
∫
As t → ∞ , x, y → 0. Area required = =
1 2 0
∫
∞
∫1
0
1 1 1 + = 24 6 8
The point is (0, 0)
y dx = ∫
1
t2
∞ 1+ t3
•
(1 − 2t 3 ) dt (1 + t 3 ) 2
=
∞
∫1
2t 5 − t 2 dt (1 + t 3 )3
∞ 2t 5 t2 dt − ∫ dt 3 3 1 (1 + t ) (1 + t 3 )3
1 ∞ = 2( ) − 1 3t 2 (1 + t 3 ) −3 dt 8 3 ∫1
Alternatively, use AB inˆ , ɶ followed by Pythagoras thm
1 −4 and 5
3 2, 1
2 x + y − 2z = 6
∞
13
12 6 2 2 2 2 2 2 − = 4 3 −2 = 3 6 + 2 + 5 = 3 65 units 10 5
1 2 iii) Area of triangle ABC = ( AC )( 65) 2 3 1 2 = (3 × 2)( 65) = 2 65 units 2 2 3 1 3 iv) l : r = −4 + λ 2 , λ ∈ℜ , Two vectors // to π2 are 5 1 1 3 −14 −1 normal to π2 is // to −4 × 2 = 14 = 14 1 5 1 14 1 −1 Equation of π2 : r ⋅ 1 = 0 , ie. − x + y + z = 0 1 To find line of intersection: −x + y + z = 0
(shown)
= −
1 3
2 2 1 = 1 1 3 −2 −6
−1 1 1 0 1 0 −1 2 The augmented matrix, M is , RREF (M) = 2 1 −2 6 0 1 0 2 x 2 + z 2 1 2 1 x−z =2 y = 2 = 2 + 0 z ie. Equation of line: r = 2 + µ 0 , µ ∈ℜ y=2 0 1 z z 0 1 [Alternatively, Cartesian equation of line: x − 2 = z , y = 2 ]
3 −2 2 = 1 − 1 (1 + t ) = 1 + 1 0 − 1 = 5 units 4 6 4 24 4 3 −2 1 i) distance between point A and the plane π1
Page 5 of 6
Page 6 of 6
1
2
AJC H2 Maths _Prelim 2010_P2 (Solutions) 1 1 2 − 1 9 + ax a 2 2 a 2 2 −1 2 = ( 9 + ax ) (1 + bx ) = 3 1 + x + x + ... (1 − bx + ...) 1 + bx 2 2! 9 18 2 a a a a2 2 = 3 1 + x − ≈ 3+ x +− − 3b x 2 x + ... (1 − bx 2 + ...) 6 648 18 216 By comparing coefficients, a = 6 and b = -2 1 9 + ax The valid range for expansion of is x < . 1 + bx 2 2
f ( x) =
4i
x3 + 3x 2 + 2 x + 1 1 1 = x+ − x 2 + 3x + 2 x +1 x + 2 4ii
x + 3x + 2 x + 1 N 1 1 = ∑ x + − ∑ x 2 + 3x + 2 x +1 x + 2 x =1 x =1 x =1 N 1 1 = ∑ x + − x +1 x + 2 x =1 N N 1 1 = ∑ x + ∑ − x+2 x =1 x =1 x + 1 N 1 1 = ( N + 1) + − 2 2 N +2 N
3
N
f ( x) = ∑
2
y= f(x)
y
f ( x) dx = sum of the area under the curve
1
N
x =1
3
Let w =
N +1
∫
f ( x) dx
1 2
0
N
3
N+1
x
1 , z
1 1 − aw 1 −a = a⇒ = a ⇒ w− = w w w a
2
=
2 1 cos 3t − 5 2
2 1 cos 3t − . 5 2
dV 1 = 300 − kV ⇒ ∫ dV = ∫ 1 dt dt 300 − kV 1 ⇒ − ln 300 − kV = t + C k ⇒ 300 − kV = e− k ( t +C )
When t = 0 , V = 0
1 & w2=0 a
⇒ 300 − 0 = Ae0 ⇒ A = 300
When t = 20 , V = 4500 , ⇒ 4500 =
If 0 < a <
a=
( )
cos 2 t + sin 2 t + 1 cos 2 2t + sin 2 2t + 1
⇒ 300 − kV = Ae − kt
The locus of w is a straight line (the perpendicular bisector of w1 = 1 1 ⇒ 0 < 2a < 1 and >1 2 2a 1 Hence 2a < 2a Therefore the 2 loci do not intersect.
1 2
π (since 0 ≤ t < π ). 3 0 −1 −1 0 −1 r 0 r 4iii When t = π , a = 1 , b = 0 AB = 0 − 1 = −1 and AC = s − 1 = s − 1 2 −1 1 2 −1 3 / 2 0 −1 1 1 2 −1 r Since A, B and C are collinear, AB = k AC ⇒ −1 = k s − 1 3/ 2 1 Looking at the z component, k= 3/2, so r = -2/3 and s = 1/3 5i
1
|z – a | = a ⇒
cos 3t −
B
2
Thus, cos 3t = −1 , ie. t =
As shown in the diagram, as the curve is concave upwards, Area of N rectangles < area bounded by the curve ∴ ∑ f ( x) <
a ⋅b = a b
to minimize cos ∠AOB , ie.
x =1 N +1
∫
cos ∠AOB =
θ
2 For maximum ∠AOB , since 0 ≤ ∠AOB ≤ π and cos θ is a decreasing function over [ 0, π] , we aim
N
∑ f ( x) = sum of N rectangles under the curve.
AP a 0.5 1 = = = AB 1 − a 2 − 0.5 3 P a o ∴θ = 19.5 A 1/2 1 1 Hence 160.5o ≤ arg z − ≤ 180o or −180o < arg z − ≤ −160.5o a a 1 As a approaches 0, arg z − → 180o a cos t cos 2t 1 1 1 a ⋅ b = sin t ⋅ − sin 2t = cos t cos 2t − sin t sin 2t − = cos(t + 2t ) − = cos 3t − ɶ ɶ 2 2 2 −1 1 2 sin θ =
1 2a
a
⇒V =
300(1 − e − kt ) (Shown). k
300(1 − e −20 k ) ⇒ 15k = (1 − e −20 k ) k
From the GC, k = 0.030293
1 a
300(1 − e −0.030293t ) ⇒ t = 30.7 0.030293 The residents will have 10.7 minutes between the 1st and 2nd alarm. 2nd alarm : when V = 6000
2a
1 2
∴ 6000 =
300 = 9903 m3 which is impossible as the canal has only a fixed volume of 0.030293 6000 m3 . The model is not valid for large values of t.
t → ∞ ⇒V →
Page 1 of 5
Page 2 of 5
6
No of ways to invite her guests = (2)7 – 1 = 127
(iv) The management must have a sampling frame (the list of all cars parked). If there are N cars, choose a random number, k, from 1 to N/200 (take the nearest integer value), then select every (N/200)th car until a sample of 200 is obtained.
No of ways to arrange a round table = 2! X 4! X 2! X 8 = 768 No of ways to select 2 facilitators and 2 teams (Without any restrictions) = 8C2 x (6C3 x 3C3) ÷ 2! = 280 No of ways to select the 2 facilitators and 2 teams with Lee family forming 1 team = 4C3 x 5C2 x 3C3 = 40
9
No. of possible formations = 280 – 40 = 240 7
(a)(i) For x on y, x = −0.3085 y + 31.13 ⇒ x = −0.309 y + 31.1 (3 s.f.) For y on x, y = −2.8526 x + +95.999 ⇒ y = −2.85 x + 96.0 (3s.f.) (ii) Since chemical Y is the controlled variable, use regression line of x on y. 0 = −0.3085 y + 31.13 ⇒ y = 100.91 The estimation is not valid as this is an extrapolation, linear relation may not hold outside the range of data. 1(b)(i) By comparing the linear product moment correlation for the 3 models, Model C is the most appropriate with the highest value of r = 0.993 as it best describes the data given. Using linear transformation w = ln x , Regression line of w on y is w = −0.026136 y + 3.8294 ⇒ w = −0.0261 y + 3.83 (3 s.f.) (ii)
(b)(iii) r = (− 0.9 ) = bd where b = −1.5 2
Since
(− 0.9)2 Hence d =
( x , z )lies on the reg line z = −1.5
− 1.5
P(3 consecutive correct | 3 correct answers) =
10
Let X be the r.v. denoting the number of orders for herbal chicken soup in 30 min. X ~ Poi (2.3) Let Y be the r.v. denoting the number of orders for herbal chicken soup in 1 hour. i.e. Y ~ Poi(4.6) Y ~ Poi(2.3 × 2)
(ii)
= −0.54
P(Y > 5) = 0.031090 ≈ 0.0311(3.s. f )
29.1 + 16.2 + 8.9 + 5.1 + 3.8 = 3.3220655 5 z = −1.5(3.3220655) + 12 = 7.0169
x=
Let T be the r.v. denoting the number of orders for herbal chicken soup in a day (8 hours). i.e. T ~ P (36.8) T ~ P (2.3 × 16) Since λ > 10 , T ~ N (36.8,36.8) approximately. P ( 20T ≥ 1.4 ( 8T + 250 ) )
350 = P ( 8.8T − 350 ≥ 0 ) = P T ≥ = P(T ≥ 39.773) 8.8 = P(T≥40) = P(T>39.5) cc = 0.328
x − (3.3220655) = −0.54( z − 7.0169)
∴ x = −0.54 z + 7.11 (3.s.f) 8
(i) Unbiased estimate of µ is x = Unbiased estimate of σ 2 is s2 = (ii)
80 + 12 = 12.4 200
(iv)
Let A be the r.v. denoting the number of orders for herbal chicken soup in 2 hours (lunch time) A ~ P(2.3 × 4) i.e. A ~ P(9.2) Let H be the r.v. denoting the number of days with more than 10 orders during lunch period from 12 to 2pm. P ( A > 10 ) = 1 − P ( A ≤ 10 ) = 1 − 0.68202 = 0.31798
200 1425 80 2 −( ) =7 199 200 200
H0 : µ = 12
X − 12 ~ N(0,1) under H0 by CLT s / 200 Using GC, p -value = 0.0325 Given that H0 is not rejected, α < 3.25 (iii) Since we will be using a one-tailed test in stead of a two-tailed test, 1 α < (3.250944) = 1.625472 ⇒ α < 1.63 2 H1 : µ ≠ 12
4.6 ) approximately 100
x + 12 ,
(iii)
x on z is
= 0.29471 ≈ 0.295(3.s. f )
Y ~ Poi (4.6) , Mean = variance = 4.6
Since n = 100 is large, by CLT, Y ∼ N (4.6,
z = −1.5 x + 12
Hence, reg line of
(0.44)3 (0.56) 2 × 3 3 = (0.44)3 (0.56) 2 × C35 10
P(negative score) = P(0 correct or 1 correct) = (0.56)5 + C15 (0.56)4 (0.44) = 0.271
(i) P ( 6 ≤ Y < 10 ) = P (Y ≤ 9 ) − P (Y ≤ 5)
Change in w = −0.026136(5) = −0.13068 ≈ −0.131 (3 s.f.) w decreases by 0.131. 2
1 5 K’∩C represents the event where Alice does not know the correct answer but she (ii) answers correctly. 1 (1 − p ) × P( K '∩ C ) 1 1 5 (iii) P(K’|C) = solving, get p =0.75. = = 1 16 P(C ) 16 p × 1 + (1 − p ) × 5 When p = 0.3, P(C) = 0.3+0.7×0.2=0.44 P(C) = p ×1 + (1 − p ) ×
(i)
Test Statistic : Z =
H ~ B ( 25, P ( A > 10 ) )
i.e. H ~ B ( 25, 0.31798)
P ( H < 14) = P ( H ≤ 13) = 0.98947 = 0.989 (3.s.f) Since the probability of having less than 14 days with more than 10 orders during the lunch period is quite high, he should close down his business. Page 3 of 5
Page 4 of 5
11
(iv) The use of the Poisson model is not suitable in this context as the number of orders during lunch and dinner period will likely be higher than the rest of the hours in a day, thus it is unlikely that the mean number of orders is the same for each 30 minute period. (i) Let A be the r.v. denoting the amount of time taken by the machine to produce a large tub of ice- cream. A ∼ N ( µ , σ 2 ) P( A < 60) = 0.88 P( A > 50) = 0.70
60 − µ PZ < = 0.88 σ 60 − µ = 1.17499
σ µ = 60 − 1.17499σ − − − −(1)
50 − µ PZ > = 0.70 σ 50 − µ = −0.52440
σ µ = 50 + 0.52440σ − − − −(2)
Equating (1) and (2), ∴σ = 5.8845 = 5.88 (3.s.f) and
µ = 53.08583 = 53.1 (3.s.f)
(ii) Let B be the r.v. denoting the amount of time taken by the machine to produce a small tub of ice- cream. B ∼ N (20, 22 )
(
Required prob = P A1 + A2 + A3 + A4 + A5 − 3 ( B1 + B2 + B3 ) > 60
)
Let T = A1 + A2 + A3 + A4 + A5 − 3 ( B1 + B2 + B3 ) .
T ∼ N ( 85.42915, 281.1367 )
(
)
P A1 + A2 + A3 + A4 + A5 − 3 ( B1 + B2 + B3 ) > 60 = P ( T > 60 ) = 1 − P ( −60 < T < 60 ) = 0.935317 = 0.935 (3.s.f) (iv) Let C be the number of times the machine takes at least 60 minutes to produce a large tub of ice-cream. C ∼ Bin ( n, P ( A ≥ 60 ) ) i.e. C ∼ Bin ( n,1 − 0.88 ) Since n is large, np > 5 and n(1-p) > 5,
C ∼ N ( 0.12n, 0.1056n )
P ( C > 20 ) < 0.2 P ( C > 20 + 0.5 ) < 0.2 (cc) 20.5 − 0.12n PZ > < 0.2 0.1056n Using GC, P ( Z > 0.84162 ) = 0.2 20.5 − 0.12n > 0.84162 ⇒ 0.12n + 0.2735 n − 20.5 < 0 0.1056n Solving -14.26 < n < 11.98 greatest n = 143. Thus,
Page 5 of 5
CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2
MATHEMATICS
9740/01
Paper 1
31 August 2010 3 hours
Additional Materials:
Answer Paper Graph Paper List of Formulae (MF 15)
READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, arrange your answers in NUMERICAL ORDER. Place this cover sheet in front and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
Name : __________________________________________ Class : ___________ Question No.
Marks
Question No.
Marks
1
/4
7
/8
2
/7
8
/ 10
3
/9
9
/ 10
4
/8
10
/ 15
5
/8
11
/ 12
6
/9
TOTAL
/ 100
------------------------------------------------------------------------------------------------------------------------This document consists of 5 printed pages. [Turn Over]
1
1.
Three sisters Audrey, Catherine and Gladys went on a shopping spree while on holiday. They bought items for their sister Eliza, who could not go due to her ‘A’ level examinations. After returning home, they could not recall the prices of the items they had bought, but they had their credit card statements with the total amounts they each had spent. The list of items that each sister had bought and the amount on their statements are summarised in the table below: Type of Item (Price) High heels ($x per pair) Facial Mask ($y per box) Handbags ($z per piece) Credit Card Statement ($)
Catherine 5 10 3 1298.20
Audrey 2 7 8 1158.30
Gladys 3 15 5 1837.70
(i) Find the values of x, y and z. (ii) Hence evaluate the total cost of their gift to Eliza, which consists of 1 pair of high heels, 5 boxes of facial mask and 2 handbags. 2.
n + 1 u n for n ≥ 1 . n2 (i) Write down the values of u 2 , u3 , u 4 and u5 .
[1]
A sequence is defined by u1 = 1 and u n +1 =
[2]
(ii) Hence make a conjecture for u n in terms of n. Prove your conjecture using mathematical induction. 3.
[3]
[5]
In an East Asia ancient culture, there is a war monument which is formed by placing concentric circular slabs of granite on top of each other. The slabs of granite are of decreasing measurements. (i) The first slab of the monument has diameter 200 cm and the diameter of each subsequent slab is three-quarters the diameter of the previous slab. Show that the total sum of the cross-sectional circular area of the slabs used will be less than 72 000 cm2, no matter how many slabs there are.
[4]
(ii) The first slab of the monument has thickness 50 cm and each subsequent slab has thickness d cm less than of that for the previous slab. Given that the maximum possible number of slabs is 14, find the largest integer value of d.
[2]
(iii) Given that d = 3, write down an expression in terms of n for the volume of the nth slab and hence evaluate the total volume of the monument which consist of 14 slabs.
[3]
[Turn Over]
2
4.
The diagram below shows the graph of y = 2 ln x + 2 − x . The two roots of the equation x − 2 ln x = 2 are denoted by α and β , where α < β . y
α
β
x
(i) Find the values of α and β , correct to 3 decimal places. A sequence of real numbers x1, x2, x3, … satisfies the recurrence relation xn+1 = ln( xn ) 2 + 2 for n ≥ 1 . (ii) Prove that if the sequence converges, it must converge to either α or β . (iii) By considering x n +1 − x n and the graph above, prove that x n +1 > x n if α < x n < β , x n +1 < x n if x n < α or x n > β . (iv) Hence deduce the value that xn converges to for x1 = 2 , giving your answer correct to 3 decimal places.
5.
ax 2 + x The curve C has equation y = where x ≠ 1 and a is a constant. x −1 (i) Show that C does not have any stationary points when − 1 < a < 0 . (ii) Sketch C when − 1 < a < 0 . Show on your diagram, the equations of the asymptotes and the coordinates of any points of intersection with the axes.
[2]
[2]
[2] [2]
[3] [5]
[Turn Over]
3
y
6.
(a)
2 −2
0
x
1
(−1, −3)
The diagram above shows the graph of y = f (x) . On separate diagrams, sketch the graphs of (i) y2 = f (x) (ii) y = f ′(x) showing clearly in each case the axial intercepts, the asymptotes and the coordinates of the turning points, where possible. (b) A graph with equation y = g(x) undergoes in succession, the following transformations: A: A reflection in the y-axis B: A translation of 4 units in the direction of the positive x-axis C: Scaling parallel to the y-axis by a factor of 2 The equation of the resulting curve is y = 2 x −3 . Determine the equation y = g(x). 7.
[3] [3]
[3]
Relative to the origin O, the points A, B and C have position vectors 1 2 − 1 2 , 1 and 2 respectively. 1 3 3 The point P lies on the line AB produced such that AP:PB = 3: 1. (i) Find the vector OP .
[2]
0 2 (ii) The line l has equation r = 1 + λ − 1 , where λ ∈ ℜ . 2 1 Determine whether l and the line through A and B are intersecting.
[3]
(iii) Find the shortest distance from C to AB. Hence or otherwise, find the area of triangle ABC.
[3]
[Turn Over]
4
8.
9.
The equation of a closed curve is (x + 2y)2 + 3(x − y)2 = 27. (i) Show, by differentiation, that the gradient of the curve at the point (x, y) may be dy y − 4x expressed in the form dx = 7y − x (ii) The normal to the curve at the point (−2, 1) cuts the x-axis at P and the y-axis at Q. Denoting the origin as O, find the area of the triangle OPQ. (iii) Find the equations of the tangents to the curve that are parallel to the y-axis. Let y = ln (1 − sin x). d2y dy2 −y y dy (i) Show that e dx = − cos x. Hence, show dx2 + dx = e − 1 (ii) Find the Maclaurin’s series for y, up to and including the term in x3. cos x up to and (iii) Using the result in part (ii), find the Maclaurin’s expansion for sin x − 1 including the term in x2. cos x (iv) Given that x is small, use binomial expansion to expand up to and sin x − 1 including the term in x2. State the range of x for the expansion to be valid.
10. (a)
Find ∫ e 2 x sin x dx .
[3] [4] [3]
[3] [3]
[1]
[3]
[4]
(b)
A curve C is defined by the parametric equations x = 2(t − sin t ) , y = 2(1 − cos t ) for − π ≤ t ≤ π . [2] (i) Sketch the curve C. (ii) Find the exact area of the region bounded by C and the line y = 4. [6] (c) The region R in the second quadrant is bounded by the y-axis, the line y = x + 5, and 2 the curve y = e x . Find the volume of the solid formed when R is rotated through 4 [3] right angles about the y-axis. 11. (a) (i) Solve the equation z 6 = −18 + (18 3 )i , expressing the solutions in the form re iθ , where r > 0 and − π < θ ≤ π . (ii) Show all the roots on an Argand diagram and write down an equation of the form x 2 + y 2 = k which contains all the roots.
[4] [3]
(b) The complex number z satisfies the relations
3π . 4 (i) Illustrate both of these relations on a single Argand diagram. (ii) Hence write down the exact area represented by the given relations. z − 3 + 2i ≤ − 2 3 + i and 0 < arg( z − 6 + 5i ) ≤
~End of Paper~
5
[4] [1]
CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2
MATHEMATICS
9740/02
Paper 2
15 September 2010 3 hours
Additional Materials: Answer Paper Graph Paper List of Formulae (MF 15) READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, arrange your answers in NUMERICAL ORDER. Place this cover sheet in front and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
Name : ____________________________________________ Class : _______________ Question No.
Marks
Question No.
Marks
1
/5
7
/ 12
2
/ 10
8
/ 10
3
/ 11
9
/6
4
/ 14
10
/8
5
/7
11
/8
6
/9
TOTAL
/ 100
-----------------------------------------------------------------------------------------------------------------------------------This document consists of 6 printed pages.
[Turn Over
Page 1
Section A: Pure Mathematics [40 marks] 1 Solve algebraically the inequality 2 x 2 + 4 x − 70 −1 ≥ 0 . x 2 + 4 x − 77 Hence, solve the inequality 2e 2 x + 4e x − 70 −1 ≥ 0 . e 2 x + 4e x − 77
[3]
[2]
2 The equations of the line l1 and the plane p1 are respectively, 1 − 1 1 2 r = 0 + µ 1 and r = s1 + t − 1 , where µ, s, t ∈ ℜ 0 1 1 1 (i)
[3]
Find the acute angle between l1 and p1.
α (ii) A second plane p2 has equation r ⋅ 1 = 1 . Given that the two planes p1 and p2 β 5 4 x − 15 intersect at the line l2: = y; z = , find the values of α and β. −2 2
[3]
(iii) The plane p3 with equation 2 x + by + z = 1 is parallel to l2. Find the value of b. Hence find the distance between l2 and p3.
[4]
3 The functions f and g are defined as follow: 2 , x g : x ln( x − 3) , f :x x+
(i)
x ∈ ℜ \ {0} x>3
[2]
Show that fg does not exist.
Let the function h be defined as follows: h : x ln( x − 3) ,
for x > a
(ii) Find the least value of a such that fh exists. Define fh in similar form.
[3]
(iii) Sketch the graph of f, indicating the turning points, asymptotes and axial intercepts, if any. State the range of f and determine if f -1 exists.
[4]
The function f has an inverse if its domain is restricted to x > k . Find the least value of k.
[2]
[Turn Over
Page 2
4 (a) Verify that y = x is a particular solution of the differential equation dy x 2 + y 2 = , x, y ≠ 0 dx 2 xy
[2]
(b) Show that the substitution y = ux reduces the differential equation dy x 2 + y 2 = dx 2 xy to the differential equation du 1 − u 2 . = x dx 2u Hence find the general solution of the differential equation dy x 2 + y 2 . = dx 2 xy
[3]
[4]
(c) Due to a rapid disease outbreak, the population of fish in a river, x (in thousands), is believed to obey the differential equation d2x = 4ae − 2t dt2 where t is the time in days, and a > 0 is a constant. Given that the entire population of fish is wiped out by the disease eventually, show that the general solution of the [3] differential equation is x = ae −2t . Explain the meaning of a, in the context of the question. Sketch the family of solution curves of the differential equation for a = 1 and 2. [2]
Section B: Statistics [60 marks] 5 (a) Tourists visiting an amusement park in Singapore can purchase a day pass. Each day pass allows one tourist to take any of the 6 designated rides only once. (i)
How many different ways can the order of the rides be chosen if a tourist goes on all 6 rides? [1]
(ii) The attendants in the park keep track of the rides each tourist has taken by marking on the ticket as shown in the diagram on the right. Filling in a square indicates that the tourist has taken that particular ride. If a tourist can choose to go on a ride or not, how many different ways can the ticket be marked?
Amusement Park Crazy Monkey Swing Merry Go Bush A Ride to Heaven Nightmare to Remember Round about the Park Foodie Food
[1]
(iii) If a tourist goes on at least one ride, find how many different selections he can make?
[1]
(b) From 10 students, including Vera and Daen, a group of 5 students from the Mathematics Society will be selected to attend an Enrichment Camp. Daen will not join the group without Vera, but Vera will join the group without Daen. In how many ways can the group be formed? [4] [Turn Over
Page 3
6 The fish-balls used by a popular noodle stall are supplied by two companies, supplier A and supplier B. On average, 1 out of 50 fish-balls from supplier A is deformed and 3 out of 100 fish-balls from supplier B are deformed. It is known that the noodle stall gets a proportion, denoted by p of their fish-balls from supplier A. (i)
Represent the situation using a probability tree.
[2]
1 (ii) Given that p = , show that the probability that a randomly chosen fish-ball from the 3 2 noodle stall is deformed will be . 75
[1]
(iii) For a general value of p, the probability that a fish-ball is supplied by B given that it is 3(1 − p ) [3] deformed is denoted by f( p). Show that f( p) = . 3− p Prove by differentiation that f is a decreasing function for 0 ≤ p ≤ 1 , and explain what [3] this statement means in the context of the question.
7 The occurrences of floods per year at a particular residential area in Singapore follow a Poisson distribution with mean 4. (i)
Find the probability that in 4 months, this particular residential area is flooded at least twice.
[2]
(ii) A random sample of 12 periods of 4 months is taken. Find the probability that in at most 5 of these 12 periods, this particular residential area is flooded at least twice.
[2]
(iii) Using a suitable approximation, find the probability that in 5 years, this particular residential area is flooded at least 11 times.
[4]
(iv) In a long term study of the flooding problem in this particular residential area, it is proposed that the residential area be observed for 40 years. The probability that in at least n years out of these 40 years, this residential area is flooded at most thrice each year is found to be less than 0.8. Using a suitable approximation, find the least value of n. [4]
[Turn Over
Page 4
8 The masses of snapper fish and pomfret fish sold by a fishmonger are normally distributed and independent of each other. The mean mass, standard deviation and selling price of snapper fish and pomfret fish are given in the following table:
Mean mass in kg Standard deviation in kg Selling price per kg in $
Snapper fish Pomfret fish 1 0.6 0.1 0.05 12 7
Find the probability that the (i)
[2]
total mass of 3 snapper fish and 2 pomfret fish is more than 4.5 kg
(ii) mass of 3 snapper fish exceeds twice the mass of a pomfret fish by more than 1.85 kg. [2] (iii) total selling price of a snapper fish and 2 pomfret fish is more than $21.
[2]
A customer buys 15 fish, out of which n are snapper fish and the rest are pomfret fish. The probability that the customer pays more than $150 is less than 0.7. Find the largest value of n.
[4]
9 CJC has 800 JC2 students in 2010 as follows:
Boys Girls Total
Arts 75 145 220
Science 320 260 580
Total 395 405 800
The Student Council (SC) wishes to conduct a survey on a sample of JC2 students to find out their preferences for Graduation Night. (a) Explain how stratified sampling can be used to select a sample of 30 students. Suggest another set of strata that may be suitable for this survey. (b) The amount that a sample of 50 students are willing to pay for Graduation Night can be summed up as follows: ∑ x = 4537, ∑ ( x − x ) 2 = 4825.62 Assume that the mean and variance for this sample are good estimators of the population mean and variance. Another sample of 60 students is surveyed. Find the probability that the mean of this second sample lies between 90 and 100.
[Turn Over
Page 5
[3]
[3]
10 A manufacturer claims that the average length of its metal rods is 14 cm. (i) A customer complains that the manufacturer understated the average length of its metal rods. A random sample of 8 metal rods is taken and the length, x cm, of each metal rod is summarised as follows: ∑ x = 113.40, ∑ x 2 = 1607.72 Test, at the 4% significance level, whether the customer’s complaint is valid. State any assumptions made in carrying out the test. Explain, in the context of the question, the meaning of ‘at the 4% significance level’.
[4] [1]
(ii) Another random sample of 9 metal rods with standard deviation 0.200 cm is taken. What range of values should the mean length of the sample be, in order for the customer’s complaint to be not valid, at the 4% significance level? Give your answer correct to 2 decimal places.
[3]
11 The table below gives the proportion of people (in %) in an occupation earning more than $5000 and the proportion of university graduates (in %) in that occupation. Occupation % of graduates, x % earning more than $5000, y
Teacher
Chemist
Accountant
Lecturer
Engineer
Electrician
Police
Plumber
Taxi Driver
97
87
75
84
52
36
22
10
8
66
65
62
45
53
43
33
18
10
A (i)
Calculate the product moment correlation coefficient between x and y.
(ii) Draw a scatter diagram for the data.
[1] [1]
One of the values of y appears to be incorrect. (iii) Indicate the corresponding point on your diagram by labelling it P and explain why the scatter diagram for the remaining points may be consistent with a model of the form y = a + b ln x .
[2]
(iv) Omitting P, calculate the least squares estimates of a and b for the model y = a + b ln x .
[2]
(v) Assume that the value of x at P is correct. Estimate the value of y for this value of x.
[1]
(vi) Comment on the use of the model in (iv) in predicting the value of y when x =100.
[1]
~End of Paper~
Page 6
Reg. No.
Name
Year 6C(
)
DUNMAN HIGH SCHOOL, SENIOR HIGH PRELIMINARY EXAMINTION 2010 Higher 2
MATHEMATICS
9740/01
Paper 1
21 September 2010 3 hours
Additional Materials:
Answer Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your Name, Class and Register Number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, attach the question paper to the front of your answer script. The total number of marks for this paper is 100.
For teachers’ use: Qn Q1 Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Total
6
8
8
9
9
10
12
14
14
100
Score Max Score
4
6
This question paper consists of 6 printed pages (including this cover page).
2 1
1− x in ascending powers of x up to and including the term in x 2 and state 2+ x the range of values of x for which the expansion is valid. [4] Expand
n
2
Prove by induction that
1
[5]
r =1 ∞
Hence state the value of
n
∑ (2r + 1)(2r + 3) = 3(2n + 3) . 1
∑ (2r + 1)(2r + 3) .
[1]
r =1
3
(i)
Show that n 2 − 4n + 5 = (n − a )2 + b , where a and b are constants to be determined. [1]
∑( N
(ii)
Show that
)
n 2 + 1 − n 2 − 4n + 5=
n =3
N 2 + 1 + ( N − 1) 2 + 1 − 5 − 2. [3]
(iii) Without the use of a graphic calculator, deduce that the sum in (ii) is strictly less than 2 N + 1. [2]
4
y b
S 2
= y x3 + 1
R
x O
(i)
y x + 1. Given that the The diagram above shows the curve with equation = two shaded areas R and S have the same value, find the value of b. [4]
(ii)
Find the volume of the solid generated when S is rotated completely about the x-axis. [4]
3
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
3 5
With respect to the origin O, the position vectors of three points A, B and C are given b 2 3 → → → and OC 5 . = by OA = 3 , OB 7 = a 1 2 (i) Find the values of a and b if A, B and C are collinear. [2] →
→
(ii)
If given instead that OA is perpendicular to OB , find a relationship between a and b. Furthermore, if angle AOC is 60ο , find possible values for a and b, giving your answers correct to the nearest integer. A student claimed that since angle AOC is 60ο , angle BOC must be 30ο. Without performing any calculation, state, with a reason, whether his claim is necessarily true. [6]
6
7
∫
et dt . (1 + 3et ) 2
(a)
Find
(b)
Write down
(c)
Evaluate
(i)
∫
4
d tan( x 2 ) ) . Hence find ( dx
[2]
∫ x sec ( x ) dx. 3
2
2
x 2 x − 3 dx without the use of the graphic calculator.
[4]
[3]
0
Explain whether the following statement is always true: “If z = x + yi, x, y ∈ , is a root of the equation P(= z ) an z n + an −1 z n −1 + ⋅⋅⋅ + a1 z += a0 0, n ∈ + , then z= x − yi is also a root.”
[1] (ii)
Solve the equation z 4 + 3 + i =0, expressing the roots in the form reiα , where
r > 0 and −π < α ≤ π . (iii) Show the roots on an Argand diagram.
[5] [1]
(iv) The points representing the roots in (iii) form a quadrilateral. Find the area of this quadrilateral.
[2]
[Turn over DHS 2010 Year 6 H2 Math Preliminary Examination
4 8
The diagram shows a hexagon PQRSTU inscribed in a circle with radius 6 cm. = UT = 2 x cm . The sides QR and UT are parallel, and QR 2x
Q
P
6
U
(i) (ii)
• O
2x
R
S
6
T
Show that A, the area of the hexagon PQRSTU, is 2 ( 6 + x ) 36 − x 2 cm 2 . [3] Using differentiation, find the value of x when A is maximum. (You need not verify that it gives a maximum value.)
[4]
Initially x = 6 cm, and the lengths of the parallel sides QR and UT are each decreasing 1 at a constant rate of cm s −1. Find the rate of change of A at the instant when 10 x = 2 cm. [3] 1 [ Area of a trapezium = × sum of the lengths of the parallel sides × height ] 2 9
(a)
y ln(1 + e x ) . It is given that= (i) (ii)
d 2 y dy dy Show that = 1 − . dx 2 dx dx Find the Maclaurin’s series for y up to and including the term in x 2 .
[3] [2]
(iii) Verify that the same result is obtained if the standard series expansions for e x and ln(1 + x) are used. [4] (b)
Given that x is sufficiently small for x3 and higher powers of x to be neglected, and that 10 tan x − 3 = cos 2 x , form a quadratic equation in x and hence find the value of x, leaving your answer in exact form. [3]
[Turn over DHS 2010 Year 6 H2 Math Preliminary Examination
5 10
(a)
The curve C has parametric equations = x eθ cos θ ,
(i)
Show that the equation of the tangent at −
0 ≤θ ≤
. 4 2 π4 e , 2 is given by 2
π
4x+
expressed as
∫
π 4 0
eθ cos 2θ dθ . Hence, evaluate the area, leaving your
answer correct to 2 decimal places.
(b)
π
2 [4] . 2 Show that the area bounded by the curve C and the x-axis can be
y e =
(ii)
= y sin θ + cos θ ,
[4]
The diagram shows the graph of y = f ( x ) , which has a turning point at A ( −2, 2 ) .
y x=2
A ( −2, 2 )
O
x
Sketch, on separate diagrams, the graphs of (i)
y = f '( x),
[3]
(ii)
y=
1 , f(x)
[3]
showing clearly all relevant asymptotes, intercepts and turning point(s), where possible.
[Turn over DHS 2010 Year 6 H2 Math Preliminary Examination
6 11
(a)
(b)
(c)
A geometric series has first term a. Find the range of values of a if the sum to 1 infinity of the series is . [3] 2 1 The rth term of a series is given by = Tr + ln 32 r. Find the sum of the first N 2r 3 terms. [4] Jon wants a gigantic cake prepared for his mother’s birthday. The cake is to 9 consist of 5 layers, where each layer has a square base with length that is of 10 the layer beneath it and a constant height of h units. The cost of the cake is proportional to its volume and the largest layer costs $200. (i)
Given that the largest layer has length a units, find the cost of the whole cake, rounded to the nearest dollar. [4]
(ii)
Jon also wants candles to be placed at each layer such that the total number of candles used is 75. If the largest layer has d2 candles and each subsequent layer has d candles less than the one directly below, find the number of candles placed at the top layer. [3]
END OF PAPER
DHS 2010 Year 6 H2 Math Preliminary Examination
Reg. No.
Name
Year 6C(
)
DUNMAN HIGH SCHOOL, SENIOR HIGH PRELIMINARY EXAMINTION 2010 Higher 2
MATHEMATICS
9740/02
Paper 2
23 September 2010 3 hours
Additional Materials:
Answer Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your Name, Class and Register Number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, attach the question paper to the front of your answer script. The total number of marks for this paper is 100.
For teachers’ use: Qn Q1 Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Total
11
11
4
7
8
9
9
11
12
100
Score Max Score
8
10
This question paper consists of 7 printed pages (including this cover page).
2 Section A: Pure Mathematics [40 marks] 1
Show that the equation ( z − 1 + i)( z* − 1 − i) =2 represents a circle with centre C(1,–1) and radius 2 and sketch the locus on an Argand diagram. On the same diagram, sketch the locus represented by z + z* = [3] 3. The points of intersection of these two loci are represented by A and B. Find, in any order, (i)
the length AB,
[1]
(ii)
the complex numbers represented by A and B.
[2]
State the cartesian equation of the perpendicular bisector of the line segment joining A and B. Explain also why the perpendicular bisector of the line segment joining any two distinct points on the circumference of the circle must pass through the centre of the circle, C. [2]
2
For a > 0 , the functions f and g are defined as follows: ax , x + a2 g : x x2 − a2 , f :x
x ≠ −a 2 ,
x∈ .
(i)
Show that the inverse of f exists.
[2]
(ii)
Define f −1 in a similar form.
[3]
(iii)
Show that the composite function fg does not exist.
[2]
(iv)
Solve the equation f( x) = f −1( x) , expressing your answer(s) in terms of a. [3]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
3 3
In an experiment, Andy and Bob attempt to devise a formula to describe how the volume of water, V m3, in a tank, changes with time at t hours. (i)
dV 1 8 Andy gives his formula as= V − 2 . Given that V = 1 when t = 0, dt 60 V
show that V = (ii)
3
t
8 − 7e 20 . Sketch this solution curve.
[4]
d 2V However, Bob believes that it is more likely to be = 12t 2 − 2. Given that dt 2 dV = 0 when t = 0 , show that the general solution for V can be expressed as dt 1 1 V = (t 2 − ) 2 + C − , where C is a constant. 2 4 Hence, or otherwise, sketch on a single diagram, two distinct members of the family of solution curves. [5]
(iii)
4
It is also given that V = 1 when t = 0 in Bob’s model. Suppose that the water in the tank does not overflow, explain, using your diagrams in parts (i) and (ii), why Andy’s model is more appropriate compared to Bob’s model. [2]
The planes Π1 and Π2 are defined by 2 Π1 : r 4 = 10, 1
1 Π2 : r 3 = 8. 1
(i)
Find the acute angle between the two planes.
[3]
(ii)
Obtain a vector equation of l1, the line of intersection of the two planes.
[4]
x−2 7−z The cartesian equation of another line, l2, is given by= = , y m, 6 3 where m is a real constant.
(iii)
If the plane Π1 and line l2 intersect at the point (6, m, 5), find the value of m. [2]
(iv)
Show that the lines l1 and l2 are perpendicular for all values of m.
[2]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
4 Section B: Statistics [60 marks] 5
A group of students plan to collect data from students’ parents for a project. (i)
Describe how a quota sample of size 80 might be obtained based on the parents’ educational qualifications. [2]
Subsequently, they decide to invite 80 parents to respond to an online survey. The table below shows the profile of educational qualifications among the students’ parents:
Father Mother Total
6
“O” Level 180 300 480
“A” Level 360 420 780
University 660 480 1140
Total 1200 1200 2400
(ii)
Explain why stratified sampling is preferred over the method in (i).
[1]
(iii)
How many mothers with an “A” Level qualification should be included in the stratified sample? [1]
To assess the level of work satisfaction in relation to the time spent at work, the human resource department of an organisation polls nine of its officers. The number of work hours per week (t) of each officer and the level of job satisfaction (x) are recorded, where a higher value of x indicates a higher level of satisfaction. The data are shown in the table below: Work hours (per week), t 20.1 22.0 24.4 25.3 28.8 36.5 40.6 46.0 55.1 Satisfaction level, x 24.5 16.3 18.6 12.5 5.2 4.7 1.4 1.8 0.8 (i) (ii)
(iii)
Draw a scatter diagram for the data and find the product moment correlation coefficient for the sample. [3] b A suggested model is of the form x= a + . Find a and b. Justify why this t model provides a better fit than a linear model between x and t. [2] Use the model in (ii) to predict the satisfaction level when an officer works 5.0 hours per week. Comment on the reliability of your prediction. [2]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
5 7
For every Monday of the week, the probability that Mylo wears a tie is 0.4. The probability that he wears a jacket is 0.2. If he wears a jacket, the probability that he wears a tie is 0.6. Find the probability that, on a randomly selected Monday, (i)
Mylo wears a tie and a jacket,
[2]
(ii)
Mylo wears neither a tie nor a jacket.
[2]
For Tuesday and Wednesday, the probability that Mylo wears a jacket is
(iii)
8
•
twice the probability he wears a jacket on the previous day if he wears a jacket on the previous day,
•
the same as the probability he wears a jacket on the previous day if he does not wear a jacket on the previous day.
By constructing a tree diagram, find the probability that Mylo wears a jacket on the third day given that he wears a jacket on exactly two of the three days. [4]
A test consists of five Pure Mathematics questions, A, B, C, D and E, and six Statistics questions, F, G, H, I, J and K. (i)
(ii)
The examiner plans to arrange all eleven questions in a random order, regardless of topic. Find the number of ways to arrange all eleven questions such that (a)
the last question is a Pure Mathematics question,
[2]
(b)
a Pure Mathematics question must be separated from another with exactly one Statistics question. [2]
Later, the examiner decides that the questions should be arranged in two sections, Pure Mathematics followed by Statistics. Find the number of ways to arrange all eleven questions such that (a)
question A is followed by question F,
[2]
(b)
questions B and K are separated by more than seven questions.
[3]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
6 9
A health food company claims that a breakthrough product that it launches, Zenobrain, has the benefit of maintaining good mental health and will help its consumers acquire a mean IQ score of not less than 115. To verify the claim, a random sample of 12 consumers is taken and their IQ scores, x, are recorded and summarised as follows:
50 and ∑ ( x − 100 ) ∑ ( x − 100 ) =
10
2
= 4008 .
(i)
Calculate unbiased estimates of the population mean and variance.
[3]
(ii)
State the null and alternative hypotheses and carry out an appropriate test at 5% level of significance. [4]
(iii)
State, with a reason, whether (a)
any assumption is needed for the test in (ii) to be valid,
[1]
(b)
the conclusion would be the same if a two-tailed test is used with the same level of significance. [1]
In a randomly chosen week, the numbers of unsolicited text messages and phone calls received by a mobile line subscriber follow independent Poisson distributions with means 5 and 3 respectively. (i)
Find the probability that the subscriber receives exactly 2 unsolicited text messages in a day. [2]
(ii)
Show the probability that the subscriber receives at most 10 unsolicited text messages or phone calls in a week is 0.816. [2]
(iii)
The subscriber decides to terminate his mobile line subscription if, in the next 10 weeks, there are more than 3 weeks where he receives more than 10 unsolicited text messages or phone calls in a week. Find the probability that he terminates his mobile line subscription. [3]
(iv)
Another subscriber intends to terminate her subscription if she receives a total of 20 or more unsolicited text messages or phone calls in the next 2 weeks. By using a suitable approximation, calculate the probability that she terminates her subscription. [4]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
7 11
(a)
The distance, X km, covered by a school athlete during a regular training session has mean 4 km. During competition season, training increases in intensity and the distance, Y km, covered during the training session increases to a mean of 6 km. Given that X and Y are independent normal distributions with same variance σ 2 , and P(Y − X > 3) = 0.4, find the probability that the athlete covers a total distance between 8 km and 12 km in two randomly chosen regular training sessions. [5]
(b)
The amount of time that a student spends online each day has mean 120 minutes and standard deviation 45 minutes. A random sample of 60 students is taken and they are surveyed on the amount of time that they spend online each day. Find the probability that (i)
the total time spent online each day by the 60 students is at least 7000 minutes, [3]
(ii)
the sample mean time spent online each day by the 60 students is within 5 minutes of the population mean time of 120 minutes. [3]
Explain whether you need to assume that the amount of time spent online by a student each day follows a normal distribution in your calculations above. [1]
END OF PAPER
DHS 2010 Year 6 H2 Math Preliminary Examination
Dunman High School 2010 Year 6 H2 Mathematics (9740) Preliminary Examination Paper 1 Suggested Solutions Qn 1
Suggested Solution 1− x 1⎛ x⎞ = (1 − x) × ⎜1 + ⎟ 2+ x 2⎝ 2⎠
−1
2 ⎞ ⎛ 1 x ⎞⎛ x ⎛ x ⎞ = ⎜ − ⎟ ⎜1 − + ⎜ ⎟ + ... ⎟ ⎟ ⎝ 2 2 ⎠ ⎜⎝ 2 ⎝ 2 ⎠ ⎠
1 1 1 1 1 − x − x + x 2 + x 2 + ... 2 2 4 4 8 1 3 3 2 = − x + x + ... 2 4 8 =
x < 1 ⇒ −2 < x < 2 2
Valid values of x: 2
∑ (2r + 1)(2r + 3) = 3(2n + 3) for n ∈ Z n
Let Pn be the proposition
1
n
+
.
r =1
When n = 1: 1
∑ (2r + 1)(2r + 3) = (2(1) + 1)(2(1) + 3) = 15 RHS = 3(2(1) + 3) = 15
LHS =
1
1
r =1
Since LHS = RHS, ∴ P1 is true. Assume Pk is true for some k ∈ Z + ,
∑ (2r + 1)(2r + 3) = 3(2k + 3) , k
i.e.
1
k
r =1
to prove Pk +1 is true, k +1
i.e.
k +1
∑ (2r + 1)(2r + 3) = 3(2k + 5) . 1
r =1
k +1
LHS =
∑ (2r + 1)(2r + 3) 1
r =1
∑ (2r + 1)(2r + 3) + (2k + 3)(2k + 5) k
=
1
1
r =1
=
k 1 + 3(2k + 3) (2k + 3)(2k + 5)
1
1
1
1
=
k (2k + 5) + 3 3(2k + 3)(2k + 5)
2k 2 + 5k + 3 3(2k + 3)(2k + 5) (2k + 3)(k + 1) = 3(2k + 3)(2k + 5) k +1 = = RHS (shown) 3(2k + 5) =
∴ Pk is true ⇒ Pk +1 is true Since P1 is true, Pk is true ⇒ Pk +1 is true , by mathematical induction,
Pn is true for n ∈
+
.
n 1 = 3⎞ 3(2n + 3) ⎛ 3⎜ 2 + ⎟ n⎠ ⎝ 1 1 ∴ as n → ∞, → 3 6 ⎛ ⎞ 3⎜ 2 + ⎟ n ⎝ ⎠
3(i)
n2 − 4n + 5 = (n − 2)2 − 4 + 5 = (n − 2)2 + 1
3(ii)
N
⎛ 2 ⎞ 2 ⎜ n + 1 − n − 4n + 5 ⎟ ⎠ n =3 ⎝
∑
=
N
⎛ 2 ⎞ 2 ⎜ n + 1 − ( n − 2) + 1 ⎟ ⎝ ⎠ n =3
∑
= 32 + 1
− 12 + 1
+ 42 + 1
− 22 + 1
+ 52 + 1 M M
− 32 + 1 M M
+ ( N − 2) 2 + 1
− ( N − 4) 2 + 1
+ ( N − 1) 2 + 1
− ( N − 3) 2 + 1
+ N 2 +1
− ( N − 2) 2 + 1
=
N 2 + 1 + ( N − 1) 2 + 1 − 5 − 2
2
4(i)
Area of R = ∫
0
−1
0 3 y dx = ∫ ⎡⎣ x3 + 1⎤⎦ dx = −1 4 1
y = 1 + x3 ⇒ x = ( y − 1) 3 b
b
2
2
1
Area of S = ∫ x dy = ∫ (y -1) 3 dy = 4 3⎡ 3 b − 1 ( ) − 1⎤⎥ ⎢ 4⎣ ⎦ Equating and solve for b:
=
4 3⎡ 3 ⎤ 3 −1 b − 1 = ( ) ⎢ ⎥ 4⎣ 4 ⎦ 3
⇒ b = 1 + 2 4 = 2. 68 (3 s.f.) 4(ii)
1
For y =b , x = ( b − 1) 3 = 1. 1892 = k (say) Volume required k 2 3 2 ⎡ 2 ⎤ = π ⎢b k − 2 (1) − ∫1 ( x + 1) dx ⎥ ⎣ ⎦
= 3. 53π (or 11.1) (unit cube) 5(i)
If A, B and C are collinear, then ⎯⎯ →
⎯⎯ →
AB = λ BC
⎛b − 2 ⎞ ⎛3− b ⎞ ⎜ ⎟ ⎜ ⎟ ⎜7 −3 ⎟ = λ ⎜5− 7⎟ ⎜2− a⎟ ⎜1 − 2 ⎟ ⎝ ⎠ ⎝ ⎠ i.e. λ = −2, a = 0, b = 4
5(ii)
⎯⎯ →
⎯⎯ →
If OA is perpendicular to OB , then ⎯⎯ → ⎯⎯ →
OA OB = 0
⎛2⎞ ⎛b⎞ ⎜ ⎟⎜ ⎟ ⎜3 ⎟ ⎜7⎟ = 0 ⎜ a ⎟ ⎜ 2⎟ ⎝ ⎠⎝ ⎠ i.e. 2b + 21 + 2a = 0
3
b
4 3⎡ ⎤ 3 y 1 − ( ) ⎢ ⎥ 4⎣ ⎦2
⎛ 2 ⎞ ⎛ 3⎞ ⎜ ⎟⎜ ⎟ ⎜ 3 ⎟ ⎜ 5⎟ ⎜ a ⎟ ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠ = cos 60ο 13 + a 2 35 2(6 + 15 + a) = 13 + a 2 35 31a 2 − 168a − 1309 = 0 a = 10 (nearest int.) or
a = −4 (nearest int.)
b = −20 (nearest int.)
b = −6 (nearest int.)
His claim is not necessarily true since points O, A, B and C may not be coplanar.
6(a)
et 1 dt = t 2 (1 + 3e ) 3
∫ =
∫ 3e (1 + 3e )
t −2
t
dt
(1 + 3et )−1 1 +c = − +c −3 3(1 + 3et ) 3
(
1
∫ x sec ( x ) dx = 2 ∫ x ⎡⎣2 x sec ( x ) ⎤⎦ dx 1 = ⎡ x tan ( x ) − 2 x tan ( x ) dx ⎤ ⎥⎦ ∫ 2 ⎢⎣
6(b)
)
d tan ( x 2 ) = 2 x sec 2 ( x 2 ) dx
2
2
2
= 6(c)
∫
4
2
2
2
2
2
1⎡ 2 x tan ( x 2 ) − ln sec ( x 2 ) ⎤ + c ⎦ 2⎣
x 2 x − 3 dx
0
=−
∫
3
x 2 ( x − 3) dx +
0
∫
3
4
x 2 ( x − 3) dx
3 4
⎡ x4 ⎤ ⎡ x4 ⎤ = − ⎢ − x3 ⎥ + ⎢ − x3 ⎥ ⎣4 ⎦0 ⎣ 4 ⎦3
27 or 13.5 2 No. The statement is not always true. It applies only for (polynomial) equation in z with real coefficients. =
7(i) 7(ii)
z 4 + 3+ i = 0 ⇒ z 4 = − 3 − i ⇒
z = 2e 4
z=2
1 4
⎛ 5π ⎞ − i⎜ ⎟ ⎝ 6 ⎠
1 5π i ( − + 2 kπ ) e4 6 1 4
∴z = 2 e
-i
5π 24
1 4
1 4
=2 e
or 2 e
i
7π 24
i
(12k −5)π 24 1 4
or 2 e
i
, k = 0,1, 2,3
19π 24
4
1 4
or 2 e
-i
17π 24
7(iii)
Im
Z2
Z3
L
Re
O Z1
Z4
7(iv)
L
The quadrilateral is a square. Let the length of each side be L 1 4 2
Pythagoras Theorem: L = 2|z| =2(2 ) = 2 2 2
8(i)
2
ON = 36 − x 2 ⎡1 ⎤ A = 2 × ⎢ × (12 + 2 x ) 36 − x 2 ⎥ ⎣2 ⎦ = 2 ( 6 + x ) 36 − x 2
2x N
Q 6
P
6
U
8(ii)
R 6
• O
2x
S
6
T
dA 2x ⎞ ⎛ 1 ⎞⎛ = 2 36 − x 2 + 2 ( 6 + x ) ⎜ ⎟ ⎜ − ⎟ dx ⎝ 2 ⎠⎝ 36 − x 2 ⎠ = = =
72 − 2 x 2 − 12 x − 2 x 2 36 − x 2
4 (18 − 3x − x 2 ) 36 − x 2 4 ( 6 + x )( 3 − x ) 36 − x 2
dA = 0 : x > 0 ⇒ x = 3 cm dx d d d 1 dx 1 ⇒ =− ( QR ) = ( 2 x ) = 2 ( x ) = − dt dt dt 10 dt 20
For maximum A ,
When x = 2,
dA 4 ( 8 )(1) = = 32 = 4 2 dx 32 5
dA dA dx = × dt dx dt ⎛ 1 ⎞ = 4 2 ×⎜ − ⎟ ⎝ 20 ⎠ =−
2 cm2 s−1 5
A is decreasing at the rate of
9(a) (i)
9(a) (ii)
9(a) (iii)
2 cm s −1. 5
y = ln(1 + e x )
⇒ e y = 1 + ex d dy = ex : ey dx dx dy ⇒ = e x− y dx d d2 y ⎛ dy ⎞ = e x− y ⎜1 − ⎟ : 2 dx dx ⎝ dx ⎠ 2 d y dy ⎛ dy ⎞ ⇒ 2 = ⎜ 1 − ⎟ (shown) dx dx ⎝ dx ⎠ dy 1 d 2 y 1 When x = 0, y = ln 2, = , = dx 2 dx 2 4 1 2 x 1 y = ln 2 + x + 4 + ... 2 2 1 1 2 = ln 2 + x + x + ... 2 8 ⎛ ⎛ ⎞⎞ x2 ln(1 + e x ) = ln ⎜ 1 + ⎜ 1 + x + + ... ⎟ ⎟ 2 ⎠⎠ ⎝ ⎝ ⎛ ⎞ x2 = ln ⎜ 2 + x + + ... ⎟ 2 ⎝ ⎠ ⎛ x x2 ⎞ = ln 2 + ln ⎜ 1 + + + ... ⎟ ⎝ 2 4 ⎠ 2
⎛ x x2 ⎞ ⎜ + ⎟ ⎛ x x2 ⎞ ⎝ 2 4 ⎠ = ln 2 + ⎜ + ⎟ − + ... 2 ⎝2 4 ⎠ x x2 ⎛ 1 ⎞ ⎛ x2 ⎞ = ln 2 + + − ⎜ ⎟ ⎜ ⎟ + ... 2 4 ⎝ 2 ⎠⎝ 4 ⎠ = ln 2 +
1 1 x − x 2 + ... (verified) 2 8
6
9(b)
10 tan x − 3 = cos 2 x
( 2x) 10 x − 3 = 1 −
2
2 ⇒ x + 5x − 2 = 0 2
∴x = 10(a) (i)
−5 ± 52 − 4(−2) −5 + 33 = (rej -ve as x is small) 2 2
x = eθ cos θ ,
y = sin θ + cos θ ,
0 ≤θ ≤
π
4
dx dy = eθ (cos θ − sin θ ), = cos θ − sin θ , dθ dθ dy dy dx = = e-θ / dx dθ dθ At (eθ cos θ ,sin θ + cos θ ), the equation of the tangent is ( y − sin θ − cos θ ) = e-θ ( x − eθ cos θ ),
Set θ =
π 6
,
π
π
π
3 1 3e 6 3e 6 3 + 1 at ( ) , the equation of the tangent is ( y − − ) = e 6 (x − ), , 2 2 2 2 2
y=e
10(a) (ii)
−
π
6x+
1 2
Area under the curve C is π
A = ∫ 4 (sin θ + cos θ ) eθ (cos θ − sin θ ) dθ 0
π
= ∫ 4 eθ (cos 2 θ − sin 2 θ ) dθ 0
π
= ∫ 4 eθ cos 2θ dθ
( shown)
0
= 0.68 (2 d.p.) 10(b) (i)
y y = f '( x)
-2
x
O
x=2 7
y
10(b) (ii)
y=
1 f(x )
A’(-2, 0.5) 2
x
O
11(a)
a 1 = 1− r 2 2a = 1 − r r = 1 − 2a ⇒ 1 − 2a < 1 ⇒ − 1 < 1 − 2a < 1 1 (since r ≠ 0) 2
0 < a < 1, a ≠
11(b)
r
N ⎛1⎞ ∑ Tr = ∑ ⎜ 2 ⎟ + ∑ 2r ln 3 r =1 r =1 ⎝ 3 ⎠ r =1 N
N
N
11(c) (i)
⎛1⎞ 1− ⎜ ⎟ 1 9 = × ⎝ ⎠ + N ( N + 1) ln 3 1 9 1− 9 N 1⎛ ⎛1⎞ ⎞ = ⎜ 1 − ⎜ ⎟ ⎟ + N ( N + 1) ln 3 8 ⎜⎝ ⎝ 9 ⎠ ⎟⎠ Volume of whole cake = a 2 h + (0.9a ) 2 h + (0.9 2 a ) 2 h + (0.93 a ) 2 h + (0.9 4 a ) 2 h = (1 + 0.9 2 + 0.9 4 + 0.96 + 0.98 )a 2 h =
11(c) (ii)
[1 − (0.9 2 )5 ] 2 a h = 3.4280a 2 h 1 − 0.9 2
Cost of whole cake = $3.4280 × 200 = $686 (nearest dollar) 5 ⎡⎣ 2(d 2 ) + (5 − 1)(−d ) ⎤⎦ = 75 2 d 2 − 2d = 15
(d − 5)(d + 3) = 0 d =5 d = −3 (rej. since d >0) or No. of candles at top layer = 52 + (5 − 1)(−5) = 5
8
Dunman High School 2010 Year 6 H2 Mathematics (9740) Preliminary Examination Paper 2 Suggested Solutions SECTION A 1
( z − 1 + i)( z * − 1 − i) = 2 2
z −1+ i = 2 z −1+ i = 2
i.e. A circle with centre (1,–1) and radius
2. (shown)
Im
A O
Re
2 D C(1,-1)
B x=1.5
(i)
AB = 2 AD = 2
(ii)
( 2)
2
⎛1⎞ −⎜ ⎟ ⎝2⎠
2
= 7 Complex numbers represented are 3 ⎛ 7 ⎞ 3 ⎛ 7 ⎞ + ⎜⎜ − 1⎟⎟ i and − ⎜⎜ + 1⎟⎟ i. 2 ⎝ 2 2 2 ⎠ ⎝ ⎠ Cartesian equation of the perpendicular bisector of the line segment joining A and B is y = − 1. Since any two distinct points on the circumference of the circle are equidistance from the centre C, hence perpendicular bisector of the line segment joining these points must pass through C.
1
2(i)
y
y=a
O x = −a
x
2
Recommended (1)From the graph above, any horizontal line y = b, b ∈ therefore f is one-one and the inverse of f exists.
cuts the graph of f at most once,
OR (2)From the graph above, any horizontal line y = b, b ≠ a , cuts the graph of f exactly once, therefore f is one-one and the inverse of f exists.
(ii)
ax , x ≠ −a 2 , 2 x+a xy + a 2 y = ax y=
x=
a2 y , a− y
y ≠ a,
a2 x , x ≠ a, or a−x Rg = [− a 2 , ∞), Df = \{−a 2 },
f −1 : x a (iii)
f −1 : x a
a3 − a 2 , x ≠ a, a−x
∴ Rg ⊄ Df , thus fg does not exist
2
(iv) Method 1(Recommended) f( x) = f −1( x) ⇒ f( x) = x ax ⇒ =x x + a2 ⇒ x 2 + a 2 x − ax = 0 ⇒
x( x + a 2 − a) = 0
⇒
x = 0 or x = a − a 2 .
Method 2 (not preferred) f( x) = f −1( x)
⇒
ax a2 x = x + a2 a − x ax(a − x) = a 2 x( x + a 2 )
⇒
ax[(a + 1) x + (a 3 − a )] = 0
⇒
ax[(a + 1) x + (a 3 − a )] = 0
⇒
x = 0 or x =
⇒
dV 1 8 1 V 3 −8 = (V − 2 ) = ( 2 ) dt 60 V 60 V 2 V dV 1 = 3 V − 8 dt 60 1 1 ln | V 3 − 8 |= t + C ' 3 60
⇒
3(i)
⇒ ⇒ ⇒
t
| V 3 − 8 |= e 20
a − a 3 −a (a 2 − 1) = = − a (a − 1) = a − a 2 . a +1 a +1
+ C ''
t 20
V 3 − 8 = Ae ,
,
C '' = 3C ' A = ± eC ''
When t = 0, V = 1, A = −7, ⇒ ⇒
3
t
V = 8 − 7e 20 V 1
8 20ln( ) or 2.67 7
t
3
(ii)
d 2V = 12t 2 − 2 2 dt dV ⇒ = 4t 3 − 2t + C1 dt dV = 0, ∴ C1 = 0 When t = 0, dt dV ⇒ = 4t 3 − 2t dt ⇒ V = t 4 − t 2 + C , C is a constant. V = t4 − t2 + C 1 1 = (t 2 − ) 2 + (C − ) 2 4
V
(I)C >
1 4
(II)C ≤
1 4
C
C
(iii)
t
1 When t = 0, V = 1, then C2 = 1 > . 4 Therefore given the above initial condition, Bob’s model corresponds to solution curve type (I) in part (ii). Therefore in Bob’s model, the volume of water approaches infinity in the long run (not realistic) whereas in Andy’s model, the volume of water reasonably diminishes to zero in the long run/after some time. Thus, Andy’s model is more appropriate than Bob’s model.
4(i)
⎛ 2 ⎞ ⎛1 ⎞ ⎜ ⎟⎜ ⎟ ⎜ 4 ⎟ .⎜ 3 ⎟ ⎜1 ⎟ ⎜1 ⎟ n1 .n1 15 ⎝ ⎠⎝ ⎠ = = cos θ = | n1 || n 2 | 21 11 21 11 ∴θ = 9.3o.
4
(ii)
⎛1 ⎞ ⎜ ⎟ d = n1 × n 2 = ⎜ −1⎟ ⎜2 ⎟ ⎝ ⎠ Set z=0, 2 x + 4 y = 10
x + 3y = 8 ⇒
x = −1, y = 3
⎛ −1⎞ ⎛1 ⎞ ⎜ ⎟ ⎜ ⎟ ∴ l1: r = a1 +α d1 = ⎜ 3 ⎟ + α ⎜ −1⎟ , α ∈ . ⎜0 ⎟ ⎜2 ⎟ ⎝ ⎠ ⎝ ⎠ Alternative
2 x + 4 y + z = 10 x + 3y + z = 8 Let z = t ∈ , ⇒
2 x + 4 y = 10 − t x + 3y = 8 − t
t t x = −1 + , y = 3− , 2 2 − 1 1 ⎛ ⎞ ⎛ ⎞ t ⎜ ⎟ ⎜ ⎟ ∴ l1: r = ⎜ 3 ⎟ + α ⎜ −1⎟ , α= ∈ 2 ⎜0 ⎟ ⎜2 ⎟ ⎝ ⎠ ⎝ ⎠ (iii) Since the point with co-ordinates (6,m.5) lies on the first plane, a d1 = D1 ⇒
⎛6 ⎞ ⎛ 2⎞ ⎜ ⎟⎜ ⎟ ⇒ ⎜ m ⎟ ⎜ 4 ⎟ = 10 ⎜ 5 ⎟ ⎜1 ⎟ ⎝ ⎠⎝ ⎠ ⇒ 12 + 4m + 5 = 10 7 ⇒m=− . 4 (iv)
⎛2 ⎞ ⎛2 ⎞ ⎜ ⎟ ⎜ ⎟ l2 : r = a 2 +β d 2 = ⎜ m ⎟ + β ⎜ 0 ⎟ , β∈ . ⎜7 ⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠ ⎛1 ⎞ ⎛ 2 ⎞ ⎜ ⎟⎜ ⎟ d1 d 2 = ⎜ −1⎟ ⎜ 0 ⎟ = 2 − 2 = 0 (independent of the value of m) ⎜ 2 ⎟ ⎜ −1 ⎟ ⎝ ⎠⎝ ⎠ Therefore lines l1 and l2 are perpendicular for all real values of m. 5
SECTION B 5(i) To obtain a quota sample of size 80: Identify and categorise the parents into mutually exclusive sub-groups according to education levels. Set a quota, i.e. a target number of respondents for each group where the total adds up to 80. Poll respondents on a first-come-first-serve basis, say, when the parents arrive at school in the morning with their children, until the number for each category is filled.
(ii)
Stratified sampling is more representative in terms of the proportion of parents’ educational qualifications in each category.
(iii)
420 × 80 = 14 2400
6(i)
x
t
From GC, r = −0.860 (ii)
From GC, regression line x = −37.612 + ie, a = −37.6 , b = 261
( 3 sig fig )
Suggested model between x and
260.56 t
1 is a better fit with |r |= 0.930 > |r|= 0.860 for the linear t
model between x and t.
(iii)
x = −37.612 +
260.56 = 78.9 5.0
t = 5 lies outside the data range of t , thus model may not be valid and estimate not likely to be reliable.
6
7(i)
Let J be the event where Mylo wears a jacket and T be the event where Mylo wears a tie. P (T J ) = 0.6
(ii)
P (T ∩ J ) = 0.6 P( J ) P (T ∩ J ) = 0.6 0.2 P (T ∩ J ) = 0.6 × 0.2 = 0.12 P (T ∪ J ) '
= 1 − P (T ∪ J ) = 1 − [ P(T ) + P ( J ) − P (T ∩ J ) ] = 1 − ( 0.4 + 0.2 − 0.12 ) = 1 − 0.48 = 0.52 (iii)
0.8
J
0.2
J'
0.4
J
J 0.4 J 0.6
0.2
0.8
0.2
J' J
J'
0.6 0.8
Mon
0.6 0.4
Tue
J' J
J'
0.2
J
0.8
J'
J' Wed
Let J n be the nth day where Mylo wears a jacket. Required Probability P ( J 3 ∩ J1 ) + P ( J 3 ∩ J 2 ) = P ( J 3 ∩ J1 ) + P ( J 3 ∩ J 2 ) + P ( J1 ∩ J 2 ) (0.2)(0.6)(0.4) + (0.8)(0.2)(0.4) (0.2)(0.6)(0.4) + (0.8)(0.2)(0.4) + (0.2)(0.4)(0.2) 0.048 + 0.064 = 0.048 + 0.064 + 0.016 0.112 = 0.128 = 0.875
=
7
8(i) (a) (b)
Number of ways = 10!(5) = 18144000
S
S
S
S
Number of ways = ( 6!)( 5!)( 3) = 259200
(ii) (a) (ii) (b)
S
S
or =(5!)6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ (3!) = 259200
Number of ways = ( 4!)( 5!) = 2880
Case One: 8 Questions (4M and 4S) between B and K B K ⎛5⎞ ⎜ ⎟ ( 4!) ⎝ 4⎠
4!
Case Two: 8 Questions (3M and 5S) between B and K B K ⎛ 4⎞ ⎜ ⎟ ( 3!) ⎝ 3⎠
5!
Case Three: 9 Questions between B and K B 4!
Pure Mathematics Questions
K
5!
Statistics Questions
Number of ways ⎛ 4⎞ ⎛5⎞ = ⎜ ⎟ ( 3!)( 5!) + ⎜ ⎟ ( 4!)( 4!) + ( 5!)( 4!) ⎝ 3⎠ ⎝ 4⎠ = 2880 + 2880 + 2880 = 8640
8
9(i)
u = x − 100
Let ∴
∑ u = 50 , ∑ u
2
= 4008
Unbiased estimate of population mean: x = u + 100 = 4.16667 + 100 = 104.17 ≈ 104 (3 s.f .) Unbiased estimate of population variance: 1⎛ 50 2 ⎞ ⎜ 4008 − ⎟ = 345.42 ≈ 345 (3 s.f .) 11 ⎝ 12 ⎠ To test H0: μ = 115 against H1: μ < 115 s2 =
(ii)
One-tail test at 5% level (α = 0.05) Use t-test since σ2 is unknown and sample size of 12 is small X − 115 under H0, T = ~ t (11). 345.42/12 From GC, p-value = 0.0342 Since p-value =0.0342< 0.05, there is sufficient evidence to reject H0 at the 5% level of significance and conclude that the mean IQ score is less than 115, hence the manufacturer’s claim is disputable.
(iii) (a) The IQ score of customers is normally distributed.
10 (i)
(ii)
(iii)
(iv)
(b) For 2-tailed test, p-value =2(0.0324) =0.0684 > 0.05. H0 will not be rejected. The conclusion would be different. Let X be the number of unsolicited text messages received in a day. X ~ Po 5 7 P( X = 2) = 0.125 (3 s.f.) Let Y be the number of unsolicited text messages or phone calls received in a week. Y ~ Po ( 8 )
( )
P(Y ≤ 10) = 0.816 (3 s.f.) (shown) Let W be the number of weeks where receives more than 10 unsolicited text messages or phone calls in a week out of 10 weeks. W ~ B (10, 0.184 ) P(W > 3) = 1 − P(W ≤ 3) = 0.0944 (3 s.f.) Let T be the total number of unsolicited text messages or phone calls received in the next 2 weeks.
9
T ~ Po (16 ) Since λ =16>10, ∴ T ~ N(16,16) approximately. P(T ≥ 20) = P(T > 19.5)
(apply c.c.)
= 0.191 (3 s.f.) 11 (a)
Y − X ~ N(2, 2σ 2 )
P(Y − X > 3) = 0.4 3− 2 P(Z > ) = 0.4 2σ From GC, 1 = 0.25335 2σ σ = 2.7910 Var( X1 + X 2 ) = 2 × 2.79102 = 3.94712 X1 + X 2 ~ N (8,3.94712 ) b(i)
P(8 < X1 + X 2 < 12) = 0.345 (3 s.f.) Let X min be the amount of time spent by a student online each day. E( X1 + X 2 L + X 60 ) = 60 × 120 = 7200
(
Var( X1 + X 2 L + X 60 ) = 60 × 452 = 90 15
)
2
Since n=60 is large, by Central Limit Theorem, 2⎞ ⎛ X1 + X 2 L + X 60 ~ N ⎜ 7200, 90 15 ⎟ approximately. ⎝ ⎠
(
)
P(X1 + X 2 L + X 60 ≥ 7000) (ii)
= 0.717 (3 s.f.) Since n=60 is large, by Central Limit Theorem, ⎛ 452 ⎞ ⎟ approximately. X ~ N ⎜ 120, ⎜ 60 ⎟⎠ ⎝ P( X − 120 < 5) = P(−5 < X − 120 < 5) = P(115 < X < 125) = 0.611 (3 s.f.) We do not need to assume that the amount of time spent online follows a normal distribution since by the Central Limit Theorem, the sample mean follows a normal distribution approximately when n is large.
10
1
Solve the inequality x 4 , ≤ x − 2 ( x − 2) 2 giving your answer in an exact form. Hence solve
2
ex 4 ≤ x x e + 2 (e + 2
2
.
[3] [2]
)
The sequence of numbers un , where n = 0, 1, 2, 3, …, is such that u0 = −2 and (n + 2)un −1 . un = 2un −1 + n + 1 Proof by induction that, for n ≥ 0, un =
3
n+2 . 2n − 1
The functions f and g are defined as follows: 2 f : x (2 x − 1) − 2, x < −1 ,
g:x ln ( x + a ) ,
4
[5]
x > −1 .
(a)
Define f −1 in a similar form.
[3]
(b)
State the value of a such that the range of g is (0, ∞) .
[1]
(c)
Show that the composite function gf exists, and find the range of gf, giving your answer in terms of a. [2]
A curve is defined by the parametric equations t t , , where t ≠ −1, 1. x= y= 2 1+ t 1− t2 (i)
Show that the tangent to the curve at any point with parameter t has equation
(1 − t )
2 3
(ii)
y= (1 + t 2 ) x − 4t 3. 3
[3]
1 . Hence determine the 2 acute angle between this tangent and the line y= x + 3 . [3] Find the gradient of the tangent to the curve at t =
2
[Turn Over
5
Robert took a study loan of $100 000 from a bank on 1st January 2010. The bank charges an annual interest rate of 10% on the outstanding loan at the end of each year. After his graduation, Robert pays the bank $x at the beginning of each month. The first payment is made on 1st January 2014. Let un denote the amount owed by Robert at the end of nth year after 2013, where n ∈ +0 . (i)
Find u0 .
[1]
(ii)
Show that un = 1.1n u0 − kx(1.1n − 1) , where k is a constant to be determined. [4]
(iii) Given that Robert owes the bank less than $1000 at the end of 2020, find the minimum value of x, giving your answer to the nearest dollar. [3]
6
7
∫
1 dt . 3 − 4t 2
[3]
(a)
Find
(b)
Use the substitution u = 5 x to find ∫ 5 x cos 2 ( 5 x ) dx .
[5]
It is given that the function y = f ( x ) has the Maclaurin’s series 1 + 4 x + ax 2 + ... and
(
satisfies 1 + x 2
b (1 + y ) , where a and b are real constants. ) ddyx = 2
(i)
Show that b = 2 and find the value of a.
(ii)
Find the series expansion of including the term in x 2 .
[4]
f( x) in ascending powers of x, up to and 4+ x [3]
(iii) State the equation of the normal to the curve y =
3
f( x) at x = 0. 4+ x
[1]
[Turn Over
8
A B C 4−r in the form . + + r −1 r r + 2 (r − 1)r (r + 2)
(i)
Express
(ii)
Hence find
n
4−r
∑ (r − 1)r (r + 2)
[2]
.
[3]
r =2
Give a reason why the series is convergent, and state its limit. n
(iii) Use your answer to part (ii) to find
[2]
3− r
∑ r (r + 1)(r + 3) .
[2]
r =2
9
On a single Argand diagram, sketch the loci given by 2 (i) z −1 − i ≥ 2 , z +1 π arg ≥ , 3 + i 12 (iii) z > z − 1 .
[7]
Hence, or otherwise, find the range of values of z − i and arg ( z − i).
[3]
(ii)
10
A file is downloaded at r kilobytes per second from the internet via a broadband connection. The rate of change of r is proportional to the difference between r and a constant. The initial value of r is 348. If r is 43, it remains at this constant value. (i)
dr Show that = k (r − 43) . dt
[2]
(ii)
Hence obtain an expression for r in terms of k and t.
[4]
The total amount of data downloaded, I kilobytes, in time t seconds, is given by dI =r . dt
(iii) Given that there is no data downloaded initially, find I in terms of k and t. [2] (iv) It is given that a file with a size of 5700 kilobytes takes 90 seconds to download. Find the value of k . [2] (v)
Explain what happens to the value of r in the long run.
4
[1]
[Turn Over
D 11 C j k
i
A
O
B The diagram above shows part of the structure of a modern art museum designed by Marcus, with a horizontal base OAB and vertical wall OADC. Perpendicular unit vectors i, j, k are such that i and k are parallel to OA and OC respectively. The walls of the museum BCD and ABD can be described respectively by the equations −1 14 5 −1 r ⋅ −5 =36 and r = 0 + λ 4 + µ 0 , where λ , µ ∈ . 6 0 0 4 (i)
Write down the distance of A from O.
[1]
(ii)
Find the vector equation of the intersection line of the two walls BCD and ABD. [3]
(iii) Marcus wishes to repaint the inner wall ABD. Find the area of this wall.
[3]
Suppose Marcus wishes to divide the structure into two by adding a partition such that it intersects with the walls BCD and ABD at a line. This partition can be described by the equation 2 x − 7 y + α z = β , where α , β ∈ . (iv) Find the values of α and β . (v)
[2]
Another designer, Jenny, wishes to construct another partition which is described by the equation 2 x − 7 y + α z = γ , where γ ≠ β . State the relationship between Jenny’s and Marcus’ partitions. [1] Deduce the number of intersection point(s) between the walls BCD, ABD, and Jenny’s partition. [1]
5
[Turn Over
12
The curves C1 and C2 have equations ( x − 2) 2 = a 2 (1 − y 2 ) and y = 1 < a < 2, respectively. Describe the geometrical shape of C1.
x2 − 4 , where x +1 [1]
(a)
State a sequence of transformations which transforms the graph of x 2 + y 2 = 1 to the graph of C1 . [3]
(b)
(i)
Sketch C1 and C2 on the same diagram, stating the coordinates of any points of intersection with the axes and the equations of any asymptotes. [6]
(ii)
Show algebraically that the x-coordinates of the points of intersection of C1 and C2 satisfy the equation
( x + 1) ( x − 2 ) 2
(iii)
2
= a 2 ( x + 1) 2 − a 2 ( x 2 − 4) 2 .
[2]
Deduce the number of real roots of the equation in part (ii).
6
[1]
[Turn Over
Section A: Pure Mathematics [40 marks] 1
A tin has a fitting cylindrical lid which overlaps its cylindrical body by 5 cm. When completely closed, it has base radius x cm and height y cm, as shown in the diagram. The body and the lid are made from thin metal sheet such that the difference in their radii is negligible. The total area of metal sheet used to make the tin and its lid is 400π cm2.
Show that the volume V cm3 of the tin is given by = V π x ( 200 − x 2 − 5 x ) .
If x varies, find the values of x and y for which V has its maximum value.
2
(i)
[6]
Solve the equation z 5 − 32 = 0, expressing the answers in the form reiθ , where r > 0 and −π < θ ≤ π .
[2]
2w + 1 Explain why the equation 0 has four roots. − 32 = w
[1]
5
(ii)
The roots of the equation are denoted by w1, w2, w3 and w4. By finding 4
show that
1
∑w i =1
is a real number.
1 , w [4]
i
2
[Turn Over
3
The nth term and the sum of the first n terms of a sequence are denoted by un and Sn respectively. Given that Sn is a quadratic polynomial in n and u1 = 100 , u2 = 90 and 10
∑u r =3
r
= 360 , find Sn in terms of n.
[4]
Hence show that this sequence is an arithmetic progression.
4
[3]
The points A, B, C have position vectors i, 2j − tk, tk relative to an origin O respectively, where t is a fixed constant. The points X and Y lie on AB and BC respectively such that AX BY µ , = = XB YC 1 − µ where µ is a parameter such that 0 < µ < 1.
5
(i)
Find the vector XY in terms of t and µ .
[2]
(ii)
Prove that O, X, Y are non-collinear.
[2]
(iii) Determine if XOY can be 90ο , justify your answer.
[3]
(iv) Find the projection vector of XY onto 4i + j.
[2]
A point P(x, y) moves in the x-y plane such that the distance from the line x = −3 is always equal to the distance from the point ( 6, −2 ) . (i)
Prove that the locus of P can be represented by a curve C with equation
( y + 2) (ii)
2
= 9 ( 2 x − 3) .
[2]
Sketch the curve C, making clear the main relevant features of C.
[2]
The region R is bounded by the curve C, the lines y = 7 and x = 2 . Find (iii) the exact area of R , showing your working clearly,
[4]
(iv) the volume of revolution formed when R is rotated through 2π radians about the x-axis. [3] 3
[Turn Over
Section B: Statistics [60 marks] 6
An airline wishes to assess its in-flight service for a specific flight and employs a marketing research company to administer a survey. The seats on this flight are divided into classes as follows:
No. of Seats
First Class 20
Business Class 80
Economy Class 300
Total 400
An employee of the company proposes using stratified sampling method to select 80 seats and ask the passengers occupying these seats to complete a questionnaire. Describe how this sample can be obtained. Suggest a practical difficulty which may be encountered in carrying out this proposal. [3] Another employee suggests using simple random sampling method to choose a sample of 80 passengers from the list of passengers who have checked in. Explain why it may not be appropriate to use this sampling method. [1]
7
A boy intends to arrange a set of coloured square tiles flat on the floor as shown below with each row being labelled. First row Second row Third row Fourth row
There are four identical blue tiles, three identical yellow tiles, two identical green tiles and one red tile. Find the number of ways to arrange the tiles if (i)
there are no restrictions,
[1]
(ii)
exactly one yellow tile and exactly two blue tiles are in the same row,
[3]
(iii) there are less than 3 yellow tiles in the fourth row .
[2]
All the ten tiles are now placed in a row. Find the number of ways he can arrange the tiles such that all the blue tiles are separated. [2]
4
[Turn Over
8
The life span, x (in hours), of a certain electronic component is known to follow a normal distribution with mean 9000 hours and standard deviation σ hours. Following a change in the manufacturing procedure, a batch of components is produced and a random sample of 10 components is taken from this batch. (a)
The life spans of the 10 components are summarized by
Σ(x – 9000) = 2010, Σ(x – 9000)2 = 911 157. Test at the 5% level of significance whether the mean life span of the components has increased after the change in the manufacturing procedure. [3] (b)
9
Let x denote the mean life spans (in hours) of the 10 components in the sample. If σ = 25, find the set of values of x so that we can conclude at the 1% level of significance that the population mean life span of the components has increased. State an assumption for the above test to be valid. [5]
An orchard owned by Mark produces oranges which have masses (in grams) that follow a normal distribution N(190, 576). Visitors to this orchard can buy the oranges at $0.10 per 100 g. Find the probability that the payment made by a visitor buying 20 oranges will differ from 2 times the payment made by another visitor buying 10 oranges by at most $0.15. [4] The orchard produces apples which are graded according to their mass. Apples with a mass exceeding 150 g are graded as 'large' while apples with a mass less than 70 g are graded as 'small'. Mark finds that the proportion of apples graded as 'large' is the same as that of the apples graded as 'small'. It is given that the mass of a randomly chosen apple from Mark's orchard follows a normal distribution with mean µ g and standard deviation 30 g. (i)
Find the value of µ .
(ii)
What is the probability that Mark will get at least 5 apples graded as 'large' when he randomly selects 65 apples from his orchard? [3]
[1]
5
[Turn Over
10 A public opinion poll surveyed a sample of 1000 voters. The table below shows the number of males and females supporting Party A, Party B and Party C.
Male Female (a)
Party A 200 250
Party B 130 300
Party C 70 50
One of the voters is chosen at random. Events A, C and M are defined as follows: A : The voter chosen supports Party A. C : The voter chosen supports Party C. M : The voter chosen is a male. Find (i) P(A M ), (ii)
P(M ' ∩ C ').
Determine whether A and M are independent. (b)
[4]
It is given that in the sample, 20% of Party A supporters, 30% of Party B supporters and 5% of Party C supporters are immigrants. (i)
One of the voters selected from the sample at random is an immigrant. What is the probability that this voter supports Party A? [2]
(ii)
Three voters are chosen from the sample at random. Find the probability that there is exactly one immigrant voter who supports Party C or exactly one female who supports Party A (or both). [4]
11 At the counselling centre CareforSociety, the average number of call-ins received in a month regarding alcohol abuse problem is denoted by λ . The probability of receiving at most 9 such calls in a week is 0.701. Assuming that there are 4 weeks in a month, (i)
show that the value of λ is 32.5, correct to 3 significant figures. State a condition under which the distribution used is valid. [3]
(ii)
by using a suitable approximation, find the probability that in a month, the number of call-ins received is more than 25 but not more than 40. [4]
The centre also has 70 support groups, each consisting of n people, which help one another to deal with alcohol abuse problems. It is known that, on average, 3 in 20 people in such support groups will be successful in correcting their alcohol abuse problem. Given that in the 70 support groups, the probability of having an average of at least 4 people per group successfully correcting their alcohol abuse problem is more than 0.7, determine the minimum value of n. [4]
6
[Turn Over
12 With the implementation of the new bus fare system, Jasmine wanted to know how the new system would affect her. She identified 12 common locations and used a map to measure the straight line distance, x km, of each location from her home. She also measured the road distance, y km, of each location from her home and the corresponding bus fare, s cents. The data are shown below. Location Straight line distance, x Road distance, y Bus fare, s (i)
A
B
7.7
3.0
8.8 121
C
D
E
F
G
H
24.1 13.2
9.3
9.0
10.4
3.5
3.3
28.0 16.1
9.4
8.9
12.5 15.8 22.5 5.0 2.2 2.8
81
181
125 121
149
137
149
I
J
K
L
17.6 4.5 2.0 2.5
173
91
71
71
By considering the values of x and y, explain why Location F should be omitted from any further analysis. State, with a reason, another location that should be omitted. [2]
Omit the data for the two locations in (i). (ii)
Use a suitable regression line to give an estimate of the straight line distance when the road distance is 20.0 km. [2]
(iii) Draw a scatter diagram of s against y. State, with a reason, which of the following models is more appropriate to describe the relationship between y and s: Model I: s= a + by 2 , Model II: s= a + b ln y , where a and b are positive constants. [3] (iv) Using the more appropriate model found in (iii), calculate the equation of the corresponding regression line. [2] (v)
Comment on the use of the regression line found in (iv) to estimate the road distance travelled if the bus fare is 170 cents. [2]
7
[Turn Over
2010 HCI Prelim Paper 1 Solutions Qtn 1.
Solutions x 4 − ≤0 x − 2 ( x − 2) 2
( x − 1) − 5 ≤ 0 x2 − 2 x − 4 ⇒ ≤0⇒ 2 ( x − 2) 2 ( x − 2) 2
⇒ ( x − 1) 2 − 5 ≤ 0, , ( x − 2) 2 is always positive for all real values of x.
⇒ 1− 5 ≤ x ≤ 1+ 5 , x ≠ 2 ex 4 ≤ x For x e + 2 (e + 2) 2
Replace x by −e x , ⇒ 1 − 5 ≤ −e x ≤ 1 + 5 ⇒ x ≤ ln 2.
(
)
5 −1
Let P(n) be the proposition un =
n+2 . 2n − 1
When n = 0, LHS of P(0) = u0 = −2 (given) 2 RHS of P(0) = = −2 −1 ∴ P(0) is true. Assume P(k) is true for some k ∈ i.e. uk =
+
∪ {0}
k +2 . 2k − 1
Show that P(k+1) is true k +3 i.e. uk +1 = . 2k + 1 When n = k + 1, LHS of P(k+1) = uk +1 =
( k + 3)( k + 2 ) =
(k + 3)uk 2uk + k + 2
2k − 1 k 2⎞ + ⎛ 2⎜ ⎟+k +2 ⎝ 2k − 1 ⎠
⎡ (k + 3)(k + 2) ⎤ = ⎢ ⎥ ⎣ (2k + 1)(k + 2) ⎦
2
[Turn Over
k +3 = RHS of P(k+1) 2k + 1 Since P(0) is true & P(k) is true ⇒ P(k + 1) is also true, hence by mathematical induction P(n) is true for all n ∈ + ∪ {0} . =
3 y = f ( x)
7 -1 x = −a
y = g ( x) x
1− a
(a)
y = (2 x − 1) − 2 2
± y + 2 = 2x − 1 1 1 y + 2 Q x < −1 − 2 2 1 1 ∴ f −1 : x a − x + 2, x>7 2 2 a=2 x=
(b)
(c)
Rf = (7, ∞ ) , Dg = (− 1, ∞ ) .
Since Rf ⊆ Dg , gf exists. f g → ( 7, ∞ ) ⎯⎯ → ( ln ( 7 + a ) , ∞ ) ( −∞, −1) ⎯⎯
Df
4(i)
Rf
Rgf
dx 1− t2 = dt (1 + t 2 )2 dy 1+ t2 = dt (1 − t 2 )2
3
[Turn Over
3
dy ⎛ 1 + t 2 ⎞ =⎜ ⎟ dx ⎝ 1 − t 2 ⎠ Equation of tangent: 3
⎛ 1+ t2 ⎞ ⎛ t t ⎞ y− = x− ⎜ 2 2 ⎟ ⎜ 2 ⎟ 1− t ⎝ 1− t ⎠ ⎝ 1+ t ⎠
(1 − t )
2 3
y = (1 + t 2 ) x − t (1 + t 2 ) + t (1 − t 2 ) 3
2
2
= (1 + t 2 ) x − 4t 3 3
(ii)
When t =
1 dy , = 27 2 dx
Let α be the acute angle between the two lines.
α B
A
Note: α = A - B tan A = 27 , tan B = 1
A = tan −1 27 = 87.879° B = tan −1 (1) = 45°
α = A − B = 42.9° Alternative Solution tan α =
27 − 1 26 = 1 + ( 27 )(1) 28
α = tan −1
26 = 42.9° 28
5(i)
u0 = $1.14 (100 000) = $146 410
(ii)
u1 = 1.1(u0 − 12 x)
u2 = 1.1[1.1(u0 − 12 x) − 12 x ] = 1.12 u0 − 1.12 (12 x) − 1.1(12 x)
4
[Turn Over
u3 = 1.1 ⎡⎣1.12 u0 − 1.12 (12 x) − 1.1(12 x) − 12 x ⎤⎦ = 1.13 u0 − 1.13 (12 x) − 1.12 (12 x) − 1.1(12 x) : : n n n −1 un = 1.1 u0 − 1.1 (12 x) − 1.1 (12 x) − ... − 1.1(12 x) = 1.1n u0 − 12 x (1.1n + 1.1n −1 + ... + 1.1) ⎛ 1.1(1.1n − 1) ⎞ ⎟ = 1.1 u0 − 12 x ⎜ ⎜ ⎟ 0.1 ⎝ ⎠ n
= 1.1n u0 − 132 x (1.1n − 1)
(iii)
n = 7 at end of 2020 1.17 u0 − 132 x(1.17 − 1) < 1000 x > $2270.30 Least x to the nearest dollar = $2271
6(a)
1 dt 3 − 4t 2 1 1 dt =− 4 t2 − 3 4 1 1 dt =− 2 4 ⎛ ⎞ 3 t2 − ⎜ ⎟ ⎝ 2 ⎠
∫
∫ ∫
⎡ t− 1⎢ 1 ln =− ⎢ 4⎢ 3 t+ ⎢⎣ =− (b)
3 2 3 2
⎤ ⎥ ⎥ +C ⎥ ⎥⎦
3 2t − 3 ln + C 12 2t + 3
u = 5x du = 5 x ln 5 dx dx 1 1 ∴ = x = du 5 ln 5 u ln 5
5
[Turn Over
∫
5 x cos 2 (5 x ) dx
= ∫ u cos 2 u ⋅
1 du u ln 5
1 cos 2 u du ∫ ln 5 1 = (1 + cos 2u ) du 2 ln 5 ∫ 1 ⎡ sin 2u ⎤ u+ = +C ⎢ 2 ln 5 ⎣ 2 ⎥⎦ =
=
7i
1 ⎡ x sin 2(5x ) ⎤ ⎢5 + ⎥+C 2 ln 5 ⎣ 2 ⎦
When x = 0, y = 1,
dy = 4. dx
(1 + x ) ddyx = b (1 + y ) 2
2
dy = b(2) = 4 dx ⇒ b = 2 (Shown)
⇒
(1 + x ) 2
d2 y dy ⎛ dy ⎞ + 2x = 2⎜ 2y ⎟ 2 dx dx ⎝ dx ⎠
d2 y = 16 dx 2 16 a= =8 2!
ii
f( x)(4 + x)
−
1 2
1 x −1 = (1 + 4 x + 8 x 2 + ...)(1 + ) 2 2 4
iii
1 3 (− )(− ) 1 1 x 2 2 ( x ) 2 + ...) = (1 + 4 x + 8 x + ...)(1 + (− )( ) + 2 2 2 4 2! 4 1 3 2 x x + ...) = (1 + 4 x + 8 x 2 + ...)(1 − + 2 8 128 1 31 963 2 = (1 + x + x + ...) 2 8 128 16 Gradient of normal = − 31 Equation of normal: 6
[Turn Over
y= 8(i)
(ii)
1 16 − x 2 31
4−r A B C = + + (r − 1)r (r + 2) r − 1 r r + 2 4 − r = Ar (r + 2) + B(r − 1)(r + 2) + C (r − 1)r A = 1, B = −2, C = 1 n 4−r ∑ r = 2 ( r − 1) r ( r + 2) n
1 2 1 − + r r+2 r =2 r − 1 1 = 1−1+ 4 1 2 1 + − + 2 3 5 1 2 1 + − + 3 4 6 1 2 1 + − + 4 5 7 M + =∑
1 2 1 − + n − 4 n − 3 n −1 1 2 1 + − + n−3 n−2 n 1 2 1 + − + n − 2 n −1 n +1 1 2 1 + − + n −1 n n + 2 1 2 1 1 1 2 1 = − + + + − + 2 3 3 n n +1 n n + 2 1 1 1 1 = − + + 6 n n +1 n + 2 1 1 ⎞ ⎛ 1 lim ⎜ − + + ⎟ = 0 , hence the series in (ii) converges. n →∞ ⎝ n n +1 n + 2 ⎠ +
(iii)
∞
4−r
1
∑ (r − 1)r (r + 2) = 6 r =2
7
[Turn Over
(iv)
3− r
n
∑ r (r + 1)(r + 3) r =2
1 3 − (n − 1) 3− n + ... + + (2)(3)(5) (n − 1)(n)(n + 2) n(n + 1)(n + 3) n +1 4−r 2 =∑ − (1)(2)(4) r = 2 ( r − 1) r ( r + 2) 1 1 1 1 1 = − + + − 6 n + 1 n + 2 n + 3 12 1 1 1 1 =− − + + 12 n + 1 n + 2 n + 3 =
9.
y Q
P 1
R (1,1)
π 4
-1
π
O
1
x
6
z −1− i 2 ≥ 2
⇒ z − (1 + i) ≥ 2
π
⎛ z +1 ⎞ ≤ arg ⎜ ⎟≤π 12 ⎝ 3+i⎠ ⇒ ⇒
π
12
π
4
+
π
6
≤ arg( z + 1) ≤ π +
≤ arg( z + 1) ≤
7π 6
π 6
1 7 7 1 = + = 2 ; PQ = 4 2 4 4 Method 2: QR is the perpendicular bisector, so PQ = 2 (radius) ⇒ z −i > 2
Method 1: QR = 2 −
π 4
≤ arg ( z − i ) <
π 2
8
[Turn Over
10(i)
(ii)
Let the constant be a. dr ∴ = k (r − a) , where k is a constant. dt dr Given r = 43 when =0, dt ∴ 0 = k (43 − a ) Since k ≠ 0, then a = 43 dr ∴ = k (r − 43) (shown) dt 1 ∫ r − 43 dr = k ∫ dt ln r − 43 = kt + C1 r − 43 = e kt + C1 r = 43 + Ae kt where A = eC1 When t = 0 , r = 348 . ∴ A = 305 . ∴ r = 43 + 305e kt
(iii)
I = ∫ r dt = ∫ (43 + 305e kt ) dt 305 kt e + C2 k When t = 0 , I = 0 . 305 ∴ C2 = − k 305 kt (e − 1) ∴ I = 43t + k Given I = 5700 and t = 90 , 305 90 k ∴ 5700 = 43(90) + (e − 1) k 305 90 k 1830 = (e − 1) k 6k = e90 k − 1 = 43t +
(iv)
Solving using GC, k = − 0.167 or k = 0 (NA) (v)
r = 43 + 305e
1 − t 6 1 − t 6
If t becomes larger, 305e → 0 , r → 43 Hence r would be reduced to a steady 43 kilobytes per second in the long run.
9
[Turn Over
11i ii
iii
OA = 14 Plane ABD ⎛ 5 ⎞ ⎛ −1⎞ ⎛ 16 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 4 ⎟ × ⎜ 0 ⎟ = ⎜ −20 ⎟ ⎜0⎟ ⎜ 4 ⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
OR ⎛14 + 5λ − μ ⎞ ⎛ −1 ⎞ ⎜ ⎟⎜ ⎟ 4λ ⎜ ⎟ . ⎜ −5 ⎟ = 36 ⎜ ⎟⎜6⎟ 4μ ⎝ ⎠⎝ ⎠
⎛ 4 ⎞ ⎛14 ⎞ ⎛ 4 ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ r. ⎜ −5 ⎟ = ⎜ 0 ⎟ . ⎜ −5 ⎟ = 56 %⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝1⎠ ⎝0⎠⎝1⎠ ⎧ 4 x − 5 y + z = 56 ⇒⎨ ⎩− x − 5 y + 6 z = 36 Using GC to solve:
−14 − 5λ + μ − 20λ + 24μ = 36 25μ = 25λ + 50 μ =λ+2 ⎛14 ⎞ ⎛ 5⎞ ⎛ −1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r = ⎜ 0 ⎟ + λ ⎜ 4 ⎟ + (λ + 2) ⎜ 0 ⎟ ⎜0⎟ ⎜ 0⎟ ⎜4⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ 4⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ ∴ r = ⎜ −8 ⎟ + γ ⎜ 1⎟ , γ ∈ % ⎜ ⎟ ⎜ 1⎟ ⎝0⎠ ⎝ ⎠
⎛12 ⎞ ⎛ 4⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ 0 ⎟ + λ ⎜ 4⎟ ⎜8⎟ ⎜ 4⎟ ⎝ ⎠ ⎝ ⎠ ⎛12 ⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ 0 ⎟ + γ ⎜ 1⎟ , γ ∈ ⎜8⎟ ⎜ 1⎟ ⎝ ⎠ ⎝ ⎠
⎛ 4+γ ⎞ ⎛ 12 ⎞ uuuv ⎜ uuuv ⎜ ⎟ ⎟ OD = ⎜ −8 + γ ⎟ ⇒ γ = 8 ⇒ OD = ⎜ 0 ⎟ ⎜ γ ⎟ ⎜8⎟ ⎝ ⎠ ⎝ ⎠ (Reason: j is zero.) % ⎛ 4⎞ uuuv ⎜ ⎟ OB = ⎜ −8 ⎟ ⎜0⎟ ⎝ ⎠ ⎛12 ⎞ ⎛14 ⎞ ⎛ −2 ⎞ uuuv ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AD = ⎜ 0 ⎟ − ⎜ 0 ⎟ = ⎜ 0 ⎟ ⎜8⎟ ⎜0⎟ ⎜ 8⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛12 ⎞ ⎛ 4 ⎞ ⎛ 8 ⎞ uuuv ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ BD = ⎜ 0 ⎟ − ⎜ −8 ⎟ = ⎜ 8 ⎟ ⎜ 8 ⎟ ⎜ 0 ⎟ ⎜8⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
10
[Turn Over
1 uuuv uuuv BD × AD 2 ⎛ 8 ⎞ ⎛ −2 ⎞ ⎛ 1 ⎞ ⎛ −1 ⎞ ⎛ 4⎞ 1⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ 8 ⎟ × ⎜ 0 ⎟ = 8 ⎜ 1 ⎟ × ⎜ 0 ⎟ = 8 ⎜ −5 ⎟ 2⎜ ⎟ ⎜ ⎟ ⎜ 1⎟ ⎜ 4 ⎟ ⎜1⎟ ⎝8⎠ ⎝ 8 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Area ABD =
iv
= 8 42 = 51.8 (3 s.f.) 2(4) − 7(−8) + α (0) = β
2(12) − 7(0) + 8α = β ⇒ β = 64, α = 5 OR The 3 planes intersect at the line ⎛ 4⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜ −8 ⎟ + γ ⎜ 1⎟ , γ ∈ % ⎜ ⎟ ⎜ 1⎟ ⎝0⎠ ⎝ ⎠ ⎛ 2 ⎞ ⎛ 1⎞ ⎜ ⎟⎜ ⎟ ⎜ −7 ⎟ . ⎜ 1 ⎟ = 0 ⎜ α ⎟ ⎜ 1⎟ ⎝ ⎠⎝ ⎠ 2 − 7 +α = 0
α =5 ⎛ 4⎞⎛ 2⎞ ⎜ ⎟⎜ ⎟ ⎜ −8 ⎟ . ⎜ −7 ⎟ = 8 + 56 = 64 ⎜0⎟⎜ 5⎟ ⎝ ⎠⎝ ⎠ β = 64 v
Their partitions are parallel to each other.
12
There is no intersection point. Ellipse
(a)
( x − 2) 2 = a 2 (1 − y 2 ) ⇒
( x − 2) 2 + y2 = 1 a2
Method 1: Sequence of transformations: 1) Scale // to x-axis by factor a. 2) Translate in the positive x-direction by 2 units.
11
[Turn Over
Method 2: Sequence of transformations:
1) Translate in the positive x-direction by
2 units. a
2) Scale // to x-axis by factor a. x −1 −4
(bi) x +1 x
2
−( x 2 + x) −x−4 − (− x − 1) −3 x = −1
y = x −1
y
2−a
−2
(bii)
2+a
2
( x − 2) + y2 = 1 a2 2
−4
y=
x
x2 − 4 x +1
x2 − 4 into ( x − 2) 2 = a 2 (1 − y 2 ) : x +1 ⎛ ⎛ x 2 − 4 ⎞2 ⎞ 2 2 ⎟ ( x − 2) = a ⎜1 − ⎜ ⎜ ⎝ x + 1 ⎟⎠ ⎟ ⎝ ⎠
Sub y =
⇒ ( x + 1) ( x − 2) 2 = a 2 ( x + 1) − a 2 ( x 2 − 4 ) --- (*) 2
2
2
(shown) Hence the x-coordinate of the points of intersection of C1 and C2 satisfy equation (*). (b) (iii)
From (ii), number of intersection points between C1 and C2 gives the number of real roots of the equation (*). From the graphs, there are 2 points of intersection between C1 and C2 . Hence 2 real roots.
12
[Turn Over
HCI Prelim H2 Mathematics P2 Solutions
2010 HCI H2 Mathematics Preliminary Examination Paper 2 Solution Qn Solutions Surface area of the tin and lid 1 = 2π x 2 + 2π xy + 10π x = 400π
y=
200 − x 2 − 5 x x
Volume of the container ⎛ 200 − x 2 − 5 x ⎞ = π x2 ⎜ ⎟ x ⎝ ⎠ = π ( 200 x − x3 − 5 x 2 )
dV = π ( 200 − 3 x 2 − 10 x ) dx dV 20 =0⇒ x= or x = −10 (rejected) dx 3 20 d 2V = π ( −6 x − 10 ) < 0 when x = 2 3 dx 20 V is maximum when x = . 3 20 55 When x = , y= 3 3 (or x = 6.67, y = 18.3). 2(i)
z 5 − 32 = 0 ⇒ z 5 = 32 ei0 = 32ei2k π
⇒ z = 2e (ii)
2 kπ i
5
where k = 0, ± 1, ± 2. 5
⎛ 2w + 1 ⎞ 5 The highest power in the equation ⎜ ⎟ = 32 is four since the terms with w are ⎝ w ⎠ canceled out. Hence the equation has only four roots. 5
1⎞ ⎛ ⎜ 2 + ⎟ = 32 w⎠ ⎝ 2 kπ i 1 ⇒ 2 + = z = 2e 5 w 2 kπ i 2 kπ i 1 ⇒ = 2e 5 − 2 = 2 e 5 − 1 w 1 1 1 1 ⇒ + + + w1 w2 w3 w4
)
(
(
= 2⎡ e ⎢⎣
2π i
5
)(
−1 + e
−2π i
5
)(
−1 + e
4π i
5
)(
−1 + e
1
−4π i
5
)
−1 ⎤ ⎥⎦
HCI Prelim H2 Mathematics P2 Solutions
2π 4π ⎡ ⎤ + 2 cos − 4⎥ = 2 ⎢ 2 cos 5 5 ⎣ ⎦ 2 π 4 π ⎡ ⎤ + cos − 2⎥ ∈ . = 4 ⎢cos 5 5 ⎣ ⎦
Or use GC, 3
1 1 1 1 + + + = −10. w1 w2 w3 w4
S n = an 2 + bn + c
U1 = S1 = a + b + c = 100 S 2 = 4a + 2b + c = 190 S10 = 100a + 10b + c = 360 + 100 + 90 = 550 Using GC, a = −5 , b = 105 , c = 0 Thus Sn = −5n 2 + 105n
U n = Sn − Sn−1
(
= −5n 2 + 105n − −5 ( n − 1) + 105 ( n − 1) 2
)
= 110 − 10n U n − U n −1
4i
= 110 − 10n − (110 − 10n + 10 ) = −10 (a constant) Hence sequence is an AP. ⎛0⎞ ⎛ 1 ⎞ ⎛1 − μ ⎞ uuuv ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ OX = μ ⎜ 2 ⎟ + (1 − μ ) ⎜ 0 ⎟ = ⎜ 2 μ ⎟ ⎜ −t ⎟ ⎜ 0 ⎟ ⎜ −t μ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 0 ⎞ ⎛0⎞ ⎛0⎞ ⎛ uuuv ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ OY = μ ⎜ 0 ⎟ + (1 − μ ) ⎜ 2 ⎟ = ⎜ 2 − 2 μ ⎟ ⎜t⎟ ⎜ −t ⎟ ⎜ −t + 2t μ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0 ⎞ ⎛1 − μ ⎞ ⎛ μ − 1 ⎞ ⎛ uuuv ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ XY = ⎜ 2 − 2 μ ⎟ − ⎜ 2 μ ⎟ = ⎜ 2 − 4 μ ⎟ ⎜ −t + 2t μ ⎟ ⎜ −t μ ⎟ ⎜ −t + 3t μ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ OR ⎛ −1⎞ uuur ⎜ ⎟ AB = ⎜ 2 ⎟ ⎜ −t ⎟ ⎝ ⎠
⎛ 0⎞ uuur ⎜ ⎟ BC = ⎜ −2 ⎟ ⎜ 2t ⎟ ⎝ ⎠
2
HCI Prelim H2 Mathematics P2 Solutions ⎛ −1 ⎞ ⎛ μ − 1 ⎞ uuur ⎜ ⎟ ⎜ ⎟ XB = (1 − μ ) ⎜ 2 ⎟ = ⎜ 2 − 2 μ ⎟ ⎜ −t ⎟ ⎜ t μ − t ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 0⎞ ⎛ 0 ⎞ uuur ⎜ ⎟ ⎜ ⎟ BY = μ ⎜ −2 ⎟ = ⎜ −2 μ ⎟ ⎜ 2t ⎟ ⎜ 2t μ ⎟ ⎝ ⎠ ⎝ ⎠
ii
⎛ μ −1 ⎞ ⎛ 0 ⎞ ⎛ μ −1 ⎞ uuur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ XY = XB + BY = ⎜ 2 − 2 μ ⎟ + ⎜ −2 μ ⎟ = ⎜ 2 − 4 μ ⎟ ⎜ t μ − t ⎟ ⎜ 2t μ ⎟ ⎜ 3t μ − t ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Suppose O, X, Y are collinear. Then uuuv uuuv OX = kOY
0 ⎞ ⎛1 − μ ⎞ ⎛ ⎜ ⎟ ⎜ ⎟ ⎜ 2μ ⎟ = k ⎜ 2 − 2μ ⎟ ⎜ −t μ ⎟ ⎜ −t + 2t μ ⎟ ⎝ ⎠ ⎝ ⎠ 1 − μ = 0 ⇒ μ = 1 (Out of range) Thus O, X, Y are not collinear. iii
0 ⎞ ⎛1 − μ ⎞ ⎛ uuuv uuuv ⎜ ⎟⎜ ⎟ OX OY = ⎜ 2 μ ⎟ ⎜ 2 − 2 μ ⎟ ⎜ −t μ ⎟ ⎜ −t + 2t μ ⎟ ⎝ ⎠⎝ ⎠ 2 2 = μ(4 – 4μ + t – 2μt ) =0 4 + t2 1 1 ⇒ μ = 0 (reject) or μ = = + 2 2 4 + 2t 2 t + 2
For all t ∈ℝ\{0}, 0 < μ < 1. Thus ∠XOY can be 90° when t ≠ 0 .
3
HCI Prelim H2 Mathematics P2 Solutions iv
⎛ μ −1 ⎞ uuuv ⎜ ⎟ XY = ⎜ 2 − 4 μ ⎟ ⎜ −t + 3t μ ⎟ ⎝ ⎠ projection vector ⎛ μ −1 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ 2 − 4μ ⎟ . ⎜ 1 ⎟ . ⎜ 1 ⎟ ⎜ −t + 3t μ ⎟ ⎜ 0 ⎟ ⎜ 0 ⎟ ⎠⎝ ⎠⎝ ⎠ =⎝ 17 ⎛ 4⎞ 4μ − 4 + 2 − 4μ ⎜ ⎟ = ⎜1⎟ 17 ⎜0⎟ ⎝ ⎠ ⎛ 4⎞ 2⎜ ⎟ = − ⎜1⎟ 17 ⎜ ⎟ ⎝0⎠
5(i)
( x − 6) + ( y + 2) 2
( y + 2)
2
2
= x+3
= ( x + 3) − ( x − 6 ) 2
2
= 9 ( 2 x − 3) (ii)
(iii)
For the equation ( y + 2 ) = 9 ( 2 x − 3) , 2
When x = 2 , y = 1 .
4
HCI Prelim H2 Mathematics P2 Solutions When y = 7 , x = 6 . Method 1: Using
( y + 2)
2
∫ x dy
= 9 ( 2 x − 3)
2 2 y + 2) ⎞ ( 3 ( y + 2) 1⎛ x= + = ⎜3+ ⎟ ⎟ 2 18 2 ⎜⎝ 9 ⎠
2 ⎛ ( y + 2) ⎜3+ ⎜ 9 ⎝
⎞ ⎟ d y − 2(6) ⎟ 1 ⎠ 7 ( y + 2)3 ⎤ ⎪⎧ 1 ⎡ ⎪⎫ 12 = ⎨ ⎢3 y + − ⎬ ⎥ 27 ⎦ 1 ⎪⎩ 2 ⎣ ⎪⎭ ⎧1 ⎫ = ⎨ ⎡⎣( 21 + 27 ) − ( 3 + 1) ⎤⎦ − 12 ⎬ ⎩2 ⎭ 2 = 10 units
R=
1 2
∫
7
Method 2: Using
( y + 2)
2
∫ y dx
= 9 ( 2 x − 3)
y = −2 + 3 2 x − 3
R = 4(7) −
[ y = −2 − 3 2 x − 3 N.A.]
∫ ( −2 + 3 6
2
)
2 x − 3 dx
3 6⎫ ⎧⎪ ⎡ ⎤ ⎪ = ⎨ 28 − ⎢ −2 x + (2 x − 3) 2 ⎥ ⎬ ⎣ ⎦ 2 ⎪⎭ ⎪⎩ = { 28 – [(–12 + 27) – (–4 + 1)]} = 10 units2
(iv)
Volume required = vol. of cylinder – (vol. generated by curve from y = –2 to y = 1) 6 2 ⎡ ⎤ = ⎢π (7) 2 (4) − π −2 + 3 2 x − 3 d x ⎥ 2 ⎣ ⎦ = 196π – 92π = 327 unit3 (3 s.f.) Use random sampling method to select a sample from each class. The number of seats from each class would be proportional to the size of each stratum.
∫(
6
)
First Class Business Class Economy Class Any 1 of the answers below: 4 16 60 Some passengers have booked a flight ticket but did not turn up or changed flight so some of the seats in the sample may not have a passenger. OR
5
HCI Prelim H2 Mathematics P2 Solutions The flight is not fully booked so the chosen seat could be empty. OR The passenger may ignore the questionnaire. It is not appropriate to use simple random sampling as passengers from different classes may have different opinions on the service. The number of passengers in the first class is very small, so the passengers from the first class may not be chosen at all using the simple random sampling method. 7(i)
No. of ways =
10! = 12600 4!3!2!
7(ii)
Case 1: The 2 blue tiles and 1 yellow tile are in the 4th row with the 4th tile being red or green. No. of ways = no. of ways with B, B, Y, G in 4th row + no. of ways with B, B, Y, R in 4th row 4! 6! 4! 6! = × + × = 3240 2! 2!2! 2! 2!2!2! Case 2: The 2 blue tiles and 1 yellow tile are in the third row. 3! 7! = 1890 No. of ways = × 2 2!2!2! 3! 3! 4! 3! 4! Total no. of ways = 3240 + 1890 – 2! 2! 2! – 2! 2! 3! = 5130 – 108 – 216 = 4806
7(iii)
No. of ways such that less than 3 yellow tiles are in the fourth row 7! = 12600 − 4C3 =12600 − 420 =12180 4!2! No. of ways 6! = × 7C4 = 2100 3!2! 2010 + 10(9000) = 9201, x= 10 2 ( x − 9000 ) ) ⎤⎥ 507147 ( 2 1 ⎡⎢ ∑ 2 = s = ∑ ( x − 9000 ) − 9⎢ 10 9 ⎥ ⎣ ⎦ H0 : μ = 9000 H1 : μ > 9000
7 last part 8(i)
x − 9000 ~ t (9) 507147 9 × 10 p–value = 0.01265 < 0.05 Test Stat: =
Since the p –value = 0.01265 < 0.05, we reject H 0 and conclude that there is sufficient evidence, at 5% level of significance, that the mean life span of the electronic component has increased.
6
HCI Prelim H2 Mathematics P2 Solutions
8(ii)
H0: μ = 9000 vs H1: μ > 9000 252 Under H0, X ~ N(9000, 10 ) = N(9000, 62.5).
X – 9000
~ N(0, 1). 62.5 Level of significance = 1% P(Z > 2.326347877) = 0.01 At the 1% significance level, reject H 0 if z ≥ 2.326347877. Test Statistic =
x – 9000
≥ 2.326347877 62.5 x ≥ 9018.391395 = 9020.
z=
Assumptions: The standard deviation of the life span remains unchanged after the change in process. 9 First part
X ~ N(190, 576) T = 0.001( X1 + ... + X20 ) − 0.001(2)( X21 + ... + X30 ) ~ N(0, 0.03456)
P( | T |≤ 0.15) = P( − 0.15 ≤ T ≤ 0.15) = 0.580 OR
A = X1 +... + X20 – 2(X21 +... + X30) ~ N(0, 34560) P( | A |≤ 9(i)
0.15 ) = P( − 150 ≤ A ≤ 150) 0.001
= 0.580 Let Y be the r.v. denoting the mass of a randomly chosen apple from Mark's orchard. Y ~ N(μ , 302 )
Since the shaded area is the same, using the symmetric property of the normal curve, μ = 110 9(ii)
Probability that Mark will get an apple graded as 'large' chosen at random = P(Y > 150) = 7
HCI Prelim H2 Mathematics P2 Solutions 0.09121128 Let A be the r.v. denoting the number of apples graded as large out of 65 randomly chosen apples. A ~ B(65, 0.09121128) P ( A ≥ 5) = 1 − P ( A ≤ 4) = 0.718
10(a)
200 1 = 400 2 250 + 300 11 (ii) P(M ' ∩ C ') = = 1000 20 (i) P(A ⎜M ) =
9 1 , P ( A M ) = ≠ P ( A) 20 2 A and M are not independent. 10(b) (i) No. of immigrants in the sample = 0.2 ( 200 + 250 ) + 0.3 (130 + 300 ) + 0.05 (120 ) = 225
P ( A) =
P(voter supports Party A given voter is an immigrant) =
0.2 × 450 = 0.4 225
(ii) Number of immigrants supporting Party C = 0.05 (120 ) =6 P(exactly one immigrant voter supporting Party C or exactly one female voter supporting Party A (or both)) = P ( exactly 1 immigrant voted for C ) + P ( exactly 1 female voted for A ) − P ( both )
=
C1 994C2 + 250C1 750C2 − 250C1 6C1 744C1 = 0.434 1000 C3
6
Alternative method: Required Probability =
6 994 993 250 750 749 6 250 744 ×3+ ×3− × 3! 1000 999 998 1000 999 998 1000 999 998
= 0.434
8
HCI Prelim H2 Mathematics P2 Solutions
11(i)
⎛λ⎞ Let X be the r.v. denoting the number of call–ins in a week. Hence X ~ Po ⎜ ⎟ . ⎝4⎠ We are looking for the λ such that P(X ≤ 9) = 0.701
P( X ≤ 9)
λ From graph, the value of λ = 32.5 (to 3 sig.fig). The condition is that the rate of call–ins received by the centre is constant throughout a month / the call–in occurs randomly / The call–ins occur in a month are independent of one another 11(ii) Let Y be the r.v. denoting the number of call–ins in a week. Y ~ Po(32.5) Since the mean is bigger than 10, hence Y ~ N (32.5, 32.5) approximately. c.c P (25 < Y ≤ 40) ⎯⎯ → P (25.5 < Y < 40.5) = 0.810
11(iii Let S be the r.v. denoting the number of successful cases out of the n people in a support ) group. 3 S ~ B (n, ) 20 Since the number of groups concerned, which is 70, is large, therefore by applying CLT, 3 ⎞ ⎛ 3 ⎞⎛ n ⎜ ⎟⎜ 1 − ⎟ 3 20 20 ⎠ S ~ N ( n, ⎝ ⎠⎝ ) approximately. 20 70 EITHER n
P ( S ≥ 4) 27 0.589 28 0.812 Hence minimum value of n is 28.
OR P ( S ≥ 4) > 0.7
9
HCI Prelim H2 Mathematics P2 Solutions
P ( S < 4) < 0.3 4 − 0.15n P(Z < ) < 0.3 0.1275n 70 4 − 0.15n < −0.5244 0.1275n 70 4 − 0.15n < (−0.5244 0.15n − (0.5244 n > 5.23912 n > 27.45 Least n = 28. 12(i)
0.1275 ) n 70
0.1275 ) n −4>0 70 or
n < −5.0899(reject)
Location F should be omitted as the road distance cannot be smaller than the straight line distance, indicating that it is an incorrect data entry.
From the scatter diagram, another location that should be omitted is location H, as it is an outlier based on the scatter diagram. 12(ii) The suitable regression line is the regression x on y: x = 0.3936554 + 0.81702935 y When y = 20.0, x = 16.7 km 12(iii s ) 180
70
y 2
30
10
HCI Prelim H2 Mathematics P2 Solutions
Since the graph of s = a + b ln y is concave downwards whereas the graph of s = a + by 2 is concave upwards, the graph of s = a + b ln y will be more suitable to describe the scatter diagram of s and y. Hence model II is more suitable. 12(iv )
The appropriate regression line of s on ln y is s = 25.9499647 + (45.24427905) ln y , i.e. s = 25.9 + (45.2) ln y (to 3 s.f.)
12(v) Since r for s and ln y is 0.992 close to 1, the linear correlation is strong between s and ln y. Furthermore, 170 cents is within the data range of the sample. Therefore the estimation using the line in (iv) is reliable. Since y is the independent variable, the line found in (iv) is also suitable for the estimation.
11
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 1 in preparation for General Certificate of Education Advanced Level
Higher 2 CANDIDATE NAME CLASS
INDEX NUMBER
MATHEMATICS
9740/01 15 September 2010
Paper 1
3 hours
Additional Materials:
Answer Paper Graph Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages. Innova Junior College
[Turn over
2
1
For any given mass of gas, the volume V cm3 and pressure p (in suitable units) satisfy the relationship V
1 n p , k
where k and n are constants. For a particular type of gas, n 2.3 . At an instant when volume is 32 cm3, the pressure is 105 units and the pressure is increasing at a rate of 0.2 units s1. Calculate the rate of decrease of volume at this instant. [4]
2
Given that the coefficient of x 2 in the series expansion of
2
1 3x
value of n, where n is a positive integer.
3
n
is 108, find the [4]
The sequence of numbers un , where n 1, 2, 3,..., is such that 9 8un1 un 7 n 8 . u1 and 8
Use the method of mathematical induction to show that un n 23n for n 1 .
(i)
Determine if the sequence converges.
[5] [1]
n
(ii)
Find
ur
in terms of n.
r 1
IJC/2010/JC2
9740/01/S/10
[2]
3
C
4
O M A
B
The diagram shows a quadrilateral OABC with OA AB and OC BC . 1 1 Points A and B have position vectors and 0 respectively. 2 1 (a) (b)
5
Find cosine of angle OAB in terms of . [2] M is the midpoint of OB and 4AM MC . By considering the area of the quadrilateral OABC, show that 5 [4] OA OC OA OB . 2
Illustrate, on a single Argand diagram, the locus of the point representing the complex number z that satisfies both the inequalities
3 arg z 3 3i 4 2
and
z 3 3i 2 .
[4]
(i)
Find the greatest and least values of z 3i .
[2]
(ii)
Find the least possible value of arg z , giving your answer in radians.
[3]
IJC/2010/JC2
9740/01/S/10
[Turn over
4 6
Show, by means of the substitution w x 2 y , that the differential equation
x
dy dx
2 y 3 xy 0
can be reduced to the form dw dx
3w .
Hence find y in terms of x, given that y
7
[2]
1 when x 2 . 2
[4]
A curve is defined parametrically by x 2 cos t ,
y 2t 1 ,
where 0 t .
8
(i)
Find the equation of the normal to the curve at the point P where t
(ii)
The normal at P meets the y-axis at N and the x-axis at M. Given that the curve meets the y-axis at Q, find the area of triangle MNQ, correct to 1 decimal place. [5]
A curve has equation (i)
(ii)
IJC/2010/JC2
3
.
[5]
x 2 ( y 4) 2 1. 4 9
Sketch the curve, stating the equations of the asymptotes and the coordinates of the vertices. [3] x 2 ( y 4) 2 The region enclosed by the curve 1 , the x-axis and the line 4 9 x 2 is rotated through 4 right angles about the y-axis to form a solid of revolution of volume V. Find the exact value of V, giving your answer in terms of . [4]
9740/01/S/10
5 9
In a medical research centre, a particular species of insect is grown for treatment of open wounds. The insects are grown in a dry and cool container, and they are left to multiply. The increase in the number of insects at the end of each week is at a constant rate of 4% of the number at the beginning of that week. At the end of each week, 10 of the insects would die due to space constraint and are removed from the container. A researcher puts y insects at the beginning of the first week and then a further y at the beginning of the second and each subsequent week. He also decides that he will not take any insect out of the container. (i)
How many insects will there be in the container at the end of the first week? Leave your answer in terms of y. [1]
(ii)
Show that, at the end of n weeks, the total number of insects in the container is
26 y 250 1.04 n 1 . (iii)
10
[4]
Find the minimum number of complete weeks for the population of the insects to exceed 13 y 125 . [4]
The functions f and g are defined as f : x 1 x
for x 1
g : x e x 1
for x 0
(i)
Define f 1 in a similar form, including its domain.
(ii)
State the relationship between f and f 1 , and sketch the graphs of f and f 1 on the same diagram. [3]
(iii)
Find the exact solutions of the equation f ( x) f 1 ( x) .
[2]
(iv)
Show that the composite function fg exists.
[2]
(v)
Given that h ( x) fg( x) for x 0 , show that h is an increasing function for [2] x 0.
IJC/2010/JC2
9740/01/S/10
[3]
[Turn over
6 11
(a)
Write 2 cos 3x cos x in the form cos px cos qx , where p and q are positive integers. [1] Hence find
12
(i)
cos 3x cos x dx ,
(ii)
the exact value of
[2]
4 0
x cos 3 x cos x dx .
[4]
(b)
State a sequence of transformations which transform the graph of y sin x to 3 the graph of y sin 2 x . [2] 2
(c)
Find the numerical value of the area of the region bounded by the curves 3 y cos 3x cos x and y sin 2 x for 0 x . [3] 2 2
2 3 4 A plane 1 has equation r. 1 9 and a line 1 has equation r 0 1 . 3 0 2
Find the coordinates of P, the point of intersection of 1 and 1 . [4] Hence, or otherwise, find the shortest distance from point A (3, 0, 0) to 1 . [2] The equations of planes 2 and 3 are given as (i)
2 : 3 :
1 2 2 r 1 s 0 t 2 , and 1 3 1 x y z , where ,
.
(ii)
Find the equation of plane 2 in the form r n = d . Explain why the planes 1 and 2 intersect. [4]
(iii)
The line of intersection of planes 1 and 2 is 2 . The line 2 has equation 3 2 r = 3 1 . 2 1
Given that the three planes 1 , 2 and 3 do not have any points in common, find the conditions satisfied by and . [3]
IJC/2010/JC2
9740/01/S/10
1
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level Higher 2
CANDIDATE NAME Civics Group
INDEX NUMBER
Mathematics
9740/02
Paper 2 16 September 2010
Additional materials:
Answer Paper Graph paper List of Formulae (MF15)
3 hours
READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages.
Innova Junior College
[Turn over
2
Section A: Pure Mathematics [40 marks]
1
Three space probes, Probe A, Probe B and Probe C with mass 600 kg, 630 kg and 900 kg respectively were sent into space to find the gravitational pull on the planets Venus, Mars and Saturn. The sum of the weights of Probe B on the three planets is 13860 N. The weight of Probe C on Saturn is 2880 N more than the weight of Probe A on Venus. The average weight of Probe A on Saturn, Probe B on Venus and Probe C on Mars is 4870 N. Find the gravitational pull on each of the planets.
[3]
Hence find the weight of Probe D, which has mass 500 kg, on planet Saturn.
[1]
[Weight (in N) = Mass (in kg) gravitational pull ms 2 ]
2
Express f ( x)
x2 6 x 1
3x 1 x 2 3
in the form A Bx C , 2 3x 1 x 3
where A, B and C are constants. Hence find the exact value of
3
[3] 1 3
f x dx .
[4]
Given that y cos 1 x 2 , show that
sin y Show that
dy 2x 0 . dx
d3 y 0 when x 0 . Hence write down the first two non-zero terms in the Maclaurin dx3
series for cos 1 x 2 .
4
[8]
The curve C has equation y ax 2a 4 (i) (ii)
4a 5 , where a is a constant. 2 x
5 or a 0 . [3] 4 Sketch curve C for the case when a 1 , indicating clearly the coordinates of the turning Given that curve C has turning points, show that a
points and the equations of any asymptotes. (iii)
Hence by sketching an appropriate graph on the same diagram, solve x x6
IJC/2010/JC2
[4]
9740/02/S/10
9 . 2x
[3]
3
5
(a)
The complex number z is given by z
3 i
3
1 pi 2
, where p 0 .
Given that z 2 , find the value of p. Show that arg z (b)
[3]
5 . 6
[3]
Find the exact roots of the equation z5 2 0
in the form rei , where r 0 and .
[5]
Section B: Statistics [60 marks] 6
A company’s director wants to obtain his employees’ views on flexible working hours. The company has 600 employees. Describe clearly how you would choose a systematic random sample of 30 employees. Describe briefly one disadvantage of this sampling method in this context. [3]
7
Box A contains 6 balls numbered 1, 2, 2, 2, 5, 7. Box B contains 4 balls numbered 1, 4, 4, 7. Box C contains 3 balls numbered 3, 4, 6. One ball is removed at random from each box. (i)
(ii)
1 . 6 [1] Find the probability that each of the three numbers obtained is greater than 3 or the sum of the three numbers obtained is 13 (or both). [4]
Show that the probability that each of the three numbers obtained is greater than 3 is
All the balls are now placed in a single container. A game is played by a single player. The player takes balls, one by one and with replacement, from the container, continuing until either a number 1 results or a prime number results. The player wins if the number on the last ball chosen is 1 and loses otherwise. Find the probability that a player wins a particular game.
IJC/2010/JC2
9740/02/S/10
[3]
[Turn over
4 8
9
Magnolia cows are milked by hand and Daisy cows are milked by machine. The time taken to milk a randomly chosen Magnolia cow may be taken to have a normal distribution with mean 30 minutes and standard deviation 2 minutes. The time taken to milk a randomly chosen Daisy cow may be taken to have an independent normal distribution with mean 5.5 minutes and standard deviation 0.5 minutes. (i)
The probability that it will take less than a minutes to milk a randomly chosen Magnolia cow is 0.85. Find a. [1]
(ii)
Using an appropriate approximation, find the probability that out of 50 randomly chosen Magnolia cows, there are more than 10 but at most 40 which take less than a minutes to milk. [4]
(iii)
Find the probability that the total time taken to milk two randomly chosen Magnolia cows exceeds eleven times the time taken to milk a randomly chosen Daisy cow by at least 3 minutes. [3]
An electronic game called ‘Wishful Thinking’ is played with 5 boxes arranged in a row. When a button is pressed, each of the five boxes displays a picture of a fruit – either an apple, orange or pear. A possible result of the game is shown below.
Box 1
Box 2
Box 3
Box 4
Box 5
Events ‘Success’ and ‘Windfall’ are defined as follows. ‘Success’ : Exactly 3 boxes display the same fruit. ‘Windfall’ : All 5 boxes display the same fruit. Find (i) (ii) (iii)
the total number of possible results, the number of ways of obtaining ‘Windfall’, the total number of ways of not obtaining ‘Success’.
[1] [1] [3]
Ten electronic game machines with distinct serial numbers are sent to 3 different game centres. In how many ways can the game machines be distributed if each centre must have at least 3 machines? [3]
IJC/2010/JC2
9740/02/S/10
5 10
Observations of a cactus graft were made under controlled environmental conditions. The table gives the observed heights x cm of the graft at t weeks after grafting. t x
11
1 2.0
2 2.4
3 2.5
4 5.1
5 6.7
6 9.4
8 18.3
10 35.1
(i)
Calculate the product moment correlation coefficient between t and x.
[1]
(ii)
Draw a scatter diagram for the data.
[2]
(iii)
Using your answer in part (i) and the scatter diagram in part (ii), explain why it is advisable to draw a scatter diagram first before interpreting the value of the product moment correlation coefficient. [1]
(iv)
Explain why the scatter diagram may be consistent with a model of the form x eat b . [1]
(v)
For the model x eat b , show that the relation between ln x and t is linear. Hence calculate the equation of the appropriate regression line. [3]
(vi)
Use the regression line in part (v) to predict the height of the cactus graft 20 weeks after grafting. Hence explain in the context of the question why it is unwise to extrapolate. [2]
A company claims that the guitar strings that the company manufactures have a tensile strength of 430 kpsi (kilo-pounds per square inch) on average. An engineer obtained a sample of 8 guitar strings and the tensile strength of each guitar string, x kpsi, is measured. The data is summarised by x 430 23,
x 430 211 . 2
Test, at the 2% significance level, whether the company has overstated the average tensile strength of a guitar string. State any assumptions that you have made. [7] The engineer will be able to conclude that the company has overstated the average tensile strength of a guitar string if he conducts the same test at % significance level. State the smallest possible integer value of . [1] The engineer takes a sample of 20 guitar strings manufactured by a rival company, whose guitar strings have tensile strength that is normally distributed with mean kpsi and standard deviation 4.7 kpsi. The null hypothesis 430 is being tested against the alternative hypothesis 430 at 5% level of significance. Find the range of values of the sample mean for which the null hypothesis is rejected, giving 2 decimal places in your answer. [3]
IJC/2010/JC2
9740/02/S/10
[Turn over
6 12
A roller-coaster ride has a safety system to detect faults on the track. State a condition under which a Poisson distribution would be a suitable probability model for the number of faults detected on the track on a randomly chosen day. [1] Faults on the track are detected at an average rate of 0.16 per day. Find the probability that on a randomly chosen day, the number of faults detected on the track is between 2 and 6 inclusive. [2] Find the probability that in a randomly chosen period of 20 days, there are not more than 4 faults detected on the track. [2] There is a probability of at least 0.15 that the mean number of faults detected on the track per day over a randomly chosen long period of n days is at least 0.2. Find the greatest value of n. [3] There is also a separate safety system to detect faults on the roller-coaster train itself. Faults are detected by this system at an average rate of 0.05 per day, independently of the faults detected on the track. Find the probability that in a randomly chosen period of 20 days, the number of faults detected on the track is at most 1 given that the total number of faults detected is 5. [4]
IJC/2010/JC2
9740/02/S/10
2010 IJC JC 2 PRELIMINARY EXAMINATION 2 Paper 1 (Solutions) 1
Given V =
1 n p k
dV 1 n −1 = np dp k 1 ⎛ pn ⎞ n⎜ ⎟ k ⎝ p ⎠ n⎛1 ⎞ = ⎜ pn ⎟ p⎝k ⎠ nV = p =
Given n = −2.3 , when V = 32, p = 105,
dp = 0.2 dt
dV dV d p = × dt d p dt Since
dV nV dp = × dt p dt
dV nV , = dp p
=
( −2.3)( 32 ) × 0.2 105
= − 0.140 (to 3 s.f.) Thus, the rate of decrease of volume at the instant is 0.140cm3s −1 . 2
2
(1 + 3x )
n
= 2 (1 + 3 x )
−n
− n ( − n − 1) 2 ( 3x ) + ... 2! n ( n + 1) 9 x 2 + ... = 1 − 3nx + 2 n ( n + 1) ⎛ ⎞ = 2 ⎜ 1 − 3nx + 9 x 2 + ... ⎟ 2 ⎝ ⎠
Consider (1 + 3 x ) = 1 + ( − n )( 3 x ) + −n
( )
2
(1 + 3x )
n
( )
⎛ n ( n + 1) 2 ⎞ = ... + 2 ⎜ 9 x ⎟ + ... 2 ⎝ ⎠
Given:
coefficient of x 2 = 108 ⇒ 9n ( n + 1) = 108
n 2 + n − 12 = 0 ( n + 4 )( n − 3) = 0 n = −4 (rejected since n ∈
Thus, value of n = 3
+
)
or n = 3
3
Let Pn denote the statement un = n + 2−3n , for n ≥ 1 . When n = 1 , 9 LHS = u1 = (given) 8 1 9 RHS = 1 + 2−3 = 1 + = = LHS 8 8 Thus P1 is true. Assume that Pk is true for some k ≥ 1 , i.e., u k = k + 2 −3 k
Want to show that Pk +1 is true, i.e., −3 k +1 uk +1 = k + 1 + 2 ( )
LHS = uk +1 1 = ( uk + 7 k + 8 ) 8 1 = k + 2−3k + 7 k + 8 8 1 = 8k + 8 + 2−3k 8 1 1 = ( 8k + 8 ) + 3 2−3k 8 2 = k + 1 + 2 −3 k − 3
( (
)
) ( )
−3 k +1 = k + 1 + 2 ( ) = RHS Thus Pk true ⇒ Pk +1 is true
Since P(1) is true, and Pk true ⇒ Pk +1 is true, by mathematical induction, un = n + 2−3n , for n ≥ 1 . 3i 3ii
un = n + 2−3n The sequence does not converge because as n → ∞ , un → ∞ . n
∑ ur = r =1
=
∑ ( r + 2−3r ) n
r =1 n
n
r =1
r =1
∑ r + ∑ 2−3r
(
n = (1 + n ) + 2−3 + 2−6 + ... + 2−3n 2 n 2−3 ⎛⎜1 − 2−3 ⎞⎟ n ⎝ ⎠ = (1 + n ) + − 3 2 1− 2
( )
)
1⎛ 1 ⎜1 − n n 8 = (1 + n ) + ⎝ 8 1 2 1− 8 n 1⎛ 1 = (1 + n ) + ⎜1 − n 2 7⎝ 8
4a
⎞ ⎟ ⎠
⎛1⎞ uuur ⎜ ⎟ O OA = ⎜ α ⎟ ⎜2⎟ ⎝ ⎠ ⎛ 1 ⎞ ⎛ −1⎞ ⎛ 2 ⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ A BA = ⎜ α ⎟ − ⎜ 0 ⎟ = ⎜ α ⎟ ⎜2⎟ ⎜ 1 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ uuur uuur uuur uuur ˆ OA • BA = OA BA cos OAB
⇒
B
⎛1⎞ ⎛2⎞ ⎜ ⎟ ⎜ ⎟ 2 2 ˆ ⎜ α ⎟ • ⎜ α ⎟ = 1 + α + 4 4 + α + 1 cos OAB ⎜2⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠
ˆ =α +4 cos OAB α2 +5 Area of quadrilateral OABC = 2 × Area of ΔAOC 1 uuur uuur = 2 × OA × OC 2 uuur uuur = OA × OC
⇒ 4b
⎞ ⎟ ⎠
2
C
4 Area of ΔAOB 1 uuur uuur = OA × OB 2
O 1
M
B A 1 uuur uuuur Area of ΔBOC = OB MC 2 uuuur uuuur uuuur 1 uuur = OB 4 AM since 4AM = MC 2 ⎛ 1 uuur uuuur ⎞ = 4 ⎜ OB AM ⎟ ⎝2 ⎠ = 4 ( area of AOB ) uuur uuur ⎛ 1 uuur uuur ⎞ = 4 ⎜ OA × OB ⎟ = 2 OA × OB ⎝2 ⎠
(
)
Thus, area of quadrilateral OABC = Area of ΔAOB + Area of ΔBOC
uuur uuur 1 uuur uuur OA × OB + 2 OA × OB 2 5 uuur uuur = OA × OB 2 uuur uuur 5 uuur uuur OA × OC = OA × OB 2
=
⇒
(Shown)
5 Im
(3,3)
3 2 1
0
2
1
Re
3
5i Im
2 (3,3)
3
2
2 1
0
1
2
3
Re
The least value of z + 3i = −3
45 − 2
The greatest value of z + 3i =
45 + 2
5ii
Im
(3,3)
3
2
2
18
1
θ 0
sin θ =
θ = sin
2
1
3
Re
2 18 −1 2
arg( z ) =
π
18
= 0.49088
− 0.49088 4 = 0.295 rad (to 3 s.f.)
6
Given: w = x2 y Differentiate w.r.t x dw dy = 2 xy + x 2 dx dx
x (1) × x :
dy dx
+ 2 y + 3 = 0 ------- (1)
x2
dy dx
+ 2 xy + 3 x 2 y = 0 dw + 3w = 0 dx dw = −3w dx
dw = −3w dx 1 1 − ∫ dw = ∫ 1 dx 3 w 1 − ln w = x + c 3 ln w = −3 x − 3c w = e −3 x − 3 c
(shown)
w = ±e−3c e−3 x w = Ae−3 x x 2 y = Ae−3 x Given that y = −
⎛ 1⎞ 22 ⎜ − ⎟ = Ae−6 ⎝ 2⎠ A = −2e6
⇒ Thus, ⇒ 7i
1 when x = 2 , 2
x 2 y = −2e6 e −3 x 2 y = − 2 e 6 −3 x x
x = 2 cos t dx = −2 sin t dt dy 2 1 ∴ = =− dx −2 sin t sin t
y = 2t − 1 dy =2 dt
Given:
Gradient of tangent at the point P is 1 1 2 − =− =− π 3 3 sin 3 2 Gradient of normal at the point P is 1 3 − = 2 ⎛ 2 ⎞ ⎜− ⎟ 3⎠ ⎝ π When t = , 3
⇒ and
2π ⎛π ⎞ y = 2⎜ ⎟ −1 = −1 3 3 ⎝ ⎠ ⎛π ⎞ x = 2 cos ⎜ ⎟ = 1 ⎝3⎠
Hence equation of normal at the point 3 ⎛ 2π ⎞
y −⎜ − 1⎟ = ( x − 1) 2 ⎝ 3 ⎠ 3 3 2π y= x− −1+ 2 2 3 y = 0.86603 x + 0.22837 y = 0.866 x + 0.228
ii
Equation of normal at P: y = To find y-intercept:
3 3 2π x− −1+ 2 2 3
3 2π −1+ 2 3 or y = 0.22837
When x = 0, y = −
To find x-intercept: 2 ⎛ 3 2π +1− ⎜⎜ 3 3⎝ 2
When y = 0, x =
⎞ ⎟⎟ ⎠
or x = −0.26370 To find y-intercept of the curve When x = 0, ⇒ t =
π 2
, y = π −1
⎡ 3 1⎛ 2π ⎤ ⎞ 2 ⎛ 3 2π Area of triangle MNQ = ⎜ π − 1 − ⎢ − −1+ +1− ⎜⎜ ⎥ ⎟⎟ 2 ⎝⎜ 2 3 2 3 ⎣ ⎦⎠ 3 ⎝ = 0.252257 = 0.3 (to 1 d.p.)
8
y
y = 4+
3 x 2
centre ( 0, 4 ) vertex ( −2, 4 )
vertex ( 2, 4 ) 0
x −2
2 y = 4−
3 x 2
y
4
x
−2
0
2
⎞ ⎟⎟ ⎠
V = π ∫ x 2 dy − π ( 2 2 ) ( 4 ) 4
0
4⎛ 4 2 ⎞ V = π ∫ ⎜ 4 + ( y − 4 ) ⎟ dy − 16π 0 9 ⎝ ⎠ 4
4 3⎤ ⎡ = π ⎢ 4 y + ( y − 4 ) ⎥ − 16π 9 ⎣ ⎦0 ⎡ 4 3 ⎛ 4 ⎛ 64 ⎞ ⎞ ⎤ = π ⎢16 + ( 0 ) − ⎜ 0 + ⎜ − ⎟ ⎟ ⎥ − 16π 9 9 ⎝ 3 ⎠ ⎠⎦ ⎝ ⎣
256 ⎡ 688 ⎤ π =π ⎢ − 16π = ⎥ 27 ⎣ 27 ⎦
9i
Number of insects in the container at the end of 1st week =1.04y − 10 Wk 1 2
Start y
End 1.04y -10
1.04 y − 10 + y
1.04 (1.04 y − 10 + y ) − 10
= 1.042 y + 1.04 y − 1.04(10) − 10 3
1.042 y + 1.04 y − 1.04(10) − 10 + y
1.04 (1.042 y + 1.04 y − 1.04(10) − 10 + y ) − 10
= 1.043 y + 1.042 y + 1.04 y −1.042 (10) − 1.04(10) − 10
1.04n y + ..... + 1.043 y + 1.042 y + 1.04 y
n
ii
−10 ⎡⎣1.04n −1 + 1.04n − 2 + ......... + 1.04 + 1⎤⎦
The total number of insects in the container are = 1.04n y + ..... + 1.043 y + 1.042 y + 1.04 y −10 ⎡⎣1.04n −1 + 1.04n − 2 + ......... + 1.04 + 1⎤⎦ =y
1.04 (1.04n − 1) 1.04 − 1
(1.04 − 10
n
− 1)
1.04 − 1
= 26 y (1.04n − 1) − 250 (1.04n − 1) = ( 26 y − 250 ) ⎡⎣1.04n − 1⎤⎦ iii
( 26 y − 250 ) (1.04n − 1) > 13 y − 125 1.04n − 1 >
1 2
1.04n >
3 2
n ln1.04 > ln
3 2
n > 10.3
Therefore the minimum number of complete weeks is 11. 10i
Let
y = f ( x) = 1 − x ,
⇒
1 − x = y2
⇒
x = 1 − y2
x ≤1
f −1 ( x ) = 1 − x 2 ⇒ Df −1 = Rf = [ 0, ∞ ) Thus, f −1 : x → 1 − x 2 , x ≥ 0 ii
The graph of f −1 is the reflection of the graph of f in the line y = x . y f 1
1
11a
f
−1
x
2 cos 3 x cos x = cos ( 4 x ) + cos ( 2 x )
1
∫ cos 3x cos x dx = 2 ∫ 2 cos 3x cos x dx
i
=
1 ( cos 4 x + cos 2 x ) dx 2∫
=
1⎛1 1 ⎞ ⎜ sin 4 x + sin 2 x ⎟ + c 2⎝4 2 ⎠
1 1 = sin 4 x + sin 2 x + c 8 4
ii
∫
π
4 0
x cos 3 x cos x dx π
π 4 1 1 1 1 = x ⎡⎢ sin 4 x + sin 2 x ⎤⎥ − ∫ 4 ⎛⎜ sin 4 x + sin 2 x ⎞⎟ dx 0 4 4 ⎣8 ⎦0 ⎝8 ⎠
π
1 ⎛ π ⎞ ⎡1 ⎛π ⎞ 1 ⎛ π ⎞⎤ ⎡ 1 ⎤4 = ⎜ ⎟ ⎢ sin 4 ⎜ ⎟ + sin 2 ⎜ ⎟ ⎥ − ⎢ − cos 4 x − cos 2 x ⎥ 8 ⎝ 4 ⎠ ⎣8 ⎝4⎠ 4 ⎝ 4 ⎠ ⎦ ⎣ 32 ⎦0 1 ⎛ π ⎞ ⎡1 ⎛ π ⎞⎤ ⎡ 1 ⎛π ⎞ 1 ⎛ π ⎞ ⎛ 1 1 ⎞⎤ = ⎜ ⎟ ⎢ sin π + sin ⎜ ⎟ ⎥ + ⎢ cos 4 ⎜ ⎟ + cos 2 ⎜ ⎟ − ⎜ + ⎟ ⎥ 4 ⎝ 4 ⎠ ⎣8 ⎝ 2 ⎠ ⎦ ⎣ 32 ⎝4⎠ 8 ⎝ 4 ⎠ ⎝ 32 8 ⎠ ⎦ =
b
1 1⎤ π − 3 ⎡ 1 + ⎢− − − ⎥ = 16 16 ⎣ 32 32 8 ⎦
π
Method 1: (Translation – Stretching) 1: Translation by
3 π in the direction of the x-axis. 2
2: Stretching parallel to the x-axis with scale factor of
1 . 2
Method 2: (Stretching − Translation) 1: Stretching parallel to the x-axis with scale factor of 2: Translation by c
3 π in the direction of the x-axis. 4
Area bounded by the curves =∫
1.0471976
0
⎡ ⎛ 3 ⎞ ⎤ ⎢sin ⎜ 2 x − 2 π ⎟ − cos 3 x cos x ⎥ dx ⎠ ⎣ ⎝ ⎦
= 0.32476 = 0.325 units2 (to 3 sf) 12i π1 :
l1 :
⎛2⎞ ⎜ ⎟ r . ⎜ −1⎟ = 9 %⎜ ⎟ ⎝3⎠ ⎛ 3⎞ ⎛ 4⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜0⎟ + λ⎜ 1 ⎟ % ⎜ ⎟ ⎜ −2 ⎟ ⎝0⎠ ⎝ ⎠
For point of intersection, ⎡⎛ 3 ⎞ ⎛ 4 ⎞⎤ ⎛ 2 ⎞ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ 0 ⎟ + λ ⎜ 1 ⎟ ⎥ . ⎜ −1 ⎟ = 9 ⎜ ⎟ ⎜ −2 ⎟ ⎥ ⎜ 3 ⎟ ⎝ ⎠⎦ ⎝ ⎠ ⎣⎢⎝ 0 ⎠
⇒ ⇒
⎛ 3⎞ ⎛ 2 ⎞ ⎛ 4⎞ ⎛2⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ . ⎜ −1 ⎟ + λ ⎜ 1 ⎟ . ⎜ − 1 ⎟ = 9 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝ 3 ⎠ ⎝ −2 ⎠ ⎝ 3 ⎠ 9−6 λ= =3 8 −1− 6
1 . 2
⎛ 3 ⎞ ⎛ 4 ⎞ ⎛ 15 ⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⇒ OP = ⎜ 0 ⎟ + 3 ⎜ 1 ⎟ = ⎜ 3 ⎟ ⎜ 0 ⎟ ⎜ −2 ⎟ ⎜ − 6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Thus, coordinates of the P are (15, 3, − 6 ) .
⎛ 15 ⎞ ⎛ 3 ⎞ ⎛ 12 ⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AP = ⎜ 3 ⎟ − ⎜ 0 ⎟ = ⎜ 3 ⎟ ⎜ −6 ⎟ ⎜ 0 ⎟ ⎜ −6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Shortest distance from A to π1 ⎛2⎞ uuur ⎜ ⎟ AP. ⎜ −1⎟ ⎜3⎟ = ⎝ ⎠ ⎛2⎞ ⎜ ⎟ ⎜ −1 ⎟ ⎜3⎟ ⎝ ⎠
l1 A (3, 0, 0) ⎛2⎞ ⎜ ⎟ ⎜ −1⎟ ⎜3⎟ ⎝ ⎠
P (15, 3, −6)
π1
⎛ 12 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ . ⎜ −1⎟ ⎜ −6 ⎟ ⎜ 3 ⎟ 24 − 3 − 18 3 3 14 ⎝ ⎠ ⎝ ⎠ = = = = = 14 14 14 4 +1+ 9 Alternative solution for shortest distance: l1
Line AN: ⎛ 3⎞ ⎛2⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜ 0 ⎟ + λ ⎜ −1 ⎟ % ⎜ ⎟ ⎜3⎟ ⎝0⎠ ⎝ ⎠ N is the point of intersection of line AN and plane π1 .
⇒
⎡⎛ 3 ⎞ ⎛ 2 ⎞⎤ ⎛ 2 ⎞ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎜ ⎟ ⎢ ⎜ 0 ⎟ + λ ⎜ −1 ⎟ ⎥ . ⎜ − 1 ⎟ = 9 ⎜ 3 ⎟⎥ ⎜ 3 ⎟ ⎢⎣⎜⎝ 0 ⎟⎠ ⎝ ⎠⎦ ⎝ ⎠ ⎛ 3⎞ ⎛ 2 ⎞ ⎛2⎞ ⎛2⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ . ⎜ −1 ⎟ + λ ⎜ − 1 ⎟ . ⎜ − 1 ⎟ = 9 ⎜0⎟ ⎜ 3 ⎟ ⎜3⎟ ⎜3⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
6 + (4 + 1 + 9)λ = 9 3 ⇒ λ= 14 ⎛ 3⎞ ⎛2⎞ ⎛ 48 ⎞ uuur ⎜ ⎟ 3 ⎜ ⎟ 1 ⎜ ⎟ Thus, ON = ⎜ 0 ⎟ + ⎜ −1⎟ = ⎜ −3 ⎟ ⎜ 0 ⎟ 14 ⎜ 3 ⎟ 14 ⎜ 9 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⇒
A (3, 0, 0) ⎛2⎞ ⎜ ⎟ ⎜ −1 ⎟ ⎜3⎟ ⎝ ⎠
P
N
π1
ii
⎛ 48 ⎞ ⎛ 3 ⎞ ⎛2⎞ uuur 1 ⎜ ⎟ ⎜ ⎟ 3 ⎜ ⎟ AN = ⎜ −3 ⎟ − ⎜ 0 ⎟ = ⎜ −1⎟ 14 ⎜ ⎟ ⎜ ⎟ 14 ⎜ ⎟ ⎝ 9 ⎠ ⎝ 0⎠ ⎝3⎠ Thus, shortest distance AN uuur 3 3 14 = AN = 4 +1+ 9 = 14 14 ⎛2⎞ ⎜ ⎟ π1 : r . ⎜ −1⎟ = 9 %⎜ ⎟ ⎝3⎠ ⎛ 1⎞ ⎛ 2 ⎞ ⎛ 2 ⎞ π2 : r = ⎜⎜1⎟⎟ + s ⎜⎜ 0 ⎟⎟ + t ⎜⎜ 2 ⎟⎟ % ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝1⎠ ⎝ −3 ⎠ ⎝ 1 ⎠ ⎛ 2 ⎞ ⎛ 2⎞ ⎛ 6 ⎞ ⎛ 3⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ A normal to plane π2 = ⎜ 0 ⎟ × ⎜ 2 ⎟ = ⎜ −(8) ⎟ = 2 ⎜⎜ −4 ⎟⎟ ⎜ −3 ⎟ ⎜ 1 ⎟ ⎜ 4 ⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
π2 :
⇒
⎛ 3 ⎞ ⎛ 1⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r . ⎜ −4 ⎟ = ⎜1⎟ . ⎜ −4 ⎟ = 3 − 4 + 2 = 1 %⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 1⎠ ⎝ 2 ⎠ ⎛ 3⎞ ⎜ ⎟ r . ⎜ −4 ⎟ = 1 % ⎜ ⎟ ⎝ 2⎠
⎛ 3⎞ ⎛2⎞ ⎜ ⎟ Since ⎜ 4 ⎟ ≠ k ⎜⎜ −1⎟⎟ for any k ∈ , ⎜ 2⎟ ⎜3⎟ ⎝ ⎠ ⎝ ⎠ the normal of planes π1 and π2 are not parallel to each other
iii
⇒ The planes are not parallel to each other. ⇒ The planes will intersect in a line. ⎛ 3⎞ ⎛2⎞ ⎜ ⎟ ⎜ ⎟ l2 : r = ⎜ 3⎟ + μ⎜ 1 ⎟ ⎜ 2⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠ π3 :
αx + y − z = β
(3, 3, 2) ⎛α⎞ ⎜ ⎟ ⇒ r. ⎜ 1 ⎟ = β ⎜ −1 ⎟ ⎝ ⎠ π1 , π2 and π3 do not have any points in common
⇒ l 2 is parallel to π3 and does not lie on π3
⎛2⎞ ⎜ ⎟ ⎜1⎟ ⎜ −1 ⎟ ⎝ ⎠
l2 ⎛α⎞ ⎜ ⎟ ⎜1⎟ ⎜ −1 ⎟ ⎝ ⎠
π3
⎛ 3⎞ ⎛ α ⎞ ⎛α⎞ ⎜ ⎟ ⇒ l 2 is perpendicular to ⎜ 1 ⎟ and ⎜⎜ 3 ⎟⎟ . ⎜⎜ 1 ⎟⎟ ≠ β ⎜ −1 ⎟ ⎜ 2 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛2⎞ ⎛α⎞ Thus, ⎜⎜ 1 ⎟⎟ . ⎜⎜ 1 ⎟⎟ = 0 ⎜ −1⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠
⇒
2 α + 1 + 1 = 0 ⇒ α = −1
⎛ 3⎞ ⎛ α ⎞ and ⎜⎜ 3 ⎟⎟ . ⎜⎜ 1 ⎟⎟ ≠ β ⎜ 2 ⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠
⇒ 3(−1) + 3 − 2 ≠ β ⇒ β ≠ −2
2010 IJC JC 2 PRELIMINARY EXAMINATION 2 Paper 2 (Solutions) 1
Let V ms −2 , M ms −2 , S ms −2 be the gravitational pull on each planet Venus, Mars and Saturn respectively. 630S + 630V + 630M = 13860 ---------- (1) 900S − 600V = 2880
---------- (2)
600S + 630V + 900M = 3(4870) = 14610 --------- (3) ⎛ 630 630 630 ⎞ ⎛ S ⎞ ⎛ 13860 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ 900 −600 0 ⎟ ⎜ V ⎟ = ⎜ 2880 ⎟ ⎜ 600 630 900 ⎟ ⎜ M ⎟ ⎜ 14610 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
From GC, M = 3.8, V = 9, S = 9.2 Hence the weight of Probe D on Saturn is ( 500 × 9.2 ) N = 4600 N 2
f ( x) =
= =
x2 − 6x + 1 ( 3x + 1) x 2 + 3
(
)
A Bx + C + 2 3x + 1 x + 3 A x 2 + 3 + ( Bx + C )( 3 x + 1)
(
)
( 3x + 1) ( x 2 + 3)
x 2 − 6 x + 1 = Ax 2 + 3 A + 3Bx 2 + 3Cx + Bx + C = ( A + 3B ) x 2 + ( 3C + B ) x + 3 A + C Comparing coefficients, A + 3B =1 B + 3C = −6 3A +C=1 From GC, A = 1, B = 0, C = −2 x2 − 6x + 1 1 2 ∴ f ( x) = = − 2 2 ( 3x + 1) x + 3 ( 3x + 1) x + 3
(
∫
−1
−3
f ( x ) dx =
∫
−1
−3
)
(
)
2 ⎤ ⎡ 1 ⎢⎣ 3 x + 1 − x 2 + 3 ⎥⎦ dx −1
2 x ⎤ ⎡1 = ⎢ ln 3 x + 1 − tan −1 ⎥ 3 3 ⎦ −3 ⎣3
1 2 2 ⎛ 1 ⎞ ⎡1 ⎛ 3 ⎞⎤ tan −1 ⎜ − tan −1 ⎜ − = ln 3 ( −1) + 1 − − ⎢ ln 3 ( −3) + 1 − ⎟ ⎟⎥ 3 3 3 ⎠ ⎣3 3 3 ⎠⎦ ⎝ ⎝ 1 2 ⎛ π ⎞ ⎡1 2 ⎛ π ⎞⎤ = ln 2 − ⎜ − ⎟ − ⎢ ln 8 + ⎜ ⎟⎥ 3 3 ⎝ 6 ⎠ ⎣3 3 ⎝ 3 ⎠⎦
1 1 2 ⎛ π ⎞ 2 ⎛π ⎞ = ln − ⎜− ⎟− ⎜ ⎟ 3 4 3⎝ 6⎠ 3⎝3⎠ 1 1 π = ln − 3 4 3 3
3
( )
y = cos −1 x 2
cos y = x dy − ( sin y ) = 2 x dx dy ( sin y ) + 2 x = 0 (shown) dx 2
2
⎛ dy ⎞ + cos y ⎜ ⎟ = −2 2 dx ⎝ dx ⎠ d3 y dy d 2 y ( sin y ) 3 + ( cos y ) ⎛⎜ ⎞⎟ 2 dx ⎝ dx ⎠ dx
( sin y )
d2 y
2 ⎛ ⎛ dy ⎞ d 2 y ⎞ ⎛ dy ⎞⎛ dy ⎞ + cos y ⎜⎜ 2 ⎜ ⎟ 2 ⎟⎟ − ( sin y ) ⎜ ⎟⎜ ⎟ = 0 ⎝ dx ⎠⎝ dx ⎠ ⎝ ⎝ dx ⎠ dx ⎠
( sin y )
3
2 ⎛ dy ⎞ d y ⎛ dy ⎞ + y 3 cos ( ) ⎜ ⎟ 2 − sin y ⎜ ⎟ = 0 3 dx ⎝ dx ⎠ dx ⎝ dx ⎠
d3 y
When x = 0 , y = cos −1 0 = dy =0 dx d2 y = −2 dx 2 d3 y =0 dx3 4
π 2
4a + 5 2− x (4a + 5)(−1)
y = − ax − ( 2a + 4 ) +
⇒
dy = −a − dx
( 2 − x )2
dy =0 dx 4a + 5 −a + =0 ( 2 − x )2
For turning points, ⇒ ⇒
a ( 2 − x ) = 4a + 5
⇒
a 4 − 4 x + x 2 − 4a − 5 = 0
2
(
)
⇒ ax 2 − 4ax − 5 = 0 Since there are turning points, there are two distinct real roots. Thus, D > 0
⇒
( −4a )2 − 4(a)(−5) > 0
⇒ ⇒
4 a 2 + 5a > 0 a ( 4a + 5 ) > 0 −
5 4
0
5 or a > 0 . (Shown) 4 Alternatively dy For turning points, =0 dx 4a + 5 ⇒ −a + =0 ( 2 − x )2
Thus, a < −
4a + 5 a Since there are turning points, there are two distinct real roots. 4a + 5 + − + >0 Thus, a 5 0 −
⇒
( 2 − x )2 =
4
5 or a > 0 . (Shown) 4 9 When a = 1 , y = − x − 6 + 2− x
Thus, a < − 4(ii)
y
y = −x − 6 +
x=2
9 2− x −1
−6
5
x
−2 −6
−14
y = −x − 6
4(iii)
9 2− x 9 x > −x − 6 + 2− x x + x+6>
⇒
y
y = −x − 6 +
x=2
y= x
9 2− x −1
−6
5
x
−2 −6
−14
y = −x − 6
From GC, the curves intersect at x = 0.823 (to 3 s.f) From graph, solution is x > 2 or x < 0.823 . 5(a)
( z =
3 −i
)
(1 + pi )
=
=
(
(
3
=
2
3 +1
)
1 + p2
3 −i 1 + pi
3
2
3
)
2
8 1 + p2 z =2
⇒
8 =2 1 + p2 4 = 1 + p2 p = 3 or − 3 (rejected)
(
)
(
3 −i
⎛ 3 −i 3 ⎞ ⎟ arg z = arg ⎜⎜ 2 ⎟ ⎜ (1 + pi ) ⎟ ⎝ ⎠ = arg
= 3arg
(
)
3
(
− arg 1 + 3i
)
(
)
2
3 − i − 2 arg 1 + 3i
)
7π ⎛ π⎞ ⎛π ⎞ = 3⎜ − ⎟ − 2 ⎜ ⎟ = − 6 ⎝ 6⎠ ⎝3⎠ For −π < arg( z ) ≤ π , 7π 5π arg( z ) = 2π − = (Shown) 6 6 5(b)
5π 6 −
7π 6
z5 = − 2 = 2eπ i
= 2e(
π + 2 kπ )i
, k = 0, ±1, ±2
By De Moivre’s Theorem 1
⎛ π + 2 kπ ⎞ ⎜ ⎟i 5 ⎠
1
⎛π ⎞ ⎜ ⎟i
1
⎛ 3π ⎞ ⎜ ⎟i 5 ⎠
1
π
1
3π
z = 210 e⎝
, k = 0, ±1, ±2
z = 210 e⎝ 5 ⎠ , 210 e⎝
6
i
i
1
1
⎛ π⎞
1
− i
⎛ 3π ⎞
1
⎜ − ⎟i ⎜ − ⎟i π i , 210 e( ) , 210 e⎝ 5 ⎠ , 210 e⎝ 5 ⎠ 1
π
1
−
3π
i
z = 210 e 5 , 210 e 5 , − 210 , 210 e 5 , 210 e 5 - Use a list with all the employees’ names arranged in alphabetical order and number them from 1 to 600 600 - Determine the sampling interval : k = = 20 30 - Randomly select the first person from the first 20 people on the list, then select every 20th person subsequently until a sample of 30 employees is obtained.
The sample obtained might be over-represented by a particular department (or gender or ethnic group) of the company which has the greatest proportion of employees, hence the systematic random sample obtained is not a good representative of the population. 7(i)
7(ii)
P(all are greater than 3) 2 3 2 1 = . . = 6 4 3 6 Let X be the event : “each of the 3 numbers is greater than 3” Y be the event : “sum of the 3 numbers is equal to 13” P(Y) = P((2,7,4), (5,4,4))
3 1 1 1 2 1 . . + . . 6 4 3 6 4 3 5 = 72 =
Required probability = P( X ∪ Y ) = P( X ) + P(Y ) − P( X ∩ Y ) 1 5 1 2 1 = + − . . 6 72 6 4 3 5 = 24 P(a player wins a particular game) 2 2 4 2 ⎛4⎞ 2 = + . + ⎜ ⎟ . + ...... 13 13 13 ⎝ 13 ⎠ 13 2 ⎞ 2⎛ 4 ⎛4⎞ 2 ⎛ 1 ⎞ .⎜ ⎜⎜ 1 + + ⎜ ⎟ + ...... ⎟⎟ = ⎜ 1 − 4 ⎟⎟ 13 13 ⎝ 13 ⎝ 13 ⎠ ⎝ 13 ⎠ ⎠ 2 = 9 Let M minutes be the time taken to milk a Magnolia cow, and D minutes be the time taken to milk a Daisy cow.
=
8
(
)
(
M ~ N 30, 22 , D ~ N 5.5, 0.52 8(i)
)
P ( M < a ) = 0.85
From GC, a = 32.073 = 32.1 (3 s.f.) 8(ii)
Let X be the number of cows which take less than a minutes to milk, out of 50 Magnolia cows. X ~ B ( 50, 0.85 ) Since n = 50 is large, np = 42.5 > 5, nq = 7.5 > 5 , ∴ X ~ N ( 42.5, 6.375 ) approx. Reqd prob = P (10 < X ≤ 40 ) = P (10.5 < X ≤ 40.5 ) (using c.c.)
8(iii)
= 0.21415 = 0.214 (3 s.f.) E ( M1 + M 2 − 11D ) = 2 × 30 − 11× 5.5 = −0.5 Var ( M1 + M 2 − 11D ) = 2 × 22 + 112 × 0.52 = 38.25 M1 + M 2 − 11D ~ N ( − 0.5, 38.25 )
Reqd prob = P ( M 1 + M 2 − 11D ≥ 3) = 0.28573 = 0.286 (3 s.f.)
9(i)
Total number of possible results = 35 = 243
9(ii)
Number of ways of obtaining ‘Windfall’ = 3
9(iii)
No. of ways to obtain a success ⎛ 5! 5 !⎞ =⎜ × 2 + ⎟ × 3 = 120 3! ⎠ ⎝ 3!2 ! Total number of ways = 243 − 120 = 123
9
Reqd no. of ways 3! = 12600 2! From GC, r = 0.92378 = 0.924 (3 s.f.) =
10(i) 10(ii)
C3 × 7C3 × 4C4 ×
10
x 35.1
2.0 10
1
t
10(iii) From (i) and (ii), we see that though r is close to +1, the scatter diagram indicates a curvilinear relationship between t and x, instead of a positive linear relationship. Thus it is advisable to draw a scatter diagram first before interpreting the value of the product moment correlation coefficient. 10(iv) As t increases, x increases but by increasing amounts. Thus the scatter diagram may be consistent with a model of the form x = e at +b .
x = e at +b ln x = at + b , where a and b are constants Thus the relation between ln x and t is linear.
10(v)
Find regression line of ln x on t. t ln x
10(vi)
1 0.69315
2 0.87547
3 0.91629
4 1.6292
5 1.9021
6 2.2407
8 2.9069
10 3.5582
From GC, eqn of regression line of ln x on t is: ln x = 0.20723 + 0.33498t ln x = 0.207 + 0.335t (3 s.f.) When t = 20, ln x = 0.20723 + ( 0.33498 )( 20 ) = 6.90683 x = 999
Height of the cactus graft 20 weeks after grafting is 999 cm. It is impossible that the cactus can grow to that height after 20 weeks. Thus it is unwise to extrapolate.
11
The t-test will be used. Assume a normal distribution for tensile strength of guitar strings.
( −23) 1⎡ s = ⎢ 211 − 7 ⎢⎣ 8
2
2
⎤ ⎥ ⎥⎦
−23 + 430 8 = 427.125
x=
= 20.696
H 0 : μ = 430 H1 : μ < 430 Level of significance = 2% X − 430 ~ t7 S n
Under H 0 :
T=
From GC,
t = −1.79 p-value = 0.0585 > 0.02
Do not reject H 0 and conclude that there is insignificant evidence at the 2% significance level that the manufacturer has overstated its claim. 11
α =6
11
σ = 4.7 H 0 : μ = 430 H1 : μ ≠ 430 Level of significance = 5% Under H 0 :
Z=
X − 430 ~ N (0,1) 4.7 20
To reject H 0 ,
x − 430 < −1.96 4.7 20 x < 427.94
12
or
or
x − 430 > 1.96 4.7 20 x > 432.06
Suitable condition • The mean number of faults detected is a constant from day to day. • The faults are detected independently of one another. Let X be the number of faults detected on the track in a day. X ~ Po ( 0.16 ) P ( 2 ≤ X ≤ 6 ) = P ( X ≤ 6 ) − P ( X ≤ 1)
= 0.011513 = 0.0115 (3 s.f.)
Let Y be the number of faults detected on the track in 20 days. Y ~ Po ( 3.2 ) P (Y ≤ 4 ) = 0.78061 = 0.781 (3 s.f.)
X=
X1 + X 2 + L X n n
0.16 ⎞ ⎛ Since n is large, by Central Limit Theorem, X ~ N ⎜ 0.16, ⎟ approx. n ⎠ ⎝ ⎛ ⎞ ⎜ 0.2 − 0.16 ⎟ ⎟ ≥ 0.15 P ( X ≥ 0.2 ) ≥ 0.15 ⇒ P ⎜ Z ≥ ⎜ 0.16 ⎟ ⎜ ⎟ n ⎝ ⎠ ⇒ P Z ≤ 0.1 n ≤ 0.85
(
)
0.1 n ≤ 1.0364 ⇒ n ≤ 107.41 Greatest value of n is 107. Let W be the number of faults detected on the train in 20 days. W ~ Po (1) From GC,
12
Y + W ~ Po ( 4.2 )
P (Y ≤ 1| Y + W = 5 ) = =
P (Y = 0 ) P (W = 5 ) + P (Y = 1) P (W = 4 ) P (Y + W = 5 ) 0.00212437 0.163316
= 0.0130 (3 s.f.)
JURONG JUNIOR COLLEGE J2 Preliminary Examination
9740/01
MATHEMATICS Higher 2
20 August 2010
Paper 1 Additional materials:
3 hours Answer Paper Graph paper List of Formulae (MF15) Cover Page
READ THESE INSTRUCTIONS FIRST Write your name and civics class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together, with the cover page in front.
This document consists of 6 printed pages.
[Turn over
2
1
(i)
(ii)
2
(i) (ii)
3
1 A Bx + C in the form , where A, B, and C are constants to be + 3 1− x 1 − x 1 + x + x2 found. [2] x+2 Hence find the coefficient of x r in the expansion of , in ascending powers 1 + x + x2 of x. [4]
Express
k k + 1 k + Show that= − , where k ∈ , k ≥ 3 . 2 3 3 n r Hence find ∑ . r =3 2
[2] [4]
A curve is defined by the parametric equations = x 3(sin 2t − cos 2t ),
= y 3(2t − cos 2t ).
Find the equations of the tangent and the normal to the curve at the point P, where t =
π 4
.
The tangent and normal to the curve at P meet the y-axis at R and S respectively. Find the area of the triangle PRS.
4
[6]
A sequence u1 , u2 , u3 , is such that u1 = 0 and
un +1 =3un + 4n ( n − 1) ,
for all n ∈ + .
(i)
Prove by induction that un =3n − 2n 2 − 1 for all n ∈ + .
(ii)
Given u0 = 0 , find
[4]
2N
∑ un in terms of N .
n =0
n 1 Given r 2 = n ( n + 1)( 2n + 1) 6 r =1
∑
[4]
3
5
Given that y = 1 + ln(1 + x) , show that 2
(i)
d 2 y dy 1 y 2 + + = 0. dx dx 2(1 + x) 2
(ii)
the Maclaurin’s series for y in ascending powers of x, up to and including the
[2]
term in x3 , is 1+
1 3 17 x − x 2 + x3 . 2 8 48
[3]
2
(iii)
3 17 1 Expand 1 + x − x 2 + x 3 in powers of x up to and including x3 , simplifying 8 48 2
your answer. Explain briefly how the result can be used as a check on the correctness of the first four terms in the series for y.
6
(a)
[3]
A geometric progression has first term 1 and common ratio r. The sum of the first four terms is less than the sum to infinity of the remaining terms. Without the use of a graphic calculator, find the range of values of r.
(b)
[4]
Adam decided to save some money each day to buy his favorite toys. On the first day, he saved one twenty-cent coin; on the second day, he saved two twenty-cent coins; and so on. Find the total amount of money saved at the end of one year. (Assume 365 days in a year).
[3]
After spending all his savings in the first year to buy his favorite toys, Adam started saving in the same manner again in the following year. As an encouragement, his mother contributed the same amount that he saved every Saturday and Sunday. Assuming the second year started on a Monday, find the total amount of money saved at the end of the second year.
7
[3]
The line l passes through the points A and B with coordinates
( 5, 14, 11) respectively. The plane p has equation
( −1,
2, 3) and
2x + 3y − 6z = −7.
(i)
Show that the line l is parallel but not contained on the plane p.
[3]
(ii)
Find the distance of the line l from the plane p.
[3]
(iii)
Find a cartesian equation of the plane which contains l and is perpendicular to p. [2]
[Turn over
4
8
(a)
The complex numbers p and q are such that p= 2 + ia , where a and b are real numbers.
q= b − i ,
Given that pq= 13 + 13i , find the possible values of a and b . (b)
Sketch the locus of z which satisfies
= z − 4 − 3i 2
and
Re(z) ≥ 4 .
(ii) Show that the greatest value of arg( z − i) is
The function f is defined by f:x
x
λ − x2
[2] [2]
(i) Find the least and greatest value of z .
9
[4]
π 4
.
[1]
for x ∈ , − λ < x < 0, where λ is a
positive constant.
10
(i)
Show by differentiation that f ( x ) increases as x increases.
[2]
(ii)
Find f −1 ( x ) , stating the domain of f −1.
[4]
(iii)
1 1 Find the value of λ such that f −1f −1 − = − . 2 2
[3]
(i)
Find the integral
(ii)
By sketching the graphs of y = 3e x and y= x + 3, or otherwise, solve the inequality
x −1
∫ 1+ 4x
2
dx.
3e x > x + 3.
[3]
[3]
Hence find x ∫−2 3e − x − 3 dx, giving your answer in an exact form. 2
[4]
5
11
(a)
y y = ( ln x )
O
2
x
A
The function f is defined by f(x) = (ln x)2 for x > 0. The diagram shows a sketch of the graph of y = f(x). The minimum point of the graph is A. (i)
State the x-coordinate of A.
[1]
(ii)
Use the substitution x = eu to show that the area of the region bounded by the x-axis, the line x = e and the curve is given by
∫
1 2 u u e du. 0
Hence, find the exact value of this area. (b)
[5]
y 3 y= − x+2 2
y =1
R
O
x y = cos x
3 Find the volume of the solid formed when the region R, bounded by the lines y = − x+ 2, 2 y = 1 and the curve y = cos x, is rotated 2π radians about the y-axis, giving your answer correct to 3 decimal places. [4]
[Turn over
6
12
The curve C has equation ax − k 2 y= 2 , x − kx + k where a, k are constants and a > 0 . (i)
Find the values of k such that C has two asymptotes.
(ii)
The diagram shows the graph of y =
[2]
5x − k 2 for some k > 0. x 2 − kx + k
y
O −k
k2 5
x
On separate diagrams, draw sketches of the graphs of 25 x − k 2 (a) y= , 25 x 2 − 5kx + k (b)
y2 =
(c)
y=
k 2 − 5x , x 2 − kx + k
x 2 − kx + k , 5x − k 2
including the coordinates of the points where the graphs cross the axes and the equations of any asymptotes.
[8]
JURONG JUNIOR COLLEGE J2 Preliminary Examination
9740/02
MATHEMATICS Higher 2
26 August 2010
Paper 2 Additional materials:
3 hours Answer Paper Graph paper List of Formulae (MF15) Cover Page
READ THESE INSTRUCTIONS FIRST Write your name and civics class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together, with the cover page in front.
This document consists of 6 printed pages.
[Turn over
2 Section A: Pure Mathematics [40 marks]
1 (i)
O
x
P
The diagram shows a rectangle inscribed in a semicircle of centre, O, and fixed radius a. The length OP is denoted by x. Show that, as x varies, the perimeter of the rectangle is a maximum when its sides are in the ratio 4 : 1. [6]
(ii) Variables x and y are related by the equation 2 y 2 + xy = x 2 − + 3 , x Given that x is increasing at the rate of
where y > 0 .
1 units s −1 , find the rate of increase of y when x is 1. 5 [4]
2 A particular solution of a differential equation is given by ( x + y ) 2 = 2 xy − ( y 2 + y)
2 3 y . Show that 3
dy = −x dx
[2]
A second, related, family of curves is given by the differential equation dy x = y2 + y dx By means of the substitution y = ux , show that the general solution for y, in terms of x, is −x , y= x+c where c is an arbitrary constant.
[3]
Sketch, on a single diagram, three distinct members of the second family of solution curves, stating clearly the coordinates of the points where the curves cross the axes and the equations of any asymptotes. [5]
3
3 Relative to the origin O, the position vectors of A, B and C are
α i + 3j + 4k, 2i – 2k and 4i + β j – k respectively. (i) Point M lies on the line segment AB such that AM : MB = 1: 2 . Given that the position vector of M is 2j + 2k, find α . [2]
3 units, find β , where β is (ii) Given that the length of projection of BC onto the line OM is 2 a positive constant. [3] 2 a = r 0 + µ 0 , µ ∈ , a ∈ . Using the values of α and β (iii) The line l has vector equation −2 1 found above, determine the value of a if l makes an angle of
π
6
radians with the plane ABC. [4]
π
4 (i) Given that z 2 = 2 and arg( −iz ) = , find w in the form a + bi , where a, b ∈ , if 4 2 z 5 [4] wz = 2 2 and arg = − π . w 6 (ii) Solve the equation z 4 + 1 − 3 i =0 , giving the roots in the form re iθ where r > 0 and − π < θ ≤ π.
[3]
Show the roots on an Argand diagram.
[2]
Describe the geometrical shape formed by the points representing the roots and justify your answer. [2]
[Turn over
4 Section B: Statistics [60 marks]
5 From the letters of the word DISTRIBUTION, find (i) the number of 4-letter code-words that can be formed if the code-word contains exactly three ‘I’s. [2] (ii) the number of code-words that can be formed using all the letters such that all the three ‘I’s are separated. [2]
6 In a badminton team of 8 players, 5 are boys and 3 are girls. Boy A and Girl B are the only 2 left-handed players in the team. In a particular practice, 4 players are chosen to play doubles. Find the probability that (i) exactly 1 left-handed player is chosen,
[2]
(ii) 2 girls are chosen given that exactly 1 left-handed player is chosen,
[3]
(iii) either Boy A or Girl B is chosen (or both).
[2]
7 The number of guitars sold by a music shop per day follows a Poisson distribution with mean λ . It is known that on 2 in 7 days, there are no guitars sold. Show that λ = 1.253 , correct to 3 decimal places.
[2]
(i) Calculate the probability that less than 4 guitars are sold in a day.
[2]
(ii) Using a suitable approximation, find the probability that, in a random sample of 100 days, there will be more than 95 days in which less than 4 guitars are sold per day.
[4]
(iii) Calculate the probability that in a period of 90 days, the mean number of guitars sold per day is more than 1.5.
[3]
5 8 The random variable X has a normal distribution with mean 15 and variance 5. The random variable T is the sum of 2 independent observations of X. (i) Find P (T > 2 + 3 X ) .
[3]
(ii) Three independent observations of X are obtained. Find the probability that exactly two of the observations have value less than 20.
[3]
The random variable Y has a normal distribution with mean µ and variance σ 2 . (iii) If σ = 22.5 , find the greatest probability of P(15.1 < Y < 29.9) , stating the value of µ .
[2]
(iv) If µ = 10 and P( X + Y > 27) = 0.25 , calculate the value of σ and state an assumption needed to carry out the calculation.
[4]
9 A sample of 60 customers is to be chosen to take part in a survey conducted by a restaurant owner.
(i) Explain briefly how the restaurant owner could use quota sampling.
[1]
(ii) The purpose of the survey is to investigate customers’ opinions about the different lunch and dinner menus.
Give a reason why a stratified sample might be preferable in this context.
[1]
Explain clearly how the restaurant owner could use stratified sampling from his list of regular customers if the ratio of regular customers for lunch and dinner is 2 : 3.
[2]
[Turn over
6 10 Water in a reservoir undergoes a purification process before it can be consumed. The effectiveness, y %, of the process for various flow rates, x m3 s-1, is shown below.
x y
1 80
2 60
4 45
6 40
8 30
10 25
20 18
30 15
40 10
The variables x and y are thought to be related by the equation e y = axb , where a and b are constants. (i) Give a sketch of the scatter diagram of y against ln x . Comment on whether a linear model would be appropriate referring to the scatter diagram. [2] (ii) Find the value of the product moment correlation coefficient between y and ln x and explain whether it supports your comment in part (i). [2] (iii) Find the least squares regression line of y on ln x and estimate the values of a and b . [3] (iv) Predict the effectiveness of the process when water flows at 0.5 m3s-1. Comment on the reliability of your prediction.
[2]
(v) Explain why in this context, the above model would not be appropriate for large values of x . [1]
11 The past records of a supermarket show that the mean amount spent per customer was $59 with standard deviation $8. The supermarket’s management suspects that the mean amount spent per customer has decreased. A random sample of 8 customers was taken and the amount spent per customer , $x, was recorded. The following result was obtained. ∑ x = 432 Stating a necessary assumption about the population, test the supermarket’s management’s suspicion at the 5% significance level. [6] To encourage customers to spend more at the supermarket, the management initiated a promotional campaign whereby each customer will receive a voucher which can be used to redeem products at the supermarket. A week after the start of the campaign, the manager of the supermarket took a sample of 9 customers and the amount spent per customer, $y, is summarised by 2 −72, 1234 . ∑ ( y − 70) = ∑ ( y − 70) = The actual mean amount spent per customer is $ µ . In a test at the 5% level of significance, the hypotheses are: Null hypothesis : Alternative hypothesis :
µ = µ0 µ ≠ µ0 .
Given that the null hypothesis is rejected in favour of the alternative hypothesis, find the set of possible values of µ0 . [6]
2010 JJC H2 Mathematics Prelim Exam P1 Solutions 1(i) 1 A Bx + C ≡ + Let 3 1 − x 1 − x 1 + x + x2 1 ≡ A (1 + x + x 2 ) + ( Bx + C )(1 − x ) Hence,
1 3 1= A+C , 2 C= 3
Using cover-up rule, A = Let x = 0 ,
Comparing coefficients of x , 0 = A + B − C , B = ∴
1 1 x+2 ≡ + . 3 1− x 3 (1 − x ) 3 (1 + x + x 2 )
1 3
x+2 3 1 = − 2 3 1+ x + x 1− x 1− x
1(ii)
= 3 (1 − x 3 ) − (1 − x ) −1
−1
= 3 (1 + x 3 + x 6 + K) − (1 + x + x 2 + K)
= 2 − x − x 2 + 2 x3 − x 4 − x5 + K if r is a multiple of 3 ⎧2 Hence, coefficients of x r = ⎨ otherwise ⎩ −1 2.
n
⎛r⎞
r =3
⎝ ⎠
⎛ n ⎞ ⎛ n − 1⎞ ⎛ n − 2 ⎞ ⎛ 4⎞ ⎛ 3⎞ ⎟+⎜ ⎟ +K + ⎜ ⎟ + ⎜ ⎟ ⎝ ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 2⎠
∑⎜ 2⎟ = ⎜ 2⎟ + ⎜
⎛ n + 1⎞ ⎛ n ⎞ =⎜ ⎟−⎜ ⎟ ⎝ 3 ⎠ ⎝ 3⎠ ⎛ n ⎞ ⎛ n − 1⎞ +⎜ ⎟−⎜ ⎟ ⎝3⎠ ⎝ 3 ⎠ ⎛ n − 1⎞ ⎛ n − 2 ⎞ +⎜ ⎟−⎜ ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ M ⎛5⎞ ⎛ 4⎞ +⎜ ⎟−⎜ ⎟ ⎝ 3⎠ ⎝ 3 ⎠ ⎛ 4 ⎞ ⎛ 3⎞ +⎜ ⎟−⎜ ⎟ ⎝ 3 ⎠ ⎝ 3⎠ ⎛ n + 1⎞ =⎜ ⎟ −1 ⎝ 3 ⎠
or
(n + 1)! −1 3!(n − 2)!
1
x = 3(sin 2t − cos 2t ) , y = 3(2t − cos 2t ) dx = 3(2 cos 2t + 2sin 2t ) dt dy = 3(2 + 2sin 2t ) dt dy 3(2 + 2sin 2t ) 1 + sin 2t = = dx 3(2 cos 2t + 2sin 2t ) cos 2t + sin 2t
3.
When t =
π 4
, at point P , x = 3 , y =
3π dy and =2 2 dx
3π = 2( x − 3) 2 3π 3π When x = 0, y = − 6 . R is (0 , −6 ) 2 2 3π −1 = ( x − 3) Equation of normal at P is y − 2 2 Equation of tangent at P is y −
When x = 0, y = As PR ⊥ PS,
3π 3 3π 3 + . S is (0 , + ) 2 2 2 2
1 1 3 1 Area of the triangle PRS = RS .(3) = ( + 6)(3) = 11 2 2 2 4
4.
(i)
Let Pn be the statement un = 3n − 2n 2 − 1 for all n ≥ 1 .
When n = 1,
LHS = u1 = 0
(By definition)
RHS = 3 2 −2 ( 2 ) − 1 = 0 2
∴ P1 is true. +
Assume that Pk is true for some k ∈
, k ≥ 1.
That is, uk = 3k − 2k 2 − 1
-----------------------(1)
We want to prove Pk +1 , ie uk +1 = 3k +1 − 2 ( k + 1) − 1. 2
LHS
= 3uk + 4 ( k )( k − 1)
= 3 ( 3k − 2k 2 − 1) + 4k ( k − 1) = 3k +1 − 6k 2 − 3 + 4k 2 − 4k = 3k +1 − 2 ( k 2 + 2k + 1) − 1 = 3k +1 − 2 ( k + 1) − 1 2
∴ Pk is true ⇒ Pk +1 is true. Since P1 is true, and Pk is true ⇒ Pk +1 is true. By Mathematical Induction, un = 3n − 2n 2 − 1 is true for all n ≥ 1 .
2
4. (ii)
∑ un = ∑ (3n − 2n2 − 1) 2N
2N
n=0
n =0
=
(
) − 2 2N ( 2N + 1) ⎡2 ( 2N ) + 1⎤ − ( 2N + 1)
1 32 N +1 − 1 2
⎣
6
⎦
32 N +1 − 1 2 N + 1 = − ⎡ 2 N ( 4 N + 1) + 3⎤⎦ 2 3 ⎣ 32 N +1 − 1 ( 2 N + 1) = − 8N 2 + 2 N + 3 2 3 Method 1 y 2 = 1 + ln(1 + x)
(
5(i)
)
dy 1 = dx 1 + x dy 1 = dx 2 y (1 + x) dy 1 y = dx 2(1 + x)
2y
Method 2 1 − ⎛ 1 ⎞ dy 1 = (1 + ln(1 + x)) 2 ⎜ ⎟ dx 2 ⎝ 1+ x ⎠ 1 = 2 y (1 + x) dy 1 y = dx 2(1 + x)
d 2 y ⎛ dy ⎞ 1 −1 y 2 + ⎜ ⎟ = (−1)(1 + x) −2 = dx ⎝ dx ⎠ 2 2(1 + x) 2 2
2
y
5(ii)
d 2 y ⎛ dy ⎞ 1 +⎜ ⎟ + =0 2 dx ⎝ dx ⎠ 2(1 + x) 2 d 3 y dy d 2 y dy d 2 y 1 y 3 + . 2 +2 − =0 2 dx dx dx dx d x (1 + x)3 when x = 0, dy 1 d 2 y 3 d 3 y 17 y = 1, = , 2 = − , 3 = dx 2 dx 4 dx 8 ∴ y ≈ 1+ = 1+
2 3 1 ⎛ 3 ⎞ ⎛ x ⎞ ⎛ 17 ⎞ ⎛ x ⎞ x + ⎜ − ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ 2 ⎝ 4 ⎠ ⎝ 2! ⎠ ⎝ 8 ⎠ ⎝ 3! ⎠
1 3 17 x − x 2 + x3 2 8 48
3
5(iii)
2
3 2 17 3 ⎞ ⎛ 1 x ⎟ ⎜1 + x − x + 8 48 ⎠ ⎝ 2 3 17 ⎞ ⎛ 1 3 17 3 ⎞ ⎛ 1 = ⎜1 + x − x 2 + x3 ⎟ ⎜1 + x − x 2 + x ⎟ 8 48 ⎠ ⎝ 2 8 48 ⎠ ⎝ 2 1 1 = 1 + x − x 2 + x 3 + ... 2 3 Given y = 1 + ln(1 + x) ⇒ y 2 = 1 + ln(1 + x) 1 1 LHS = y 2 = 1 + x − x 2 + x3 + ... 2 3 RHS = 1 + ln(1 + x)
1 1 ⎛ ⎞ = 1 + ⎜ x − x 2 + x 3 + ... ⎟ 2 3 ⎝ ⎠ Since LHS = RHS, the first four terms in the series for y is correct. 6(a) S 4 < S∞ − S 4 2S 4 < S∞ 2 (1 − r 4 )
1 1− r 1− r 4 2 (1 − r ) < 1 <
r4 >
(Since 1 − r > 0 )
1 2
⎛ 2 1 ⎞⎛ 2 1⎞ ⎜⎜ r − ⎟⎟ ⎜⎜ r + ⎟>0 2 ⎠⎝ 2 ⎟⎠ ⎝ r2 < −
1 (no solution) 2
r2 >
or
1 2
1 1 ⎛ 4 ⎞⎛ 4 ⎞ 1 1 ⎛ ⎞ ⎛ ⎞ ⎜ r − ⎜ ⎟ ⎟⎜ r + ⎜ ⎟ ⎟ > 0 ⎜ ⎝ 2 ⎠ ⎟⎜ ⎝ 2 ⎠ ⎟ ⎝ ⎠⎝ ⎠ 1
6(b)
1
⎛ 1 ⎞4 ⎛ 1 ⎞4 −1 < r < − ⎜ ⎟ or ⎜ ⎟ < r < 1 . ⎝2⎠ ⎝2⎠ Total sum = 20 + 2 ( 20 ) + 3 ( 20 ) +K 365 ( 20 ) ⎛ ( 365 )(1 + 365 ) ⎞ = ( 20 ) ⎜ ⎟ 2 ⎝ ⎠ = 1335900 cents
Total sum = 1335900 + 6 ( 20 ) + 7 ( 20 ) + 13 ( 20 ) + 14 ( 20 ) + ... + 363 ( 20 ) + 364 ( 20 )
(
= 1335900 + 20 13 + (13 + 14 ) + (13 + 2 (14 ) ) + K (13 + 51(14 ) )
)
⎛ 52 ⎞ = 1335900 + 20 ⎜ ⎟ (13 + 727 ) ⎝ 2 ⎠ = 1720700 cents
4
⎛ 5 ⎞ ⎛ −1⎞ uuur ⎜ ⎟ ⎜ ⎟ AB = ⎜14 ⎟ − ⎜ 2 ⎟ ⎜ 11 ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛6⎞ ⎛ 3⎞ ⎜ ⎟ ⎜ ⎟ = ⎜12 ⎟ = 2 ⎜ 6 ⎟ ⎜8⎟ ⎜ 4⎟ ⎝ ⎠ ⎝ ⎠
7(i)
⎛ 3⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 6 ⎟ • ⎜ 3 ⎟ = 6 + 18 − 24 ⎜ 4 ⎟ ⎜ −6 ⎟ ⎝ ⎠ ⎝ ⎠ =0 ⇒ l is parallel to p.
⎛ −1⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ • ⎜ 3 ⎟ = −2 + 6 − 18 ⎜ 3 ⎟ ⎜ −6 ⎟ ⎝ ⎠ ⎝ ⎠ = −14 ≠ −7 ∴ l is parallel but not contained on the plane p. 7(ii)
Method 1 A
l
d C Let C(c,0,0) be a point on π . ⎛c⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ • ⎜ 3 ⎟ = −7 ⎜ 0 ⎟ ⎜ −6 ⎟ ⎝ ⎠ ⎝ ⎠ c=−
π
7 2
⎛ 2⎞ ⎜ ⎟ 3 ⎛ ⎛ −7 / 2 ⎞ ⎛ −1⎞ ⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟ ⎟ −6 d = ⎜⎜ 0 ⎟ − ⎜ 2 ⎟⎟• ⎝ ⎠ 49 ⎜⎜ 0 ⎟ ⎜ 3 ⎟⎟ ⎠ ⎝ ⎠⎠ ⎝⎝
⎛ −5 / 2 ⎞ ⎛ 2 ⎞ 1⎜ ⎟ ⎜ ⎟ = ⎜ −2 ⎟ • ⎜ 3 ⎟ = 1 7⎜ ⎟ ⎜ ⎟ ⎝ −3 ⎠ ⎝ −6 ⎠
5
Method 2 Let m be the line perpendicular to p and passing through A. ⎛ −1⎞ ⎛ 2⎞ ⎜ ⎟ Vector equation of line m : r = ⎜ 2 ⎟ + μ ⎜⎜ 3 ⎟⎟ , μ ∈ ⎜3⎟ ⎜ −6 ⎟ ⎝ ⎠ ⎝ ⎠
Let F be the foot of perpendicular of A to p.
⎛ −1 + 2μ ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2 + 3μ ⎟ • ⎜ 3 ⎟ = −7 ⎜ 3 − 6μ ⎟ ⎜ −6 ⎟ ⎝ ⎠ ⎝ ⎠ 2 ( −1 + 2μ ) + 3(2 + 3μ ) − 6(3 − 6μ ) = −7
μ=
1 7
⎛ −1⎞ ⎛ 2⎞ uuur ⎜ ⎟ 1 ⎜ ⎟ OF = ⎜ 2 ⎟ + ⎜ 3 ⎟ ⎜ 3 ⎟ 7 ⎜ −6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ −1⎞ ⎛ 2 ⎞ ⎛ −1 ⎞ uuur ⎜ ⎟ 1 ⎜ ⎟ ⎜ ⎟ AF = ⎜ 2 ⎟ + ⎜ 3 ⎟ − ⎜ 2 ⎟ ⎜ 3 ⎟ 7 ⎜ −6 ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 2⎞ 1⎜ ⎟ = ⎜ 3⎟ 7⎜ ⎟ ⎝ −6 ⎠ uuur 1 2 2 AF = 2 + 3 + (−6) 2 7 =1
∴ Distance of the line l from the plane p=1 unit 7(ii)
Let the plane required be p1 .
⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ −48 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Normal of p1 = ⎜ 6 ⎟ × ⎜ 3 ⎟ = ⎜ 26 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ −6 ⎠ ⎝ −3 ⎠ ⎛ −1⎞ ⎛ −48 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ • ⎜ 26 ⎟ = 48 + 52 − 9 = 91 ⎜ 3 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ −48 ⎞ ⎜ ⎟ ∴ Equation of p1 : r • ⎜ 26 ⎟ = 91 ⇒ −48 x + 26 y − 3z = 91 ⎜ −3 ⎟ ⎝ ⎠
6
8(a)
pq = 13 + 13i ( 2 + ia )( b − i ) = 13 + 13i 2b − 2i + abi + a = 13 + 13i (a + 2b) + i (ab − 2) = 13 + 13i Comparing real and imaginary parts, - (1) a + 2b = 13 ab − 2 = 13 - (2) 15 (2): b = a 15 Subst. b = into (1): a ⎛ 15 ⎞ a + 2 ⎜ ⎟ = 13 ⎝a⎠ a 2 − 13a + 30 = 0
( a − 3)( a − 10 ) = 0 a = 3 or 10 3 b = 5 or 2 8b(i) y Locus of z
R (4,5)
Q (4, 3)
3
A(0, 1)
P (4, 1)
O
Least z = OP
Greatest z = OQ
= 42 + 12
8b(ii)
x
4
= 17 Max arg( z − i ) = ∠PAR
= 42 + 32 + 2 =7
⎛4⎞ = tan −1 ⎜ ⎟ ⎝4⎠ =
π 4
7
9(i)
f ( x) =
x
λ − x2 ⎛1⎞
λ − x2 − x ⎜ ⎟ ( λ − x2 ) ⎝2⎠ f ' ( x) = λ − x2
−1
2
( −2 x )
=
λ
(λ − x )
2 32
>0
f is strictly increasing, thus f(x) increases as x increases. 9(ii) domain of f −1 = range of f = ( −∞, 0 ) . x
=
y
λ − x2
y 2 ( λ − x2 ) = x 2 ( y 2 + 1)
x2
λ y2
= = ±
x
λ y2 1+ y2
Since − λ < x < 0, that is, y < 0, x = That is, f −1 ( x) = 9(iii)
f −1f −1 ( x) =
λx 1 + x2 λx
λy 1+ y2
.
, for x < 0.
1 + ( λ + 1) x 2
λ ( −0.5 ) 1 + ( λ + 1)( −0.5 )
2
= −0.5
4λ 2 − λ − 5 =
( 4λ − 5)( λ + 1) Since λ > 0, λ =
=
0 0
5 . 4
Alternatively: f ( −0.5 ) = −0.5
( −0.5) 2 λ − ( −0.5 ) 1
λ=
=
0.5
=
λ − 0.25
5 4
8
x −1
10(i)
∫ 1 + 4x
2
dx
1 8x 1 dx −∫ dx 2 ∫ 8 1 + 4x 1 + 4 x2 1 1 = ln (1 + 4x2 ) − tan −1 (2x) + c 8 2 =
(ii)
y
y = x+3
3 y = 3e x -2.82
O
x
From the graphs, x < −2.82 or x > 0 . For x < −2.82 or x > 0 , 3e x > x + 3 i.e. 3e x − x − 3 > 0 if x < −2.82 or x > 0 For −2.82 < x < 0 , 3e x < x + 3 i.e. 3e x − x − 3 < 0 if −2.82 < x < 0 x ∫−2 3e − x − 3 dx
2
= ∫−2 3e x − x − 3 dx + ∫0 3e x − x − 3 dx 0
= [3e x −
2
x2 x2 − 3 x]0−2 + [3e x − − 3 x]02 2 2
= −1 − 3e−2 + 3e2 − 11 (need to check positive or negative within each modulus to
remove the modulus sign) = 1 + 3e−2 + 3e2 − 11 = 3e−2 + 3e2 − 10
9
11a(i) (ii)
x-coordinate of A = 1 Area of region bounded =
e
∫1
(ln x) 2 dx
dx = eu du When x = 1, u = 0 & When x = e, u = 1 Let x = eu ⇒ ln x = u,
∴ Area of region bounded =
1
∫0 u
2 u
Area of region bounded =
e du 1
∫0 u
2 u
e du
= [u 2 eu ] 0 − ∫ 0 2ueu du 1
1
1
1
= e − [ ⎡ 2ueu ⎤ − ∫ 2eu du ] ⎣ ⎦0 0 1
= e – 2e + ⎡ 2eu ⎤ ⎣ ⎦0 = − e + 2e − 2 = e − 2 11(b) 3 Let P be the point of intersection of y = − x + 2 and y = cos x . 2 Using GC, the coordinates of point of intersection = (0.94031, 0.58954) Volume formed about the y –axis =π
1
∫ 0.58954
2
⎡2 ⎤ −1 2 ⎢⎣ 3 (2 − y )⎥⎦ − (cos y ) dy
= π (0.0907) = 0.285 (3 d.p) 12(i) Horizontal asymptote is y = 0 . Discriminant of x 2 − kx + k = 0 ⇒ k 2 − 4k = 0 ⇒ k ( k − 4 ) = 0 . Therefore when k = 0 or 4, C has two asymptotes.
(ii)(a)
y
O
k 2 25
x
−k
10
(ii)(b)
y
k x O
k2 5
− k
(c)
O −1 k
k2 5
1 k2 k y = x+ − 5 25 5
11
2010 JJC Prelim P2 Solutions 1(i)
a
y
x Let the length and the width of the rectangle be 2x and y The perimeter S = 2y + 4x ……(1) y 2 = a 2 − x 2 …………………(2) Subst (2) into (1) S = 2 a2 − x2 + 4 x dS 2x =− +4=0 dx a2 − x2 2x − +4=0 2 a − x2 x =2 a2 − x2
4 2 a 5 1 ⇒ y2 = a 2 5 4 1 ⇒ x= a a and y = 5 5 d 2s 2a 2 = − 3 < 0 dx 2 (a 2 − x 2 ) 2 x2 =
Length : width = 2x : y = 4:1
1
1(ii)
2 y 2 + xy = x 2 − + 3 ……(1) x Differentiate w.r.t x dy dy 2 2 y + x + y = 2x + 2 dx dx x dy 2 (2 y + x) = 2 x + 2 − y …….(2) dx x Subst. x = 1 into (1): y 2 + y = 2 ⇒ y= 1 and y ≠ − 2 (rejected) dy (2): =1 dx dy dy dx = . dt dx dt 1 = (1)( ) 5 1 = units s −1 5 2 2 ( x + y ) 2 = 2 xy − y 3 3 2 ( x + y ) 2 − 2 xy = − y 3 3 2 y 2 + x2 = − y3 3 dy dy 2 y + 2 x = −2 y 2 dx dx dy ( y + y 2 ) = − x (shown) dx Alternative: dy dy ⎛ dy ⎞ 2( x + y ) ⎜ 1 + ⎟ = 2 x + 2 y − 2 y 2 dx dx ⎝ dx ⎠ dy ( y + y 2 ) = − x (shown) dx
2
2
dy = y2 + y dx use y = ux x
dy d u x+u = dx dx 2 ⎛ du ⎞ ∴ x ⎜ x + u ⎟ = ( ux ) + ux ⎝ dx ⎠ du x + u = u2 x + u dx 1 du =1 u 2 dx 1 ∫ u 2 du = ∫ 1 dx 1 − = x + c , where c is an arbitrary constant u x − = x+c y x c = −1 + y=− x+c x+c
2 Family of solution curves: c = −1, y = −1 −
1 x −1
c = 0, y = −1
c =1
1 c = 1, y = −1 + x +1
y
c = −1
x
y = −1
c=0
x = −1
x =1
3
3(i) Using ratio theorem, uuur uuur uuuur 2OA + OB OM = 3 uuuur uuur uuur 3OM − OB OA = 2 ⎡ ⎛ 0 ⎞ ⎛ 2 ⎞⎤ 1 ⎢ ⎜ ⎟ ⎜ ⎟⎥ = ⎢3 ⎜ 2 ⎟ − ⎜ 0 ⎟ ⎥ 2 ⎜ ⎟ ⎜ ⎟ ⎢⎣ ⎝ 2 ⎠ ⎝ −2 ⎠ ⎥⎦
1 A
2 M
C
⎛ −1 ⎞ ⎜ ⎟ =⎜ 3 ⎟ ⎜4⎟ ⎝ ⎠ α = −1 3(ii)
⎛4⎞ ⎛ 2⎞ uuur ⎜ ⎟ ⎜ ⎟ BC = ⎜ β ⎟ − ⎜ 0 ⎟ ⎜ −1⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛2⎞ ⎜ ⎟ = ⎜β ⎟ ⎜1⎟ ⎝ ⎠
uuur 3 Length of projection of BC onto the line OM = 2 uuur uuuur BC • OM 3 = uuuur 2 OM
⎛ 2 ⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ ⎜ β ⎟•⎜1⎟ ⎜ 1 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠
=
3 2
12 + 12 β +1 = 3
β 2 + 2β − 8 = 0 β = 2 or − 4 (rejected) ∴β = 2 Alternatively, β + 1 = 3 β + 1 = 3 or β + 1 = −3 β = 2 or − 4 (rejected) ∴β = 2
4
3(iii)
⎛ 2 ⎞ ⎛ −1⎞ ⎛1⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = ⎜ 0 ⎟ − ⎜ 3 ⎟ = 3 ⎜ −1 ⎟ ⎜ −2 ⎟ ⎜ 4 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 4 ⎞ ⎛ −1⎞ ⎛ 5 ⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AC = ⎜ 2 ⎟ − ⎜ 3 ⎟ = ⎜ −1 ⎟ ⎜ −1⎟ ⎜ 4 ⎟ ⎜ −5 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛1⎞ ⎛5⎞ ⎜ ⎟ ⎜ ⎟ Normal of plane ABC = ⎜ −1 ⎟ × ⎜ −1 ⎟ ⎜ −2 ⎟ ⎜ −5 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 3⎞ ⎜ ⎟ = ⎜ −5 ⎟ ⎜ 4⎟ ⎝ ⎠ ⎛a⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ • ⎜ −5 ⎟ = ⎜1⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠
⎛a⎞ ⎜ ⎟ ⎜0⎟ ⎜1⎟ ⎝ ⎠
⎛ 3⎞ π ⎜ ⎟ 5 sin − ⎜ ⎟ 6 ⎜ 4⎟ ⎝ ⎠
⎛1⎞ 3a + 4 = a 2 + 1 50 ⎜ ⎟ ⎝2⎠
( 6a + 8 )2 = 50 ( a 2 + 1) 7 a 2 − 48a − 7 = 0 a = 7 or -
1 (rejected since angle is acute, 3a +4>0) 7
∴a = 7
5
4(i)
z2 = 2 ⇒ z = 2
arg ( −iz ) =
π 4
⇒ arg ( −i ) + arg( z ) =
π 4
⎛ π⎞ −⎜− ⎟ 4 ⎝ 2⎠ 3π = 4
arg( z ) =
π
wz = 2 2 w z =2 2
∴w =2 ⎛ z2 ⎞ 5 arg ⎜ ⎟ = − π ⎜ w⎟ 6 ⎝ ⎠ 5 2 arg( z ) − arg( w) = − π 6 ⎛3 ⎞ 5 arg( w) = 2 ⎜ π ⎟ + π ⎝4 ⎠ 6 7 = π 3 =
π
(principle value) 3 ⎡ ⎛π ⎞ ⎛ π ⎞⎤ w = 2 ⎢cos ⎜ ⎟ + i sin ⎜ ⎟ ⎥ ⎝ 3 ⎠⎦ ⎣ ⎝3⎠ = 1 + 3i
6
4(ii) 4
z + 1 − 3i = 0 4
z = −1 + 3i ⎛ 2π ⎞ + 2 kπ ⎟ i ⎜ ⎠ = 2e⎝ 3
z
1 ⎛ π + kπ ⎞ i ⎜ ⎟ = 2 4 e⎝ 6 2 ⎠ 1 π (1+ 3k )i = 24 e 6
, k = 0, ±1, −2
1 π 1 2π 1 π 1 5π − i − i i i = 24 e 6 , 24 e 3 , 24 e 3 , 24 e 6
y
z3 z2 π 6 −
z4
π
x
3
z1
The points form a square since the diagonals are perpendicular and of equal length.
7
5(i)
The different letters left are D S T R B U O N 4! No of different code words = 8C1 = 32 3!
5(ii)
To permute the remaining 9 letters 1st,
6(i) 6(ii)
9! ways. 2
To slot in the 3Is, 10C3 ways. Thus number of code words that can be formed is 9! = 10C3 2 = 21772800 2 C 6C 4 P(1 left-handed player) = 81 3 = 7 C4 P(G R G L BR BR )+P(G R G R BL BR ) P(exactly 1 left-handed) 2 C1 4 C2 + 2 C2 4 C1 8 C4 = 4 7 8 2 = 35 = 4 5 7
P(2 G given exactly 1 left-handed)=
6(iii) P(Boy A or Girl B is chosen) = 1 – P (Both Boy A and Girl B are not chosen) 6 C 11 = 1− 8 4 = C4 14
Alternative: P(Boy A or Girl B is chosen) = P(Boy A is chosen) + P(Girl B is chosen) −P(Boy A and Girl B are chosen) 7 7 C C 6C = 8 3+8 3−8 2 C4 C4 C 4 1 1 3 11 = + − = 2 2 14 14
8
7
Let X denote the number of guitars sold by a music shop in a day. X P0 ( λ ) 2 7 2 ⇒ e− λ = 7 2 ⇒ λ = − ln = 1.253 7 P ( X < 4) = P( X ≤ 3) = 0.96145 ≈ 0.961 Given that P( X = 0) =
7(i)
7(ii) Let Y be the number of days out of 100 in which at least 4 guitars were sold per day. Y B(100,1 − 0.96145) = B(100, 0.03855) Since n is large and np < 5 , we use a Poisson approximation. Y P0 (3.855) approx P(more than 95 days in which less than 4 guitars were sold per day) = P( at most 4 days in which at least 4 guitars were sold per day ) = P(Y ≤ 4) = 0.657 (to 3 sf) 7(iii)
X
P0 (1.253)
For a large sample of size 90, by Central Limit theorem, 1.253 ⎞ ⎛ X N ⎜ 1.253, ⎟ approx . 90 ⎠ ⎝
(
P X > 1.5 = 0.0182
).
Alternative : X 1 + K + X 90
(
)
P0 (1.253 × 90 ) = P0 (112.77 )
P X > 1.5 = P ( X 1 + K + X 90 > 135 ) = 1 − P ( X 1 + K + X 90 ≤ 135 ) = 0.0183
8(i)
X N (15,5) Let T = X 1 + X 2 E (T − 3 X ) = E ( X 1 + X 2 ) − 3E ( X ) = 15 + 15 − 3(15) = −15 Var (T − 3 X ) = Var ( X 1 + X 2 ) + 9Var ( X ) = 5 + 5 + 32 (5) = 55 T − 3 X N (−15,55) P (T > 2 + 3 X ) = P(T − 3 X > 2) = 0.0109
9
8(ii)
P ( X < 20) = 0.9873 ≈ 0.987 Probability = 3[ P( X < 20) ] P( X > 20) 2
= 3(0.9873) 2 (1 − 0.9873) = 0.0371 Alternatively, use binomial distribution. Let W be r.v. “no. of observations with value less than 20 out of 3” W B(3, 0.987) P (W = 2) = 0.0371 8(iii)
Y
N ( μ , 22.52 ) 15.1 + 29.9 45 = = 22.5 2 2 Greatest P(15.1 < Y < 29.9) = 0.258
For greatest probability, μ =
8(iv) (iv)
X N (15,5) Y N (10, σ 2 ) X + Y N (15 + 10, 5 + σ 2 ) = N (25, 5 + σ 2 ) P ( X + Y > 27) = 0.25 ⎛ 27 − 25 ⎞ P⎜Z > ⎟ = 0.25 5 +σ 2 ⎠ ⎝ ⎛ ⎞ 2 P⎜Z > ⎟ = 0.25 5 +σ 2 ⎠ ⎝
⎛ 2 P⎜Z < 5 +σ 2 ⎝ 2 5 +σ 2 σ = 1.95
⎞ ⎟ = 0.75 ⎠ = 0.6745
Assumption : The random variables X and Y are independent of each other.
10
9(i) To obtain a quota sampling of 60, divide the diners into two subgroups : male and female. Select the first 30 males and 30 females who leaves the restaurant. Or any other relevant answers. 9(ii) By drawing random samples according to the proportion in each stratum, lunch and dinner, the sample will be a better representation of the population. Draw random samples from each stratum with sample size proportional to the size of the strata as follows : Lunch Dinner Number of 2 3 × 60 = 36 × 60 = 24 customers to be 5 5 sampled 10(i)
e y = axb y = ln a + b ln x The scatter diagram is plotted with y against ln x . y
ln x
From the scatter diagram, the points lie close to a straight line, so the linear model is appropriate. 10(ii) From GC, since r ≈ −0.982 which is very close to -1, it supports the claim in part (i).
10(iii)
y = 73.3 − 18.4 ln x a = 6.89 × 1031 (3sf )
b = −18.4 (3sf) x = 0.5, y = 73.3 − 18.4 ln(0.5) = 86.0 86.0% (3sf) Since x = 0.5 is out of the given data range of 1 ≤ x ≤ 40 , the prediction is unreliable. 10(v) For large values of x , the model gives y < 0 . So the model is not valid for large values of x .
10(iv)
11
11 Assume that the amount spent per customer follows a normal distribution. Let X be the actual amount spent per customer per visit. and μ be the actual mean amount spent per customer per visit. H 0 : μ = 59 H1 : μ < 59 82 X −μ ) and test statistic Z = ~ N(0, 1) σ 8 n 432 where x = = 54, μ = 59, σ = 8, n = 8. 8 At 5% level of significance, we use a left-tailed z- test and reject H0 if p-value < 0.05. From the GC, p − value = 0.0385 . Since p − value = 0.0385 < 0.05 , we reject H 0 and conclude that at 5% level of significance there is sufficient evidence to suggest that the mean amount spent per customer per visit has decreased in recent months. Under H 0 , X ~ N(59,
11
H 0 : μ = μ0 H1 : μ ≠ μ0 ∑ ( y −70) y= + 70 = 62 9 2 −72 ) ⎤ ( 1 ⎡ 2 s = ⎢1234 − ⎥ = 82.25 9 − 1 ⎣⎢ 9 ⎦⎥ Under H 0 , test statistic T =
Y − μ0 ~ t(8) s n where y = 62, μ = μ0 , s = 82.25, n = 9
At 5% level of significance, we use a 2-tailed t- test and reject H0 if T ≥ 2.306 Since H 0 is rejected at 5% level of significance, y − μ0 y − μ0 < −2.306 or > 2.306 s s n n
μ0 > 62 + 2.306
{μ0 ∈
82.25 82.25 or μ0 < 62 − 2.306 9 9 : μ0 < 55.0 or μ0 >69.0 }
12
Class
Adm No
Candidate Name:
2010 Preliminary Examination II Pre-university 3 MATHEMATICS 9740 PAPER 1
9740/1
Thursday
16 September
3 hours
Additional materials: Cover page Answer papers List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your name, admission no. and class in the spaces at the top of this page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This question paper consists of 6 printed pages. [Turn over
2
1
(i) Express f ( r ) =
1 in partial fractions. r − 2r 2
n
(ii) Hence find
∑r r =3
2
2 in terms of n. − 2r
[2]
[3]
(There is no need to express your answer as a single algebraic fraction.) ∞
(iii) Deduce the value of
∑r r =3
2
2
2 . − 2r
[2]
Functions f and g are defined by
f : x → e x + 10,
for x ∈ℜ,
g : x → x 2 − 3 x,
for x ∈ℜ.
(i) Sketch the graphs of f and g on the same diagram, indicating clearly the equations of any asymptotes and the coordinates of any turning points. Hence, or otherwise, solve
g ( x) = f ( x) . (ii) Show that the composite function gf exists and find an expression for gf ( x ) .
[4] [2]
(iii) If the domain of g is restricted to the set { x ∈ℜ : x ≥ a} , find the least value of a for which g −1 exists. Hence, find g −1 and state its domain.
3
[4]
The fourth, ninth and nineteenth term of an arithmetic progression are consecutive terms of a geometric progression. (i) Show that the common ratio of the geometric progression is 2.
[3]
(ii) The twentieth term of the arithmetic progression is 63. Find its first term and common difference. [3] (iii) The sum of the first n terms of the arithmetic progression is denoted by S n . Using the results in (ii), find the least value of n for which S n exceeds 200.
[3]
3
4
The curve C has equation a 2 ( x + 1) − b 2 y 2 = 1 2
where a and b are positive constants. 3 Given that the curve passes through the point − , 0 and the equations of its 2 asymptotes are = [4] y 2 x + 2 and y = −2 x − 2 , show that a = 2 and b = 1 .
Hence sketch C, stating the equations of any asymptotes and the coordinates of any points of intersection with the axes. [3]
5
The equation of a curve C is 2 x 3 − 3 xy + y 3 = p , where p is a constant. Find
dy . dx
[2]
It is given that C has a tangent which is parallel to the y-axis. Show that the y-coordinate of the point of contact of the tangent with C must satisfy 2 y6 − 2 y3 − p = 0. 1 Hence show that p ≥ − . 2
[3]
Find the values of p in the case where the line x = 4 is a tangent to C.
[3] 1 2 [2]
It is given instead that C has a tangent which is parallel to the x-axis. Show that p ≥ − in this case also.
6
A disease is spreading through a population of N individuals. It is given that the rate of increase of the number of infected individuals at any time is proportional to the product of the number of infected individuals and the number of uninfected individuals at that time. At any time t, x is the number of infected individuals. Given that initially only one person is infected, show that x = positive constant.
N eN k t , where k is a N −1 + eN k t [7]
[Turn over
4
7
f ( x ) ln ( 2 x + 1) , find f ( 0 ) , f ′ ( 0 ) , f ′′ ( 0 ) and f ′′′ ( 0 ) . Hence obtain the (i) Given that = first three non-zero terms in the Maclaurin’s series for f ( x ) .
[5]
(ii) Hence, or otherwise, show that the first three non-zero terms in the expansion of ln ( 2 x 2 + 3 x + 1) are ax + bx 2 + ax 3 , where a and b are constants to be found.
8
[3]
1 The diagram below shows the graph of y = f ( x ) with a vertical asymptote x = − . The 2 points A ( −2, 0 ) and B ( 0, 2 ) are the point of inflexion and the minimum point respectively.
y y = f (x)
B
A
x O
x= −
1 2
Sketch, on separate diagrams, the graphs of x (i) = y f +1, 2
(ii)
y=
1 . f ( x)
[3]
[3]
5
9
Given that the plane π : r ⋅ n = d and the line l : r = a + λb intersect at a point, d − a ⋅n show that λ = . [2] b ⋅n Find
10
(i)
the value of λ when l lies in π,
[2]
(ii)
the position vector of the point of intersection of l and π.
[2]
(a)
Given that the equation z 4 − z 3 − 9 z 2 + 29 z − 60 = 0 has a root of the form z = 1 + ki , where k is a non-zero real number, find the possible values of k. Hence solve the equation z 4 − z 3 − 9 z 2 + 29 z − 60 = 0.
[5]
(b) In an Argand diagram, the point P represents the complex number z such that
z − 2 − 2i ≤ 2 and
3π π ≤ arg ( z − 2 − 2i ) ≤ . 4 4
(i) Sketch the locus of P.
[3]
(ii) Hence, or otherwise, show that
11
(a)
2+ 2 ≤ arg ( z ) ≤ tan −1 . 4 2− 2
π
[3]
Use the substitution x = cos θ to find the exact value of
∫
π π /2
sin θ dθ . 1 + cos 2 θ
(b) (i) Find ∫ e x sin 2 x dx . (ii) Hence find the exact value of 5∫
[5] [4]
π /2 0
e x sin 2 x dx .
[2]
[Turn over
6
12 y P (3, 1)
C
S Q ( 8 , 0)
R (4, 0)
O
x
The diagram shows the curve C with parametric equations
1 1 x =+ 2t , y =− 2t , t > 0 . t t (i)
Find gradient of the curve at the point where t = 1.
[3]
(ii)
Show that the cartesian equation of C is x 2 − y 2 = 8.
[2]
Two points, P and Q, lie on the curve C with coordinates
( 3,1)
and
(
8, 0
)
respectively. Point R ( 4, 0 ) lie on the x-axis. The region S is bounded by the lines QR and PR and the arc PQ of the curve C. (iii) Find the exact value of the volume of revolution when S is rotated completely about the x–axis. [3]
End of Paper
Class
Adm No
Candidate Name:
2010 Preliminary Examination II Pre-University 3 MATHEMATICS Higher 2
9740/02
Friday
17 September 2010
3 hours
Additional materials: Cover Page, Writing paper List of Formulae (MF 15)
READ THESE INSTRUCTIONS FIRST Write your name, admission number and class in the spaces at the top of this page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This question paper consists of 6 printed pages. [Turn over] 1
Section A: Pure Mathematics [40 marks]
1
The rth term of a sequence is given by ur = r(3r + 1), r = 1, 2, 3, …. n
(i)
Write down the values of
∑u r =1
r
for n = 1, 2, 3, and 4.
[2]
n
(ii)
Make a conjecture for a formula for ∑ u r , giving your answer in the form r =1
nf(n), where f(n) is a function of n.
[1] n
(iii) Use the method of mathematical induction to prove your conjecture for ∑ u r . r =1
[4]
2
(i)
Solve the equation z 5 = 32 , expressing the solutions in the form re iθ , where [3] r > 0 and − π < θ ≤ π .
(ii) Show all the solutions on an Argand diagram.
[2] 5
w (iii) Show that the roots of the equation w5 − 32 − 1 = 0 are 2 kπ where k = 0, ±1, ±2 . 1 − i cot 5
3
[4]
12 9 (a) The points M and N have position vectors 6 and p respectively, where p q 3 and q are constants. (i) The straight line which passes through M and N has equation 9 1 [3] r= 6 + λ 3 where λ ∈ R . Find the values of p and q. 3 0 (ii)
The line MN is extended to point S such that N divides MS in the ratio 3:2. Find the position vector of S. [2] [Turn over]
2
1 2 (b) The planes p1 and p2 have equations r −3 = 7 and r −1 = 4 respectively, a 1 where a is a constant. Both planes pass through the point (1, b,3) and meet in the line l where b is a constant. (i) (ii)
4
(a)
Find the values of a and b. Hence find the vector equation of line, l.
[3] [3]
A conical water tank with its axis vertical and downwards has a radius of r m at the top and is
5 r m high (see diagram). If water flows into the tank 2
at a rate of 25 m3s −1 , find the rate at which the depth of water is increasing when the water is 15 m high.
[3] r
5 r 2
(b)
A chocolate maker is interested in using containers in the shape shown below to package her chocolates. The container is made up of an open cylinder of height h cm and radius r cm, with a hollow hemispherical cover of radius r cm. In order to minimise production cost in this economic recession, the chocolate maker wants to use containers with the least surface area while maintaining the volume of each container at 500 cm3. The material used to construct the container costs $0.02 per cm2. 1000 5 2 (i) Show that the surface area, S is given by [4] S = + πr . r 3 (ii) Find the radius that gives the minimum surface area. [4] (iii) Find the minimum cost of one container, leaving your answer to 2 decimal places. [2] 4 2 [Volume of sphere, V = π r 3 ; Surface area of sphere, S = 4π r ] 3 r h
3
Section B: Statistics [60 marks] 5 The Millennia Institute Library is planning to organize a campaign to promote reading among students. The librarian conducted a survey on 50 randomly chosen students to collect ideas for the campaign. However, she realized that most of these students happen to be from a certain level. (i)
Advise the librarian on a more appropriate sampling method.
[3]
(ii)
Give an advantage and disadvantage of the sampling method selected in part (i). [2]
6
7
Two players M and N regularly play each other in chess. When M has the first move in a game, the probability of M winning the game is 0.4 and the probability of N winning the game is 0.2. When N has the first move in a game, the probability of N winning the game is 0.3 and the probability of M winning the game is 0.2. Any game that is not won by either player ends in a draw. (a)
To decide who has the first move in a game, a fair coin is tossed. The player who gets heads will make the first move. Find the probability that the game ends in a draw. [2]
(b)
Find the probability that N had made the first move given that the game ends in a draw. [3]
(c)
To make the game more interesting, M and N change the procedure for deciding who has the first move in the game. As a result of the new procedure, the probability of M making the first move in a game is p. Find the value of p such that M and N have an equal chance of winning the game. [2]
Two families are invited to a party. The first family consists of a man and both his parents while the second family consists of a woman and both her parents. The two families sit at a round table with two other men and two other women. Find the number of possible arrangements if (i) there is no restriction, [1] (ii)
the men and women are seated alternately,
[2]
(iii) members of the same family are seated together and the two other women must be seated separately, [3] (iv) members of the same family are seated together and the seats are numbered. [2]
4
8
In a particular study, a medical student recorded the age and blood pressure of 8 men. The regression line of y on x is determined to= be y 1.144 x + 79.914 . Unfortunately, the student then lost the record of one of the men. He urgently needs the missing data, b in his report. Age (x) Blood Pressure (y)
49 133
58 148
44 b
75 166
37 123
57 147
63 154
66 152
(i)
Show that b is 130.
[3]
(ii)
Draw a scatter diagram for the data in the table.
[2]
(iii) Find the linear product moment correlation coefficient between y and x.
[1]
(iv) Explain why a linear product moment correlation coefficient that is close to 1 alone is not sufficient to conclude a strong linear relationship between two variables. [1] (v)
Use the given regression line to estimate the age of a man when his blood pressure is 150, giving your answer to the nearest integer.
(vi) Explain whether this choice of the regression line is appropriate
9
[1] [1]
Durians and mangoes are sold by mass. The masses, in kg, of durians and mangoes are modelled as having independent normal distributions with means and standard deviations as shown in the table.
Durians Mangoes
Mean 1.6 0.3
Standard Deviation 0.2 0.05
Durians are sold at $8 per kg and mangoes at $3 per kg. (i)
Find the mass, m that will be exceeded by 80% of the durians.
[1]
(ii)
Find the probability that the total mass of 3 randomly chosen durians and 4 randomly chosen mangoes exceeds 6.5 kg. [3]
(iii) The mean mass of n randomly chosen durians is D kg. Given that P( D < 1.45) = [3] 0.0122 , find the value of n. (iv) Find the probability that the total selling price of 3 randomly chosen durians and 4 randomly chosen mangoes is less than $45. [3]
5
10 There is an outbreak of an infection caused by a new strain of a virus in City X. The probability, p, of a randomly chosen person being infected is 0.2. Find the probability that, in a random sample of 12 people chosen from City X, at least 4 are infected. [3] After a certain time, the virus mutates and the value of p increases to 0.6. By using a suitable approximation, find the probability that, in a random sample of 100 people, there are at least 60 but not more than 80 who are infected. [4] The virus then mutates to a much more deadly form and the value of p is now 0.97. By using a suitable approximation, find the probability that, in a random sample of 100 people, at most 5 are not infected.
[3]
11 A beverage producer claims that each packet of soya bean milk he produces contains 250 ml of the drink. A consumer group took a sample of 50 packets and recorded the volume, x in ml of each packet. The results are summarized by
∑ x =12349 , ∑ x
2
=3054283 .
(i)
Find the unbiased estimates of the population mean and variance.
[2]
(ii)
Test, at 5% significance level, whether the producer is overstating the mean volume. [4]
(iii) In conducting the test in part (ii), explain if there is a need to assume that the volume of a packet of soya bean milk follows a normal distribution. [1] (iv) In conducting the test in part (ii), explain the meaning of ‘a significance level of 5%’ in the context of the question. [1] (v)
Given that the population variance of the volumes of packets of soya bean milk is 85, find the range of values of the sample size for which the producer’s claim will not be rejected at a significance level of 5%. [3]
End of Paper
6
2010 MI PU3 Prelim Exam II 9740/02 H2 Mathematics Paper 1 Suggested Solutions
1
(i) f ( r ) =
1 1 A B = = + r − 2r r ( r − 2 ) r r − 2 2
By cover-up rule, A = − 1 , B = 1 2
∴ f (r ) = −
2
1 1 + 2r 2 ( r − 2 )
2 1 1⎤ ⎡3 = lim ⎢ − − ⎥ n →∞ r =3 r − 2r ⎣ 2 n −1 n ⎦ 3 = 2 ∞
(iii)
2
∑
2
(ii) n 2 1 n ⎡ 1 1⎤ = 2 × ∑ f (r ) = 2 × ∑ ⎢ − 2 r = 3 ⎣ r − 2 r ⎥⎦ r =3 r − 2 r r =3 1 = 1− 3 1 1 + − 2 4 1 1 + − 3 5 1 1 + − 4 6 + ... 1 1 + − n−4 n−2 1 1 + − n − 3 n −1 1 1 + − n−2 n 3 1 1 = − − 2 n −1 n n
∑
2
(i)
f ( x) = e x + 10
y
g ( x) = x 2 − 3 x
11
A
y = 10
3
0
x
(3/2, -9/4) For (ii)
g ( x) = f ( x) , by GC, x ≈ −2.02 Rf = (10, ∞) ⊆ Dg = ℜ
∴ gf exists. (shown) gf ( x ) = ( e x + 10 ) − 3 ( e x + 10 ) 2
1
(iii)
From the graph of g above, g −1 exits if x ≥ Hence, the least value of a =
3 . 2
3 . 2
2
3⎞ 9 ⎛ Let y = g ( x) = ⎜ x − ⎟ − 2⎠ 4 ⎝ 3 3 9 ∴ x = + y + , since x ≥ 2 2 4
∴ g −1 ( x ) = 3
(i)
3 9 ⎡ 9 ⎞ + x + , Dg −1 = ⎢ − , ∞ ⎟ 2 4 ⎣ 4 ⎠
Let the AP with T1 = a, common difference = d. T4, T9, T19 are consecutive terms of a GP :
⇒r=
T19 T9 a + 18d a + 8d = ⇒r= = T9 T4 a + 8d a + 3d
⇒ ( a + 18d )( a + 3d ) = ( a + 8d ) ⇒ 10d 2 = 5ad 2
5a 1 = a ---------(1) 10 2 ⎛1 ⎞ a + 8⎜ a ⎟ a + 8d ⎝ 2 ⎠ = 2 (shown) ∴r = = a + 3d ⎛1 ⎞ a + 3⎜ a ⎟ ⎝2 ⎠
Since d ≠ 0, d =
(ii)
Given T20 = a + 19d = 63 ---------(2)
⎛1 ⎞ a ⎟ = 63 ⎝2 ⎠
Substitute (1) into (2), a + 19 ⎜
∴a = 6 & d = 3
(iii)
Sn =
n [ 2(6) + (n − 1)3] > 200 2
3n 2 + 9n − 400 > 0 n < −13.14 (NA) or n > 10.14 Hence, least n = 11
2
4
a
2
( x + 1)
2
−b y 2
2
( x + 1) =1⇒ ⎛1⎞ ⎜ ⎟ ⎝a⎠
Asymptotes : y = ±
2
2
−
y2 ⎛1⎞ ⎜ ⎟ ⎝b⎠
2
=1
a ( x + 1) b
Comparing with y = 2 x + 2 and y = −2 x − 2 , a = 2 ⇒ a = 2b ------(1) b
And the curve passes through the point
a
2
3 ⎛ ⎞ + 1⎟ ⎜ − 2 ⎝ ⎠
⎛ 3 ⎞ ⎜ − ,0 ⎟ ⎝ 2 ⎠
,
2
− 0 = 1
⇒ a = 2 (shown), since a is positive.
Substitute into (1), b = 1 (shown) y
y = − 2x −2
(0, √3)
(−1.5, 0)
(−0.5, 0)
y = 2x + 2
x
(0, −√3)
5
2 x 3 − 3 xy + y 3 = p ⇒ 6 x 2 − 3x
-------(1)
dy dy − 3y + 3y2 =0 dx dx dy y − 2 x 2 ⇒ = 2 dx y −x
If tangent is parallel to the y-axis, y 2 = x . Substitute into (1),
2 ( y 2 ) − 3 ( y 2 ) y + y3 − p = 0 3
⇒ 2 y 6 − 2 y 3 − p = 0 (shown) For the point of contact of the tangent with C,
3
b 2 − 4ac = ( −2 ) − 4 ( 2 )( − p ) ≥ 0 2
1 ∴ p ≥ − (shown) 2 When x = 4 is a tangent to C, y 2 = 4 ⇒ y = ±2 When y = 2, p = 112 . When y = −2, p = 144 If the tangent is parallel to the x-axis, y = 2 x 2 . Substitute into (1),
2 x3 − 3x ( 2 x 2 ) + ( 2 x 2 ) − p = 0 ⇒ 8 x 6 − 4 x3 − p = 0 3
For the point of contact of the tangent with C,
b 2 − 4ac = ( −4 ) − 4 ( 8 )( − p ) ≥ 0 2
1 ∴ p ≥ − (shown) 2 6
dx = kx ( N − x ) , k is a positive constant. dt 1
∫ x( N − x) dx = ∫ kdt 1 ⎡1 1 ⎤ + dx = kt + C ∫ ⎢ N ⎣ x N − x ⎥⎦ ln x − ln( N − x) = Nkt + NC ⎛ x ⎞ ln ⎜ ⎟ = Nkt + NC ⎝N−x⎠ x = Ae Nkt , A = e NC N−x When t = 0, x = 1,
1 =A N −1
x 1 Nkt = e N − x N −1
⇒ x ( N − 1 + e Nkt ) = Ne Nkt
∴x=
Ne Nkt N − 1 + e Nkt (shown)
4
7
(i)
f ( x ) = ln ( 2 x + 1)
⎫ ⎪ ⎪⎪ −2 −2 f ′′( x) = −2 ( 2 x + 1) (2) = −4 ( 2 x + 1) ⎬ −3 −3 ⎪ f ′′′( x) = −4(−2) ( 2 x + 1) (2) = 16 ( 2 x + 1) ⎪ ⎪⎭ f ′( x) =
2 −1 = 2 ( 2 x + 1) 2x +1
f (0) = 0 ⎫ f ′(0) = 2 ⎪⎪ ⎬ f ′′(0) = −4 ⎪ f ′′′(0) = 16 ⎪⎭
1 2 1 x f ′′(0) + x3 f ′′′(0) + ... 2 3! 1 2 1 3 = 0 + 2 x + x (−4) + x (16) + ... 2 6 8 ∴ f ( x) = 2 x − 2 x 2 + x3 + ... 3 f ( x)
= f (0) + xf ′(0) +
(ii)
ln ( 2 x 2 + 3 x + 1) = ln(2 x + 1)( x + 1) = ln(2 x + 1) + ln( x + 1) 8 1 1 = 2 x − 2 x 2 + x 3 + x − x 2 + x 3 + ... 3 2 3 5 = 3 x − x 2 + 3 x 3 + ... 2
⇒ a = 3, b = −
5 2
8
(i) ⎫ A ( −2, 0 ) → ( −4, 0 ) → A′ ( −4,1) ⎪ ⎪ B ( 0, 2 ) → ( 0, 2 ) → B′ ( 0,3) ⎬ ⎪ 1 Asymptote : x = − → x = −1 ⎪ 2 ⎭
5
y ⎛x⎞ y = f ⎜ ⎟ +1 ⎝2⎠
B′ A′ x
x = −1 (ii)
A ( −2, 0 ) → asymptote at x = −2
⎫ ⎪ 1 1 ⎪ Asymptote at x = − → x − intercept at x = − 2 2 ⎪ ⎬ ⎛ 1⎞ ⎪ B ( 0, 2 ) , min. pt. → ⎜ 0, ⎟ , max. pt. ⎪ ⎝ 2⎠ ⎪ "New" asymptote : y = 0 ⎭ y
(0, ½)
(- ½, 0)
0
x
x=-2
6
9
r ⋅n = d ------ (1) r = a + λb ------(2)
Substitute (2) into (1),
( a + λb ) ⋅ n = d ⇒ a ⋅ n + λb ⋅ n = d ∴λ =
d − a⋅n b ⋅n
(Shown)
(i)
If l is in π , a is on π , a ⋅ n = d . b . n = 0
Hence, λ has infinitely many solutions.
(ii)
When l and π intersect at one point, λ =
d − a ⋅n , from above. b ⋅n
Substitute this into the equation of the line l, the position vector
⎛ d − a⋅n ⎞ ⎟b ⎝ b ⋅n ⎠
required is r = a + ⎜ 10
(a) Given that z = 1 + ki is a root, so substitute into the given equation
(1 + ki ) − (1 + ki ) 4
3
− 9 (1 + ki ) + 29 (1 + ki ) − 60 = 0 2
1 + 4ki − 6k 2 − 4k 3i + k 4 − (1 + 3ki − 3k 2 − k 3i ) − ( 9 + 18ki − 9k 2 ) + 29 + 29ki − 60 = 0
Comparing the real or imaginary parts on both sides, 4k − 4k 3 − 3k + k 3 − 18k + 29k = −3k 3 + 12k = 0 ⇒ k = ±2 or k = 0 (N.A.) OR,
−6k 2 + k 4 + 3k 2 − 9 + 9k 2 − 31 = k 4 + 6k 2 − 40 = 0 By GC,
⇒
k = ±2
Hence, ( z − (1 − 2i ) ) ( z − (1 + 2i ) ) = z 2 − 2 z + 5 is a factor of the given equation z 4 − z 3 − 9 z 2 + 29 z − 60 = ( z 2 − 2 z + 5 )( z 2 + z − 12 ) = 0
(z
2
− 2 z + 5 ) = 0 or
(z
2
+ z − 12 ) = 0
∴ z = 1 ± 2i , z = −4 or z = 3
7
(b) (i)
arg ( z − 2 − 2i ) =
3π 4
arg ( z − 2 − 2i ) =
y A
B
0
4
D α C α
2
π
Note : α = π/4
x
2
(ii) min. (arg z) = argument of any complex numbers along CD
= tan −1
=
π
2 2
4
At A, x = 2 − 2 and y = 2 + 2
. -----(1)
max. (arg z) = argument of a,
where A ≡ a
AB -----(2) OB ⎛ 2+ 2 ⎞ = tan −1 ⎜⎜ ⎟⎟ ⎝ 2− 2 ⎠ = tan −1
Hence,
11
⎛ 2+ 2 ⎞ ≤ arg ( z ) ≤ tan −1 ⎜⎜ ⎟⎟ (Shown) 4 ⎝ 2− 2 ⎠
π
Let x = c o s θ ⇒
(a)
, x = 0 . When θ = π , 2 −1 −1 sin θ θ = d dx 2 0 1+ x 1 + cos 2 θ
When θ =
∫
π π /2
π
dx = −s i nθ dθ
x=−1.
∫
−1
= − ⎡⎣ t a n − 1 x ⎤⎦ 0
=
π
4
8
(b)(i)
∫e
x
sin 2 x dx = e x sin 2 x − ∫ e x .2 cos 2x dx = e x sin 2 x − 2 ⎡ e x cos 2 x + ∫ e x .2sin 2 x dx ⎤ ⎣ ⎦
5∫ e x sin 2 x dx = e x sin 2 x − 2e x cos 2 x + C 1 ∴ ∫ e x sin 2 x dx = e x [sin 2 x − 2 cos 2 x ] + C 5 (b)(ii)
5∫
π /2
0
12
(i)
(e
x
π /2 1 sin 2 x ) dx = 5 × ⎡⎣e x ( sin 2 x − 2cos 2 x ) ⎤⎦ 0 5 = 2 ( eπ /2 + 1)
dy 1 dx 1 = 2+ , = 2− 2 dt dt t t2 1 2+ 2 dy dy dx t = ÷ = dx dt dt 2 − 1 t2 d y 2 +1 When t = 1, = =3 2 −1
dx
(ii)
1 1 x = 2t + ------ (1) ; y = 2t − ------ ( 2) t t (1) + (2), x + y = 4t ------ ( 3)
(1) – (2), x − y = 2
------ ( 4 ) t ( x + y )( x − y ) = 8
(3) × (4),
⇒ x 2 − y 2 = 8 (Shown) 3
(iii) Volume of the solid = π ∫ y 2 dx + volume of cone
= π
8 3
∫
8
( x2 − 8) dx + 13 π (12 ) (1)
= 4 ( 4 8 − 1 1) π or 3
(
)
4 8 2 − 1 1 π (exact value) 3
9
2010 MI PU3 Prelim Exam II 9740/02 H2 Mathematics Paper 2 Suggested Solutions
1
(i) 1
∑u r =1
2
3
4
r =1
r =1
r =1
r
= 1× 22 = 4, ∑ ur = 2 × 32 = 18, ∑ ur = 3 × 42 = 48, ∑ ur = 4 × 52 = 100
r
= n(n + 1) 2
(ii) n
∑u r =1
(iii) n
Let Pn be the statement that
∑u r =1
r
= n(n + 1) 2 for n ∈ Z +
Prove P1 is true: LHS = (1)(3 ×1 + 1) = 4 RHS = (1)(1 + 1) 2 = 4 P1 is true Assume Pk is true for some k ∈ Z + i.e. k
∑u r =1
r
= k (k + 1) 2
k +1
Prove Pk +1 is true i.e.
∑u r =1
r
= (k + 1)(k + 2) 2
LHS k
= ∑ ur + (k + 1)(3k + 3 + 1) r =1
= k (k + 1) 2 + (k + 1)(3k + 4) = ( k + 1) ( k 2 + k + 3k + 4 ) = ( k + 1) ( k 2 + 4k + 4 ) = ( k + 1)( k + 2 )
2
= RHS Since P1 is true and Pk is true ⇒ Pk +1 is true, by the Principle of Mathematical Induction, Pn is true for n ∈ Z +
1
2
(i)
z 5 = 32 = 32ei (2 kπ ) ⎛ 2 kπ ⎞ i⎜ ⎟ 5 ⎠
z = 2e ⎝
where k = ±2, ±1, 0 Im (z) Z2
(ii) Z3
2π 5
2π 5
2π 5 −
2π 5
Z4
Z1
Re (z)
2π 5
Z5
(iii) 5
⎛w ⎞ w − 32 ⎜ − 1⎟ = 0 ⎝2 ⎠ 5
5
⎛ ⎞ ⎜ w ⎟ ⎜ w ⎟ = 32 ⎜ −1 ⎟ ⎝2 ⎠ ⎛ 2 kπ ⎞ ⎟ 5 ⎠
i⎜ w = 2e ⎝ w −1 2
w = we w=
− 2e
⎛ 2 kπ ⎞ i⎜ ⎟ ⎝ 5 ⎠
⎛ 2 kπ ⎞ i⎜ ⎟ 5 ⎠
2e ⎝ e
⎛ 2 kπ ⎞ i⎜ ⎟ ⎝ 5 ⎠
−1 2e
w= e
w=
⎛ 2 kπ ⎞ i⎜ ⎟ ⎝ 5 ⎠
⎛ kπ ⎞ i⎜ ⎟ ⎝ 5 ⎠
⎛ 2 kπ ⎞ i⎜ ⎟ ⎝ 5 ⎠
⎛ i⎛⎜ k5π ⎞⎟ i⎛⎜ − k5π ⎞⎟ ⎞ ⎜e ⎝ ⎠ −e ⎝ ⎠ ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ kπ ⎞ i⎜ ⎟ ⎝ 5 ⎠
2e ⎛ i⎛⎜ k5π ⎞⎟ ⎞ 2 Im ⎜ e ⎝ ⎠ ⎟ ⎜ ⎟ ⎝ ⎠
2
⎛ kπ ⎞ ⎛ kπ ⎞ cos ⎜ ⎟ + i sin ⎜ ⎟ 5 ⎠ 5 ⎠ ⎝ ⎝ w= ⎛ kπ ⎞ i sin ⎜ ⎟ ⎝ 5 ⎠ ⎛ kπ ⎞ cot ⎜ ⎟ ⎝ 5 ⎠ w = 1+ i ⎛ kπ ⎞ w = 1 − i cot ⎜ ⎟ (shown) ⎝ 5 ⎠
3
(a)(i)
⎛12 ⎞ ⎛ 9 ⎞ ⎛1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ p ⎟ − ⎜6⎟ = λ ⎜3⎟ ⎜q ⎟ ⎜3⎟ ⎜ 0⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 3 ⎞ ⎛1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ p − 6⎟ = λ ⎜3⎟ ⎜q −3 ⎟ ⎜0⎟ ⎝ ⎠ ⎝ ⎠ λ =3 q=3
p − 6 = (3)(3)
p = 15
(a)(ii)
uuuur uuur uuur 2OM + 3OP ON = 5 uuur 2 uuuur 5 uuur OP = − OM + ON 3 3 ⎛9⎞ ⎛12 ⎞ ⎛ 14 ⎞ uuur 2⎜ ⎟ 5⎜ ⎟ ⎜ ⎟ OP = − ⎜ 6 ⎟ + ⎜ 15 ⎟ = ⎜ 21⎟ 3⎜ ⎟ 3⎜ ⎟ ⎜ ⎟ ⎝ 3⎠ ⎝3⎠ ⎝3⎠
(b)(i)
⎛ 1 ⎞ ⎛1 ⎞ ⎜ ⎟⎜ ⎟ ⎜ 1 ⎟ ⎜ −3 ⎟ = 7 ⎜ 3⎟ ⎜ a ⎟ ⎝ ⎠⎝ ⎠ 1 − 3 + 3a = 7
Point (1, b,3) is on both planes: ⎛1⎞ ⎛2 ⎞ ⎜ ⎟⎜ ⎟ ⎜ b ⎟ ⎜ −1 ⎟ = 4 ⎜ 3 ⎟ ⎜1 ⎟ ⎝ ⎠⎝ ⎠ 2−b+3= 4 b =1
a=3
(b)(ii)
3
Line l is perpendicular to the normals to both planes:
⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ 0⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −3 ⎟ × ⎜ −1⎟ = ⎜ 5 ⎟ ⎜ 3 ⎟ ⎜1 ⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛1⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ l : r = ⎜ 1 ⎟ + α ⎜ 1 ⎟ where α ∈ ⎜ 3⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠
4
⎛1⎞ ⎛ 0⎞ ⎜ ⎟ ⎜ ⎟ or r = ⎜ 1 ⎟ + α ⎜ 5 ⎟ where α ∈ ⎜ 3⎟ ⎜ 5⎟ ⎝ ⎠ ⎝ ⎠
(a) 2
1 1 ⎛2 ⎞ 4 V = π r 2 h = π ⎜ h ⎟ h = π h3 3 3 ⎝5 ⎠ 75 dV 4 = π h2 dh 25 dV dV dh = dt dh dt At h = 15m: dh 4 25= π (15) 2 dt 25 dh 25 = ≈ 0.221 ms -1 dt 36π (b)(i)
2 V = π r 2 h + π r 3 = 500 3 1500 − 2π r 3 h= 3π r 2 500 2 h= 2 − r πr 3
S = 2π rh + π r 2 + 2π r 2 ⎛ 500 2 ⎞ = 2π r ⎜ 2 − r ⎟ + 3π r 2 3 ⎠ ⎝πr 1000 4 2 = − π r + 3π r 2 3 r 1000 5 2 = + πr 3 r
4
(b)(ii)
1000 5 2 + πr 3 r 1000 10 dS = − 2 + πr 3 dr r
S=
For minimum surface area:
1000 10 dS = − 2 + πr = 0 3 dr r
1000 10 = πr 3 r2 300 r3 =
π
r = 4.570781497 r ≈ 4.57 cm d 2 S 1000 10 = 3 + π > 0 (verified r = 4.57 gives a minimum surface area) dr 2 r 3 (b)(iii)
1000 5 + π (4.570781497) 2 = 328.1714518 4.570781497 3 Costminimum = (328.1714518)(0.02) ≈ $6.56 S min =
5
(i)
Use stratified sampling. Divide the student population into different strata for example PU1, PU2 and PU3 (or any other logical strata) and select students randomly from each strata according to the strata’s proportion to the population until 50 students have been selected.
(ii)
Advantage: Good representation of student population. Disadvantage: Time-consuming
6
Win
0.4 0.4 0.5
M starts first
Draw 0.2 Lose 0.2
0.5
Win 0.5
M starts second
Draw 0.3 Lose
(a) P (game ends in draw) = (0.5)(0.4) + (0.5)(0.5) = 0.45
5
(b) P ( N made the 1st move game ends in draw ) P ( N made the 1st move ∩ game ends in draw) P(game ends in draw) 0.5 × 0.5 = 0.45 = 0.556 =
(c)
P ( M wins game) = P ( M loses game)
0.4 p + 0.2(1 − p) = 0.2 p + 0.3(1 − p) 1 p = ≈ 0.333 3 7
(i)
Number of arrangements = (10 − 1)! = 362880
(ii)
Number of arrangements = (5 − 1)!× 5! = 2880
(iii) Number of arrangements = (4 − 1)!× 3!× 3!× 4 P2 = 2592 (iv) Number of arrangements = (6 − 1)!× 3!× 3!× 10 = 43200
8
(i)
x=
∑ x = 449 = 56.125
8 n y = 1.144(56.125) + 79.914 = 144.121 b + 1023 = 144.121 8 b ≈ 130 (ii)
6
(iii)
Using GC, r = 0.9888 ≈ 0.989
(iv)
A r value that is close to 1 alone does not lead to the conclusion that there is a strong linear relationship between 2 variables because a r value close to 1 could be a result of outliers.
(v)
150 = 1.144 x + 79.914 x ≈ 61 . The age of the man is 61.
(vi)
9
Age is the independent variable (can be precisely controlled) and the estimate from the regression line of y on x is appropriate. (Also accept r is close to 1, using any of the 2 regression lines will result in a close estimate. Thus, the estimate is appropriate).
Let D be the mass of a durian D ~ N (1.6, 0.22 ) Let M be the mass of a durian M ~ N (0.3, 0.052 )
(i) Let P( D > m) = 0.8 Using GC, m = 1.43167 ≈ 1.43 kg.
(ii) D(3) + M (4) ~ N (3 × 1.6 + 4 × 0.3, 3 × 0.22 + 4 × 0.052 ) = N (6, 0.13)
P ( D(3) + M (4) > 6.5) = 0.08275892 ≈ 0.0828
⎛ 0.22 ⎞ (iii) D ~ N ⎜ 1.6, ⎟ n ⎠ ⎝ P ( D < 1.45) = 0.0122
⎛ ⎞ ⎜ 1.45 − 1.6 ⎟ P⎜Z < ⎟ = 0.0122 0.2 ⎟ ⎜ ⎜ ⎟ n ⎠ ⎝ 1.45 − 1.6 = −2.2507717 0.2 n n≈9
7
(iii) 8 D(3) + 3M (4) ~ N (8 × 4.8 + 3 ×1.2, 82 × 3 × 0.22 + 32 × 4 × 0.052 ) = N (42, 7.77) P ( 8 D(3) + 3M (4) < 45 ) ≈ 0.859
10
Let X be the number of people infected with the disease X
B(12, 0.2)
P ( X ≥ 4) = 1 − P( X ≤ 3) = 1 − 0.794568 ≈ 0.205
Let X be the number of people infected with the disease X B(100, 0.6) np = 60 > 5; n(1 − p) = 40 > 5
X
N (60, 24) approximately
P (60 ≤ X ≤ 80) cc
= P(59.5 ≤ X ≤ 80.5) = 0.541
Let Y be the number of people NOT infected with the disease Y
B(100, 0.03)
np = 3 < 5; n > 50, p < 0.1, Y
Po(3) approximately
P (Y ≤ 5) = 0.916
11
(i) Unbiased estimates of the population mean , X = Unbiased estimates of the population variance,
8
∑X n
=
12349 = 246.98 ≈ 247 50
2 ⎛ ∑ ( x) ⎞ ⎤ 1 ⎡ 2 ⎢∑ ( x ) − n ⎜ ⎥ s = ⎜ n ⎟⎟ ⎥ = 88.30571429 ≈ 88.3 n −1 ⎢ ⎝ ⎠ ⎣ ⎦ 2
(ii) H 0 : μ = 250
H1 : μ < 250 Use z-test, p-value = 0.01153 (z = -2.272) p-value = 0.01153 < 0.05, reject H 0 At 5% significance level, the producer is overstating the mean volume.
(iii) n is large (n = 50). By Central Limit Theorem, x-bar will
approximately follow a normal distribution. The z-test can still be conducted. There is no need to assume that the volume of a packet of soya bean follow a normal distribution. (iv) The significance level of 5% means that 0.05 is the probability that
we conclude the producer has overstated the mean volume when the mean volume is actually 250 ml. (v) σ 2 = 85 (given) Level of significance = 0.05. Use z-test. Do not reject H 0 if z =
246.98 − 250 > −1.6448536 85 / n
n < 25.215 n ≤ 25
9
2 1
Given that
r −1 r 2 , f (r ) = − + r ! ( r − 1) ! ( r − 2 ) ! use the method of differences to prove that N
∑ f (r )=
2−
r =2
N +1 . N!
[3]
Hence, give a reason why the series is convergent and state the sum to infinity. [2]
2
3
( )
d x2 +1 . e dx
[1]
(i)
Find
(ii)
Hence, find
(iii)
Find the exact value of
∫x e
3 x 2 +1
dx .
[3]
∫
1 0
(x e
3 x 2 +1
)
+ e 2 dx .
[2]
A sequence of positive real numbers x1 , x2 , x3 ,... satisfies the recurrence relation xn +1 =
3 xn 2 + xn
for n ≥ 1. (i)
Given that as n → ∞, xn → α, find the exact value of α .
(ii)
Show that this sequence is
[2]
(a)
strictly increasing for 0 < xn < α,
[2]
(b)
strictly decreasing for xn > α .
[2]
[Turn over MJC/2010 JC2 Preliminary Examination/9740/01
3
4
(i)
Express f ( x ) =
x−4 in partial fractions. ( x + 1)( 3x + 2 )
Hence, expand f ( x ) in ascending powers of x, up to and including the
5
term in x3 .
[4]
(ii)
State the range of values of x for which this expansion is valid.
[1]
(iii)
Find the coefficient of x n in this expansion.
[2]
On a single Argand diagram, sketch the following loci. (i)
z − 5 = 3 + 7i ,
[1]
(ii)
z − 6 − 3i =
[1]
z − 4 + 3i .
Two complex numbers that satisfy the above equations are represented by p and q, where Re( p ) < Re(q ) . By using the cartesian equations of the loci, find p and q. Hence, determine the value of arg( p − q ) .
6
[5]
Newton’s Law of Cooling states that the rate at which the temperature of a body falls is proportional to the amount by which its temperature exceeds that of its surroundings. At time t minutes after cooling commences, the temperature of the body is θ o C . Assuming that the room temperature remains constant at 30 o C and the body has an initial temperature of 90 o C , show that θ= 30 + 60e − kt , where k is an arbitrary constant.
[5]
Given that it takes 8 minutes for the temperature of the body to drop from 90 o C to 55 o C , determine how much more time is needed for the body to cool to 35 o C , leaving your answer to one decimal place.
MJC/2010 JC2 Preliminary Examination/9740/01
[3]
4
7
(a)
1 A geometric progression has first term a and common ratio − . The first 2 two terms of the geometric progression are the first and fourth terms respectively of an arithmetic progression. Find the sum of the first n evennumbered terms of the arithmetic progression in terms of a and n.
(b)
[4]
A customer borrows $50000 from a bank at the beginning of a month. In the middle of the month, the customer pays $x to the bank. On the last day of every month, the bank adds interest at a rate of 3.5% of the remaining amount owed after payment has been made. This repayment process is repeated every month until the loan is repaid in full. (i)
Find, in terms of x, the amount owed at the beginning of the third month.
(ii)
[1]
Show that, if the repayment of the loan is completed upon the nth
1750(1.035n −1 ) payment, then x ≥ . 1.035n − 1
8
[4]
The diagram shows the graph of y = f ( x ) , where f ( x ) is a cubic polynomial and C is a maximum point. y
A ( −2, 75 )
C (1, 12 ) B ( 0, 3) 0
x
[Turn over MJC/2010 JC2 Preliminary Examination/9740/01
5 It is also given that when y = 3 , x = 0, α or β , where α < 0 < β . On separate diagrams, sketch the graphs of (i)
= y f (1 − 2 x ) ,
[3]
(ii)
= y2 f ( x) − 3 .
[3]
Determine the equation of the curve.
9
[4]
A line l passes through the points A and B with coordinates ( 0, −1, 2 ) and (1, 0,1) respectively. (i)
Find the angle between OA and the line l.
[2]
(ii)
Hence, find the shortest distance from the origin to the line l.
[1]
A plane π 1 has equation r ( i + 2 j + 3k ) = 4. (iii)
Show that the line l lies on the plane π 1.
[2]
A second plane π 2 contains the line l and is perpendicular to the plane π 1. (iv)
Find a vector equation of π 2 .
[2]
A third plane π 3 is perpendicular to both the planes π 1 and π 2 , and is at a perpendicular distance of (v)
3 units from the origin.
Find possible vector equations of π 3 .
MJC/2010 JC2 Preliminary Examination/9740/01
[3]
6 10
The functions f, g and h are defined by x∈ ,
f : x cos x
for
1 g : x ln 1+ x
for − 1 < x ≤ 1,
h:x
1 ln for −1 < x < 1, 1+ x −1 x= for 1.
(i)
Show that the composite function fg exists.
(ii)
Find the series expansion for fg( x) up to and including the term in x3 . [4]
(iii)
Write down the value of h −1 (−1) . Hence, find h −1 ( x) , expressing your
[2]
answer in the form
h −1 : x
p( x) for c for
x > a, x = −1,
where a and c are constants, and p is a function of x.
11
The complex numbers z and w are given by = z
[4]
3 + i and w =−1 + i 3 .
(a)
Find the set of values of the positive integer n for which z n − ( z * ) = 0 . [4]
(b)
Sketch an Argand diagram, with origin O, showing the points Z, W and P
n
representing the complex numbers z, w and z + w respectively. Show that OWPZ is a square. By considering the argument of z + w , deduce that tan
[3] 5π = 2+ 3 . 12
[3]
[Turn over MJC/2010 JC2 Preliminary Examination/9740/01
7 2 x 2 − 3x + 1 . x2 − 4
12
The curve C has equation y =
(i)
Prove, using an algebraic method, that C cannot lie between 7−3 5 7+3 5 . and 8 8
(ii)
Sketch the curve C, showing clearly all asymptotes, axial intercepts and turning points.
(iii)
[3]
[4]
R is the point on C where x = −5 . The tangent and normal to the curve at R cut the y-axis at P and Q respectively. Show, to 3 significant figures, that the coordinates of P and Q are ( 0,5.15 ) and ( 0, −9.31) respectively. Hence, determine the area of triangle PQR.
MJC/2010 JC2 Preliminary Examination/9740/01
[5]
Section A: Pure Mathematics [40 marks] x+7 ≤ 1. 4 + 3x − x 2
1
Without using a calculator, solve the inequality
[4]
2
Three non-zero and non-parallel vectors p, q and r are such that p × q = 3p × r . λp , where λ is a scalar. Show that q − 3r = [2] 5 It is also given that p is a unit vector, q = 5 , r = 2 and the angle between q and r is cos −1 . 6 By considering ( q − 3r ) ⋅ ( q − 3r ) , find the exact values of λ . [4]
∑ 3( 2 n
3
Prove by induction
r =1
r
3 r ) 3 ( 2n +1 ) − n ( n + 1) − 6 for n ∈ + . −= 2
∑ 3( 2 n
Hence, find the least value of n for
r
r =5
4
(a)
[5]
− r ) > 5800 .
[3]
The variables x and y are related by = y
Find the value of
x + ln ( xy )
3
dy when both x and y are equal to 1. dx
[4]
(b)
45
Water is poured at a constant rate of 1 cm3s -1 into a cone of semi-vertical angle 45 with its axis vertical and vertex downwards (see diagram). At the beginning, the cone is partially filled with 30 cm3 of water. Find the rate at which the depth of water is increasing after 2 minutes.
[6]
[Turn Over
3
5
(i)
y x x2 y 2 The region R is bounded by the ellipse 2 + 2 = 1 as shown in 1 and the line − = b a a b
the diagram below. y b
y x − = 1 b a
O
a
R
−a
−b
x
x2 y 2 + = 1 a 2 b2
0 b 1 Show that the area of R is − ab + ∫ a 2 − x 2 dx . Hence, by substituting x = a sin θ , − a a 2
find the exact area of R in terms of a and b.
(ii)
[8]
Find the volume of revolution formed when R is rotated completely about the y-axis. Give your answer in the form ka 2b , where k is a constant to be determined.
[4]
Section B: Statistics [60 marks] 6
There are 800 students in PPQZ Secondary School where 65% of the students are boys. A random sample of 200 students is selected to find out their preferred canteen stalls. (a)
Student A suggests selecting the sample by taking the first 120 boys and the first 80 girls who visit the canteen during recess time. Identify the sampling method used. State, in the context of the question, one disadvantage of this method used. [2]
(b)
Student B suggests using systematic sampling using the following steps: (1) Label all the students from 1 to 800. (2) Using a sampling interval of 5, he randomly selects an integer from 1 to 5 to determine the first student to be chosen and selects every 5th student thereafter until 200 students are chosen. Identify the mistake that student B made and write down the correct step.
MJC/2010 JC2 Preliminary Examination/9740/02
[2]
7
In a computer game played by a single player, the player has to find, within a fixed time, the path through a maze shown on the computer screen. On the first occasion that a particular player plays the game, the computer shows a simple maze, and the probability that the player 5 succeeds in finding the path in the time allowed is . On subsequent occasions, the maze 7 shown depends on the result of the previous game. If the player succeeded on the previous occasion, the next maze is harder, and the probability that the player succeeds is one third of the probability of success on the previous occasion. If the player failed on the previous occasion, another simple maze is shown and the probability of the player succeeding is 5 again . 7 The player plays three games. Find the probability that
8
(i)
the player succeeds in exactly one of the games,
[2]
(ii)
the player succeeds in at least one of the games,
[2]
(iii)
the player has exactly two failures given that the player has at least one success.
[2]
(a)
Find the number of teams of 8 players that can be selected from a group of 13 players if at least two of the three tallest players and at most one of the two shortest players are to be included. [4]
(b)
Secret codes can be sent using 6 available letters X, Y, Y, Z, Z, Z. (i) How many different 5-letter secret codes can be sent?
[3]
(ii) How many different 5-letter secret codes will begin and end with Y?
[1]
[Turn Over
5 9
The time t minutes spent queuing at an ice-cream bar by each customer in a random sample of 50 customers was measured and it was found that
∑ t = 213.5 and ∑ ( t − t )
2
= 44.105 where
t is the mean time spent at the ice-cream bar by each customer in the sample.
Find unbiased estimates of the population mean and variance.
[2]
Test, at 1% level of significance, whether the mean time spent is less than 4.5 minutes.
[4]
Determine the smallest level of significance of the test where the null hypothesis is rejected. [1] State, giving a reason, whether any assumptions about the population are needed in order for the test to be valid. [1]
10
On average, a hospital and a police station receive 36 and 15 calls respectively in a three-week period. Calls are received at random times. The number of calls received by the hospital may be assumed to be independent of the number of calls received by the police station. (i)
Find the probability that the hospital and the police station receive a total of more than 11 calls in a randomly chosen week. [3]
(ii)
Using a suitable approximation, find the probability that out of 50 randomly chosen weeks, the number of weeks in which the hospital and police station receive a total of more than 11 calls is greater than 45. [4]
(iii)
A nurse records the number of calls received by the hospital per week for 100 randomly chosen weeks. She then calculates the average number of calls received per week based on her data. Another nurse independently goes through the same procedure. Find the probability that the sum of the two averages obtained by the nurses is at most 23. [3]
MJC/2010 JC2 Preliminary Examination/9740/02
11
(a)
King Crabs and Snow Crabs are sold by weight. The masses, in kg, of King Crabs and Snow Crabs are modelled as having normal distributions with means and variances as shown in the table. King Crabs Snow Crabs
Mean Mass 1.65 1.10
Variance 0.71 0.34
King Crabs and Snow Crabs are sold at $40 and $45 per kg respectively.
(b)
(i)
Find the probability that the average mass of 3 King Crabs and 2 Snow Crabs is less than 1.5 kg. [3]
(ii)
Find the probability that the total selling price of a King Crab and a Snow Crab exceeds $140. [3]
(iii)
State an assumption for your working in (i) and (ii) to be valid.
[1]
The masses of bars of chocolate are normally distributed with mean µ kg and standard deviation σ kg. It is known that 20% of the bars of chocolate have masses which differ from µ kg by at most m kg. Find the probability that a randomly chosen bar of chocolate has a mass which exceeds µ kg by at least 3m kg. [5]
[Turn Over
7 12
The table below shows the number of monthly new car licences, x, issued by the government and the selling price of a car, $y in year 2009. A student wants to investigate how the selling price of a car depends on the number of monthly new car licences issued by the government.
Month
Jan
Feb
Mar
Apr
May
June
July
Aug
Sep
Oct
Nov
Dec
x
1432
1339
1664
1774
2055
2275
2221
2173
1749
1854
1360
2012
y
42000
48000
37000
36000
32850
31800
31800
32050
36000
34500
46000
33000
(i)
Plot a scatter diagram for the data and explain, in the context of the question, if a linear model is appropriate in the long run. [3]
(ii)
State, with a reason, which of the following model is most appropriate to fit the data points.
y ae x + b (A)= y ax 2 + b (B)= (C) y= a +
b x
[2]
(iii)
For the model chosen in (ii), calculate the product moment correlation and comment on its value. [2]
(iv)
Use an appropriate regression line to estimate the selling price of the car when the monthly number of new car licences issued is (a) 1300 and (b) 2000. Comment on the reliability of your answers. [4]
(v)
The student concluded that the decrease in the number of new car licences issued causes the selling price of the car to rise. State, with a reason, whether you agree with this conclusion. [1]
MJC/2010 JC2 Preliminary Examination/9740/02
H2 MATHS (9740) JC2 PRELIMINARY EXAM 2010 PAPER 1 SUGGESTED SOLUTIONS Qn 1
Solution Method of Differences N N ⎛ r −1 r 2 ⎞ − + f (r ) = ∑ ⎜⎜ ⎟ ∑ ( r − 1)! ( r − 2 )! ⎟⎠ r =2 r =2 ⎝ r ! ⎡1 2 2 =⎢ − + ⎣ 2! 1! 0! 2 3 2 + − + 3! 2! 1! 3 4 2 + − + 4! 3! 2! 4 5 2 + − + 5! 4! 3! M + +
N −3 N −2 2 − + ( N − 2 ) ! ( N − 3) ! ( N − 4 ) !
+
N −2 N −1 2 − + ( N − 1)! ( N − 2 )! ( N − 3)!
+
N −1 N 2 ⎤ − + ⎥ N ! ( N − 1) ! ( N − 2 ) !⎦
= −2 + 2 + 2 +
N −2 N N −1 − + ( N − 1)! ( N − 1)! N !
N ( N − 2 ) − N 2 + ( N − 1) N! N +1 = 2− N! = 2+
As N → ∞,
N +1 → 0 , hence the series is convergent. N! ∞
∑ f (r ) = 2 r =2
Qn 2 (i)
(ii)
Solution Integration (by part)
( )
2 d x2 +1 = 2 xe x +1 e dx
∫ x e dx 1 = ∫ ( 2 xe ) ( x ) dx 2 3 x 2 +1
x 2 +1
2
(
)
2 1 2 x2 +1 x e − ∫ 2 xe x +1dx 2 2 1 2 x2 +1 x2 +1 e x +1 2 = x e −e +c = ( x − 1) + c 2 2
=
(
(iiii)
∫
1 0
(x e
)
3 x 2 +1
)
+ e 2 dx 1
2 2 ⎤ ⎡1 = ⎢ ⎡ x 2 e x +1 − e x +1 ⎤ + e 2 x ⎥ ⎦ ⎣ ⎦0 ⎣2 1 1 ⎡ ⎤ ⎡ ⎤ = ⎢ ( e 2 − e 2 ) + e 2 ⎥ − ⎢ ( −e ) ⎥ 2 2 ⎦ ⎣ ⎦ ⎣ = e 2 + 0.5e 0
Qn n 3 (i))
Solution Recurrrence Relatioon As n → ∞, xn → α & xn +1 → α
⇒α=
3α 9α 2 ⇒ α2 = 2+α 2+α
⇒ α 2 ( 2 + α ) − 9α 2 = 0 ⇒ α 2 ( α − 7 ) = 0 ⇒ α = 0 or α = 7. Thus α = 7 since α > 0. (ii))
od 1 [Graph hical] Metho 3xn xn +1 − xn = − xn 2 + xn Sketch h y = xn +1 − xn =
3xn 2 + xn
− xn
α
From the t graph: For 0 < xn < α : xn +1 − xn > 0 ⇒ xn +1 > xn Thereffore the sequuence is stricctly increasinng. For xn > α : xn +1 − xn < 0 ⇒ xn +1 < xn Thereffore the sequuence is stricctly decreasing. [Shownn]
Metho od 2 [Graph hical] 3x Sketch hy=x &y= 2+ x
xn xn +1
α
xn +1 xn
From the t graph: For 0 < xn < α : xn +1 > xn Thereffore the sequuence is stricctly increasinng. For xn > α : xn +1 < xn Thereffore the sequuence is stricctly decreasing. [Shown] od 3 [Algebrraic] Metho 3x − x 2 + xn 3 xn xn +1 − xn = − xn = n n 2 + xn 2 + xn =
(
xn 3 − 2 + xn
)
2 + xn
For 0 < xn < α :
2 + xn > 0, xn > 0 & 3 − 2 + xn > 0 sincee 2 + xn < 9 = 3 Thus: xn +1 − xn > 0 ⇒ xn +1 > xn Thereffore the sequuence is stricctly increasinng. For xn > α
2 + xn > 0, xn > 0 & 3 − 2 + xn < 0 sincee 2 + xn > 9 = 3 Thus: xn +1 − xn < 0 ⇒ xn +1 < xn Thereffore the sequuence is stricctly decreasing. wn] [Show
Qn n 4 (i)
Solutioon Binomiial Expansioon x−4 A B = + Let f ( x ) = ( x + 1)(( 3x + 2 ) x + 1 3x + 2 ⇒ x − 4 = A ( 3 x + 2 ) + B ( x + 1)
c up rulee, Using cover 2 2 ⎛1⎞ For x = − , − − 4 = B ⎜ ⎟ ⇒ B = −14 3 3 ⎝3⎠ For x = −1 , −1 − 4 = A ( −1) ⇒ A = 5
∴f ( x) =
5 14 − x + 1 3x + 2
5 14 − x + 1 3x + 2 −1 −1 = 5 (1 + x ) − 14 ( 3 x + 2 )
f ( x) =
⎡ ⎛ 3x ⎞ ⎤ −1 = 5 (1 + x ) − 14 ⎢ 2 ⎜1 + ⎟ ⎥ 2 ⎠⎦ ⎣ ⎝ = 5 (1 + x ) − −1
14 ⎛ 3 x ⎞ ⎜1 + ⎟ 2⎝ 2 ⎠
−1
−1
= 5 (1 − x + x 2 − x 3 +K)
2 3 ⎛ ⎞ ⎛ 3x ⎞ ( −1)( −2 ) ⎛ 3x ⎞ ( −1)( −2 )( −3) ⎛ 3x ⎞ −7 ⎜1 + ( −1) ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + K ⎟⎟ ⎜ 2! 3! ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ ⎠ 2 3 ⎛ 3x 9 x 27 x ⎞ − + K⎟ = 5 (1 − x + x 2 − x 3 +K) −7 ⎜1 − + 2 4 8 ⎝ ⎠ 11 43 149 3 = −2 + x − x 2 + x +K 2 4 8
(ii)
Expansion of (1 + x )
−1
⎛ 3x ⎞ Expansion of ⎜1 + ⎟ 2 ⎠ ⎝
is valid for −1 < x < 1 −1
is valid for −
2 2
Therefore, the range of values of x for the expansion of f(x) to be valid is 2 2 −
(iii)
Coefficient of x = ( −1) 5 + ( −1) n
n
Qn 5
n +1
n n n ⎡ ⎛3⎞ ⎛3⎞ ⎤ 7 ⎜ ⎟ = ( −1) ⎢5 − 7 ⎜ ⎟ ⎥ ⎝2⎠ ⎝ 2 ⎠ ⎦⎥ ⎣⎢
Solution Complex Numbers (Loci) Im z − 5 = 3 + 7i
p 6 + 3i
Re
5 4 − 3i
q z − 6 − 3i = z − 4 + 3i
Cartesian equation of (i):
( x − 5) 2 + y 2 = 42
---- (1)
Gradient of line segment joining 6 + 3i and 4 − 3i =
(
3− − 3 6−4
)=
3
Cartesian equation of (ii): y=−
1 ( x − 5) 3
---- (2)
Using GC to solve (1) & (2), we get
p = 1.5359 + 2i q = 8.4641 − 2i
⎛ 4 ⎞ 5π ∴ arg( p − q ) = π − sin −1 ⎜ ⎟ = = 2.62 (3 s.f.) ⎝8⎠ 6 Alternative:
p − q = −6.9282 + 4i
∴ arg ( p − q ) = π − tan −1
4 = 2.62 (3s.f.) 6.9282
Or using GC, ∴ arg ( p − q ) = 2.62 (3s.f.) Qn 6
Solution Differential Equations dθ = − k (θ − 30 ) , k > 0 dt 1 ∫ θ − 30 dθ = − ∫ k dt ln θ − 30 = − kt + C
θ − 30 = e − kt +C
θ − 30 = ±e− kt eC = Ae− kt (where A = ±eC ) θ = 30 + Ae− kt when t = 0, θ = 90 90 = 30 + A A = 60 ∴θ = 30 + 60e− kt (shown)
when t = 8, θ = 55 55 = 30 + 60e −8 k 5 12 1 5 k = − ln 8 12
e −8 k =
when θ = 35, ⎛1 5 ⎞ ⎜ ln ⎟t 12 ⎠
35 = 30 + 60e⎝ 8 ⎛1 5 ⎞ ⎜ ln ⎟t 12 ⎠
1 12 t = 22.707 ∴ additional time needed = 22.707 − 8 = 14.7 min (1d.p.) e⎝ 8
Qn 7 (a)
=
Solution AP and GP
1 ar = a + 3d ⇒ d = − a 2 Sum of first n even-numbered terms =
n⎡ ⎛1 ⎞ ⎤ 2 ⎜ a ⎟ + (n − 1) ( −a ) ⎥ ⎢ 2⎣ ⎝2 ⎠ ⎦
n [ 2a − an] 2 an ( 2 − n ) = 2 Amount owed at the beginning of the third month =
(b)(i)
= ⎣⎡( 50000 − x ) (1.035) − x ⎦⎤ (1.035) (b)(ii)
= 50000(1.0352 ) − (1.035 + 1.0352 ) x Amount owed at the beginning of the nth month = 50000(1.035n −1 ) − (1.035 + 1.0352 + L + 1.035n−1 ) x For the repayment to be completed during the nth payment,
50000(1.035n −1 ) − (1.035 + 1.0352 + L + 1.035n −1 ) x ≤ x
50000(1.035n−1 ) ≤ (1 + 1.035 + 1.0352 + L + 1.035n−1 ) x 1.035n − 1 x 1.035 − 1 1.035n − 1 x ≤ 0.035
≤
Thus x ≥
1750(1.035n−1 ) . 1.035n − 1
Qn 8 (i)
Solution Transformation of Graphs + System of Linear Equations
y
C ' ( 0, 12 )
⎛3 ⎞ A ' ⎜ , 75 ⎟ ⎝2 ⎠
⎛1 ⎞ B '⎜ , 3⎟ ⎝2 ⎠
0
x
y = f (1 − 2 x )
(ii)
(
A" −2, 72
y
)
C " (1, 3) B " ( 0, 0 )
α
β
0 C "' (1, − 3 )
(
A"' −2, − 72
)
Let the curve be y = ax3 + bx 2 + cx + d . Since the points ( −2, 75 ) , ( 0,3) and (1,12 ) lie on the curve. Using ( 0,3) , d = 3 Using ( −2, 75 ) , a ( −2 ) + b ( −2 ) + c ( −2 ) + d = 75 −8a + 4b − 2c = 72 LLL (1) Using (1,12 ) , a + b + c + d = 12 a + b + c = 9 LLL (2) dy Since (1,12 ) is a maximum point, = 0. dx 3ax 2 + 2bx + c = 0 3a + 2b + c = 0 LLL (3) 3
2
Using GC to solve (1),(2) and (3), a = −8, b = 7, c = 10 Thus the equation of the curve is y = −8 x3 + 7 x 2 + 10 x + 3 .
x
Qn 9
Solution Vectors (line and planes)
(i)
Vector equation of the line l ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ λ∈R r = ⎜0⎟ + λ ⎜ 1 ⎟, ⎜1⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎝ ⎠ Angle between OA and the line l ⎛0⎞⎛1⎞ ⎜ ⎟⎜ ⎟ ⎜ −1 ⎟ . ⎜ 1 ⎟ ⎜ 2 ⎟ ⎜ −1 ⎟ −3 3 ⎝ ⎠⎝ ⎠ cos θ = = = 5 3 15 15 o o θ = 39.232 ≈ 39.2 (1 d.p.)
(ii)
Let the shortest distance from the origin to the line l be x. x sin θ = 5 x = 1.41( 3 s.f.)
(iii)
Since ⎛1⎞ ⎜ ⎟ ⎜0⎟ . ⎜1⎟ ⎝ ⎠
⎛1⎞ ⎜ ⎟ ⎜ 2 ⎟ = 4 ⇒ A point on l lies on π 1 ⎜ 3⎟ ⎝ ⎠ ⎛ 1 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ and ⎜ 1 ⎟ . ⎜ 2 ⎟ = 0 ⇒ l is parallel to π 1 ⎜ −1 ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠
The line l lies on the plane (shown) Alternative Solution (1): Since ⎛ 0 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ -1⎟ . ⎜ 2 ⎟ = 4 ⇒ Point A on l lies on π 1 ⎜ 2 ⎟ ⎜ 3⎟ ⎝ ⎠ ⎝ ⎠ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ . ⎜ 2 ⎟ = 4 ⇒ Point B on l lies on π 1 ⎜1⎟ ⎜ 3⎟ ⎝ ⎠ ⎝ ⎠ ∴ line l lies on π 1 Alternative Solution (2): Since ⎡ ⎛1⎞ ⎛ 1 ⎞ ⎤ ⎛ 1 ⎞ ⎛1 + λ ⎞ ⎛ 1 ⎞ ⎢ ⎜ ⎟ ⎜ ⎟⎥ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎢ ⎜ 0 ⎟ + λ ⎜ 1 ⎟ ⎥ . ⎜ 2 ⎟ = ⎜ λ ⎟ . ⎜ 2 ⎟ = 1 + λ + 2λ + 3 − 3λ = 4 ⎜ −1⎟ ⎥ ⎜ 3 ⎟ ⎜ 1 − λ ⎟ ⎜ 3 ⎟ ⎢⎣ ⎜⎝ 1 ⎟⎠ ⎝ ⎠⎦ ⎝ ⎠ ⎝ ⎠⎝ ⎠
∴ line l lies on π 1
(iv)
(v)
⎛ 1 ⎞ ⎛1⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ normal of π 2 = ⎜ 1 ⎟ × ⎜ 2 ⎟ = ⎜ −4 ⎟ ⎜ −1 ⎟ ⎜ 3 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 5 ⎞ ⎛1⎞ ⎛ 5 ⎞ π 2 : r . ⎜⎜ −4 ⎟⎟ = ⎜⎜ 0 ⎟⎟ . ⎜⎜ −4 ⎟⎟ ⎜ 1 ⎟ ⎜1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 5⎞ π 2 : r . ⎜⎜ −4 ⎟⎟ = 6 ⎜1⎟ ⎝ ⎠ ⎛ 1 ⎞ ⎛ 5 ⎞ ⎛ 14 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ normal of π 3 = ⎜ 2 ⎟ × ⎜ −4 ⎟ = ⎜ 14 ⎟ = 14 ⎜ 1 ⎟ ⎜ 3 ⎟ ⎜ 1 ⎟ ⎜ −14 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 ⎛ ⎞ ⎜ ⎟ normal of π 3 = director vector of l = ⎜ 1 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎛1⎞ π 3 : r . ⎜⎜ 1 ⎟⎟ = d ⎜ −1 ⎟ ⎝ ⎠
perpendicular distance from origin to π 3 =
d =± 3 ∴ d = ±3 3 ⎛1⎞ or π 3 : r . ⎜⎜ 1 ⎟⎟ = 3 ⎜ −1 ⎟ ⎝ ⎠
d d = 3 ⎛1⎞ ⎜ ⎟ ⎜1⎟ ⎜ −1 ⎟ ⎝ ⎠
⎛1⎞ π 3 : r . ⎜⎜ 1 ⎟⎟ = −3 ⎜ −1 ⎟ ⎝ ⎠
Qn Solution 10 Functions + Maclaurin’s Series (i) R g = [− ln 2, ∞) and D f = Since R g ⊆ D f , the composite function fg exists. (ii)
1 ⎞ ⎛ fg( x) = cos ⎜ ln ⎟ ⎝ 1+ x ⎠ = cos ( − ln (1 + x ) ) ⎛ x 2 x3 ⎞ ≈ cos ⎜ − x + − ⎟ 2 3⎠ ⎝ 2
4
OR
⎛ x 2 x3 ⎞ ⎛ x 2 x3 ⎞ − + − − + − ⎟ x x ⎜ ⎟ ⎜ 2 3⎠ ⎝ 2 3⎠ ⎝ ≈ 1− + −L 2! 4!
⎛ 1 ⎡⎛ x2 ⎞ x 2 ⎞ ⎛ x3 ⎞ ⎛ x3 ⎞ = 1 − ⎢⎜ − x + ⎟ − 2 ⎜ − x + ⎟ ⎜ ⎟ + ⎜ ⎟ 2 ⎢⎝ 2⎠ 2 ⎠⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ ⎣ 1 = 1 − ( x 2 − x3 ) + L 2 1 1 = 1 − x 2 + x3 + L 2 2 2
(iii)
2
⎤ ⎥ +L ⎥⎦
h(1) = −1 so h −1 (−1) = 1 .
Let y = − ln(1 + x ) . Then x = e − y − 1 . For −1 < x < 1 , h( x ) = − ln(1 + x ) so h( x ) > − ln 2 . Thus h −1 ( x) = e− x − 1 for x > − ln 2 . Thus e− x − 1 for x > − ln 2, −1
h :xa
1
x = −1.
for
Qn 11 (a)
Solution Complex Numbers π π⎞ ⎛ z = 3 + i = 2 ⎜ cos + i sin ⎟ 6 6⎠ ⎝ nπ nπ ⎞ ⎛ + i sin z n = 2 n ⎜ cos ⎟ 6 6 ⎠ ⎝ n nπ ⎞ ⎛ z n − ( z * ) = 2i Im( z n ) = 2i ⎜ 2n sin ⎟=0 6 ⎠ ⎝ nπ sin =0 6 nπ = kπ , k ∈ Z + 6 The set of values of n is {n : n = 6k , k ∈ Z + }
(b)
z+w=
(
) (
3 −1 + i 1+ 3
)
Im
(
)
W
i 3
i 1+ 3
P
i
Z Re
−1
O
3 −1
3
Note that OWPZ is a parallelogram, 3 2π = arg( w) = π − tan −1 1 3
2π π π − = 3 6 2 OZ= z = 2 ∠WOZ =
OW= w = −1 + i 3 = 2 Since ∠WOZ =
π
and OZ = OW , OWPZ is a square. (shown)
2
arg( z + w) = arg( z ) + ∠POZ =
Also, arg( z + w) = tan −1
∴
5π 3 +1 = tan −1 12 3 −1
5π 3 +1 ∴ tan = = 12 3 −1
Qn 12 (i)
(
π 6
+
π 4
=
5π 12
3 +1 3 −1
)
3 +1
2
3 −1
=
4+2 3 = 2 + 3 (deduced) 2
Solution Graphing + tangent/normal 2 x 2 − 3x + 1 Let y = x2 − 4 yx 2 − 4 y = 2 x 2 − 3 x + 1
( 2 − y ) x 2 − 3x + ( 4 y + 1) = 0 For real roots of x, Discriminant ≥ 0 2 ( −3) − 4 ( 2 − y )( 4 y + 1) ≥ 0
(
)
9 − 4 8 y + 2 − 4 y2 − y ≥ 0 16 y 2 − 28 y + 1 ≥ 0 For 16 y 2 − 28 y + 1 = 0 y=
28 ±
( −28 ) − 4 (16 ) 2 (16 ) 2
28 ± 720 32 7±3 5 = 8 ∴16 y 2 − 28 y + 1 ≥ 0 =
y≤
7−3 5 8
or
y≥
7+3 5 8
Hence, C cannot lie between
7−3 5 7+3 5 and 8 8
(ii)
y y=
2 x 2 − 3x + 1 x2 − 4
y=2
( 0.764, 0.0365) ( 5.23,1.71) 1 1 2 − 4
O
x = −2
(iii)
At x = −5,
y=
1
x
x=2
22 7
dy = 0.40136 dx x =−5 Equation of tangent at x = −5 is
Using GC,
22 = 0.40136 ( x + 5 ) 7 y = 0.40136 x + 5.149657 When x = 0, y = 5.149657 = 5.15 (to 3 s.f) The coordinates of P are ( 0,5.15 ) (shown) y−
Equation of normal at x = −5 is
22 −1 = ( x + 5) 7 0.40136 y = −2.4915 x − 9.3148 When x = 0, y = −9.3148 = −9.31 (to 3 s.f) The coordinates of Q are ( 0, −9.31) (shown) y−
1 ( 5.15 + 9.31)( 5 ) 2 = 36.2 units2
Area of PQR =
H2 MATHS (9740) JC2 PRELIMINARY EXAM 2010 PAPER 2 SUGGESTED SOLUTIONS Qn 1
Solution Equations and Inequalities x+7 ≤1 4 + 3x − x 2 x+7 −1 ≤ 0 4 + 3x − x 2 x + 7 − 4 + 3x − x 2 ≤0 4 + 3x − x 2 x + 7 − 4 − 3x + x2 ≤0 − ( x2 − 3x − 4 )
(
)
x2 − 2 x + 3 ≥0 x 2 − 3x − 4 Since x 2 − 2 x + 3 = ( x − 1)2 + 2 > 0 for x ∈ , x 2 − 3x − 4 > 0 ( x − 4 )( x + 1) > 0
+
x < −1 or x > 4
Qn 2
+
−
−1
4
Solution Vectors p × q = 3p × r p × ( q − 3r ) = 0 p // ( q − 3r ) q − 3r = λ p , where λ is a scalar.
( q − 3r ) . ( q − 3r ) = λ p.λ p q − 6q.r + 9 r = λ 2 p 2
2
2
2 2 ⎛5⎞ 52 − 6 × 5 × 2 ⎜ ⎟ + 9 ( 2 ) = λ 2 (1) ⎝6⎠ λ 2 = 11
λ = ± 11 Qn 3 (i)
Solution Mathematical Induction Let Pn be the statement:
∑ 3( 2
3 − r ) = 3 ( 2n +1 ) − n ( n + 1) − 6, ∀n ∈ + . 2 r =1 Check P1 , n
r
∑ 3( 2 1
LHS of P1 :
r =1
r
− r ) = 3 ( 21 − 1) = 3
RHS of P1 : 3 ( 21+1 ) −
3 (1)(1 + 1) − 6 = 12 − 3 − 6 = 3 = LHS 2 ∴ P1 is true.
Assume Pk true for some k ∈ + .
∑ 3(2
3 − r ) = 3 ( 2k +1 ) − k ( k + 1) − 6 ----(*) 2 r =1 k +1 3 Prove Pk +1 true, i.e. ∑ 3 ( 2r − r ) = 3 ( 2k + 2 ) − ( k + 1)( k + 2 ) − 6 2 r =1 LHS of Pk +1 : k
r
⎡ k r − = 3 2 r ( ) ⎢∑ 3 ( 2r − r )⎤⎥ + 3 ⎡⎣2k +1 − ( k + 1)⎤⎦ ∑ r =1 ⎣ r =1 ⎦ 3 = 3 ( 2k +1 ) − k ( k + 1) − 6 + 3 ⎡⎣ 2k +1 − ( k + 1) ⎤⎦ 2 3 = 3 ( 2k +1 ) − k ( k + 1) − 6 + 3 ( 2k +1 ) − 3 ( k + 1) 2 3 = 3 ( 2k +1 ) + 3 ( 2k +1 ) − k ( k + 1) − 3 ( k + 1) 2 ⎛1 ⎞ = 2 ⎡⎣3 ( 2k +1 ) ⎤⎦ − 3 ( k + 1) ⎜ k + 1⎟ ⎝2 ⎠ 3 = 3 ( 2k + 2 ) − ( k + 1)( k + 2 ) − 6 = RHS 2 ∴ Pk true ⇒ Pk +1 true k +1
Since P1 is true and Pk true ⇒ Pk +1 true , therefore, by Mathematical Induction, Pn is true ∀n ∈ + . (iii)
∑ 3(2
r
− r ) > 5800
∑ 3(2
r
− r ) − ∑ 3 ( 2r − r ) > 5800
n
r =5 n
r =1
4
r =1
3 3 ⎡ ⎤ ⎡ ⎤ n +1 4 +1 ⎢⎣3 ( 2 ) − 2 ( n )( n + 1) − 6 ⎥⎦ − ⎢⎣3 ( 2 ) − 2 ( 4 )( 5 ) − 6 ⎥⎦ > 5800 3 ⎡ ⎤ n +1 ⎢⎣3 ( 2 ) − 2 ( n )( n + 1) − 6 ⎥⎦ − 60 > 5800 3 3 ( 2 n +1 ) − ( n )( n + 1) − 66 > 5800 2 3 3 ( 2 n +1 ) − ( n )( n + 1) > 5866 2 Using GC, using table method 3 ⎡ ⎤ When n = 9, ⎢3 ( 2n +1 ) − ( n )( n + 1) ⎥ = 2737<5866 2 ⎣ ⎦ 3 ⎡ ⎤ When n = 10, ⎢3 ( 2n +1 ) − ( n )( n + 1) ⎥ = 5979>5866 2 ⎣ ⎦ ∴ Least n = 10
Qn 4 (a)
Solution Application of Differentiation (Rate of change)
y = x + ln ( xy )
3
= x + 3ln ( x ) + 3ln ( y )
dy 1 3 3 dy = + + dx 2 x x y dx when x = 1 and y = 1, dy 7 =− dx 4 (b)
1 V = π r 2h 3 1 3 = π h (from diagram r = h) 3 dV = π h2 dh dh dh dV . = dt dV dt 1 = 2 (1) πh After 2 min, amt of water poured into cone = 2 × 1× 60
= 120 cm3 Amt of water in cone after 2 min = 120 + 30 = 150 cm3 1 ∴ π r 2 h = 150 3 1 3 π h = 150 3 h=
3
dh = dt
450
π 1
⎛ 450 ⎞ π⎜3 ⎟ ⎝ π ⎠ = 0.0116
Qn 5
2
Solution Integration (area and volume) y
(i)
b R -a
a -b
x
Area of R = Area of quadrant-Area of triangle b2 x2 1 dx − ab 2 −a a 2 0 b 1 = − ab + ∫ a 2 − x 2 dx (shown) − a 2 a =∫
0
b2 −
x = a sin θ dx = a cos θ dθ ⎛ 0 = a sin θ ⇒ θ = 0 ⎞ ⎜ ⎟ ⎜ −a = a sin θ ⇒ θ = − π ⎟ ⎝ 2⎠ 0 b 1 2 2 ∫ −a a a − x dx − 2 ab b 0 1 2 = ∫ π a 2 − ( a sin θ ) ( a cos θ ) dθ − ab − a 2 2
a 2b 0 1 1 − sin 2 θ ( cos θ ) dθ − ab π ∫ − a 2 2 0 1 = ab ∫ π cos 2 θ dθ − ab − 2 2 0 1 + cos 2θ 1 dθ − ab = ab ∫ π − 2 2 2 =
sin 2θ ⎤ 1 ⎡1 = ab ⎢ θ + − ab ⎥ 4 ⎦ −π 2 ⎣2 0
2
⎡ ⎛ 1 ⎛ π ⎞ sin ( −π ) ⎞ ⎤ 1 = ab ⎢0 + 0 − ⎜ ⎜ − ⎟ + ⎟ ⎥ − ab 4 ⎢⎣ ⎝2⎝ 2 ⎠ ⎠ ⎥⎦ 2 ⎛π ⎞ 1 = ab ⎜ ⎟ − ab ⎝4⎠ 2 ab = (π − 2 ) units 2 4
(ii)
Volume generated b 1 = π ∫ x 2 dy − π a 2b 0 3 2 2 b 1 a y = π ∫ a 2 − 2 dy − π a 2b 0 3 b b
⎡ a2 y3 ⎤ 1 = π ⎢ a 2 y − 2 ⎥ − π a 2b 3b ⎦ 0 3 ⎣ ⎡ 2 a 2b3 ⎤ 1 2 = π ⎢a b − 2 ⎥ − π a b 3b ⎦ 3 ⎣ ⎡ a 2b ⎤ 1 2 = π ⎢ a 2b − − πa b 3 ⎥⎦ 3 ⎣ ⎛2 ⎞ 1 = π ⎜ a 2 b ⎟ − π a 2b ⎝3 ⎠ 3 1 = π a 2b 3
Qn 6 (a)
Solution Sampling Methods Quota sampling. It will not be representative of the cohort as it misses out those who do not visit the canteen. OR It is not random.
(b)
Step (2) is incorrect. The interval size should be 4 students. He should pick the random start student from the 1st-4th student and select every 4th student thereafter.
Qn 7 (i)
Solution Probability
S
F
5 21 S
5 63
S
(ii)
58 63
F
Set: S = Success; F = Failure
2 7
5 7
5.
5 7
16 21
5 7
2 7
F
S
F
2 7
5 21
16 5 21 7
2 7
S
F S F S F ⎛ 5 ⎞ ⎛ 16 ⎞ ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 5 ⎞ ⎛ 16 ⎞ ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 5 ⎞ Required Prob = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 7 ⎠ ⎝ 21 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 21 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ 380 = 1029 Required Prob = 1 − P ( Player fails in all three games ) ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 2 ⎞ 335 = 1− ⎜ ⎟⎜ ⎟⎜ ⎟ = ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ 343
(iii)
Let Event A = Player has exactly two failures Let Event B = Player has at least one success P ( A ∩ B) Required Prob = P ( A | B ) = P ( B)
⎛ 380 ⎞ (i) ⎜⎝ 1029 ⎟⎠ = = (ii) ⎛ 335 ⎞ ⎜ ⎟ ⎝ 343 ⎠ 76 = 201
Qn 8 (a)
Solution Permutations and Combinations No of teams 3 tallest, no shortest ⎛ 3 ⎞⎛ 8 ⎞ ⎜ ⎟⎜ ⎟ = 56 ⎝ 3 ⎠⎝ 5 ⎠ 3 tallest, 1 shortest ⎛ 3 ⎞⎛ 2 ⎞⎛ 8 ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ = 140 ⎝ 3 ⎠⎝ 1 ⎠⎝ 4 ⎠
2 tallest, no shortest 2 tallest, 1 shortest
⎛ 3 ⎞⎛ 8 ⎞ ⎜ ⎟⎜ ⎟ = 84 ⎝ 2 ⎠⎝ 6 ⎠ ⎛ 3 ⎞⎛ 2 ⎞⎛ 8 ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ = 336 ⎝ 2 ⎠⎝ 1 ⎠⎝ 5 ⎠
Total no of teams= 56 + 140 + 84 + 336 = 616 b(i)
No of different 5-letter secret codes = No of codes using {X,Y,Z,Z,Z}+ No of codes using {X,Y,Y,Z,Z}+No of codes using {Y,Y,Z,Z,Z} 5! 5! 5! = + + 3! 2!2! 2!3! = 20 + 30 + 10 = 60
(ii)
Y {Z, Z, Z} Y
or
Y {X, Z, Z} Y
No of 5-letter secret codes that will begin and end with Y = 1+ 3 = 4
Qn 9
Solution Hypothesis Testing
Unbiased estimate of μ is t =
∑ t = 213.5 = 4.27 n
50 2 1 ⎡ t −t ⎤ Unbiased estimate of σ 2 is s 2 = ∑ ⎦⎥ n − 1 ⎣⎢ 1 = [ 44.105] 49 = 0.90010 ≈ 0.900 (3 s.f)
( )
⎛ σ2 ⎞ Since n = 50 is large, by Central Limit Theorem, X N ⎜ μ , ⎟ ⎝ 50 ⎠ approximately. H0: μ = 4.5 H1: μ < 4.5
X −μ S n Level of significance: 1% Critical Region: Reject H0 if p -value < 0.01 Test Statistic: Z =
Assuming H0 is true, from the GC, p -value = 0.0432439 . Since p -value = 0.0432 > 0.01 , we do not reject H0 and conclude that there is no significant evidence, at 1% level, that the population mean time spent is less than 4.5 minutes. In order for H0 to be rejected in favour of H1, we require 0.0432439 < 0.01α ⇒ α > 4.3244 4.33% is the smallest level of significance resulting in the rejection of H0.
It is not necessary to assume that the population follows a normal distribution for the test to be valid because since n = 50 is large, by Central Limit Theorem, ⎛ σ2 ⎞ X N ⎜ μ , ⎟ approximately. ⎝ 50 ⎠
Qn 10 (i)
Solution Binomial + Poisson (incl. CLT) Let H be the number of calls received by the hospital in a week. Let S be the number of calls received by the police station in a week. Then H ~ Po (12 ) , S ~ Po ( 5 ) .
Also, H + S ~ Po (17 ) .
P ( H + S > 11) = 0.91533 ≈ 0.915
(ii)
Let W be the number of weeks, out of 50, in which the hospital and police station receive a total of at most 11 calls. Then W ~ B ( 50,1 − 0.91533 ) . i.e. W ~ B ( 50, 0.08467 ) . Since n = 50 is large, p = 0.08467 is small, and np = 4.2335 < 5 , W ~ Po ( 4.2335 ) approximately. From GC, P (W ≤ 4 ) = 0.58332
≈ 0.583 (iii)
Since n = 100 is large, by the Central Limit Theorem, 12 ⎞ ⎛ H N ⎜ 12, ⎟ approximately ⎝ 100 ⎠ 12 12 ⎞ ⎛ + H1 + H 2 N ⎜12 + 12, ⎟ ( approximately ) 100 100 ⎠ ⎝ P ( H1 + H 2 ≤ 23) = 0.020613
≈ 0.0206
Qn Solution 11 Normal Distribution (a)(i) Let the weight of 1 King Crab and 1 Snow Crab be K and S respectively. K ~ N (1.65, 0.71) S ~ N (1.10, 0.34 )
K1 + K 2 + K3 + S1 + S2 5 1 E (T ) = ⎡⎣3E ( K ) + 2E ( S ) ⎤⎦ 5 1 = ⎣⎡3 (1.65 ) + 2 (1.10 ) ⎦⎤ 5 = 1.43
Let T =
1 ⎡3Var ( K ) + 2Var ( S ) ⎤⎦ 52 ⎣ 1 = ⎡3 ( 0.71) + 2 ( 0.34 ) ⎤⎦ 25 ⎣ = 0.1124 ∴ T ~ N (1.43, 0.1124 ) Var (T ) =
P ( T < 1.5 ) = 0.583
(ii)
Let the selling price of 1 King Crab and 1 Snow Crab be X and Y respectively.
(
)
X = 40 K ~ N 40 × 1.65, 402 × 0.71 ⇒ X ~ N ( 66,1136 )
(
)
Y = 45S ~ N 45 ×1.10, 45 × 0.34 ⇒ Y ~ N ( 49.5, 688.5 ) 2
Let the total selling price of an King Crab and a Snow Crab be C = X + Y C ~ N ( 66 + 49.5,1136 + 688.5 ) ⇒ C ~ N (115.5,1824.5 ) P ( C > 140 ) = 0.283
(iii)
The weight of all crabs are independent of one another. NOTE: Weight of snow crab and weight of king crab is independent of each other is insufficient, students need to bring out that the weight of all crabs are independent of one another.
(b)
Let X be the mass of a bar of chocolate, X ~ N ( μ , σ 2 ) . P( X − μ ≤ m) = 0.2 P( Z ≤ m ) = 0.2 σ
P( − m ≤ Z ≤ m ) = 0.2 σ
σ
P( Z ≤ − m ) = 0.4 m
σ
σ
= 0.25335
P( X − μ ≥ 3m) = P( Z ≥ 3m ) σ = P( Z ≥ 0.76004) = 0.224
Qn 12 (i)
Solution Correlation & Regression y 48000
31800 1339
2275 x
The linear model is not valid in the long run, as it is impossible to have the selling price of the car to be $0 or negative when the number of new car licenses increases to a certain number.
(ii)
Model C is the most appropriate, as when x increases, y decreases at a decreasing rate.
(iii)
r = 0.97817 = 0.978 (3 s.f.) It indicates a strong positive linear correlation between y and 1 / x .
(iv)
Regression line: 50012000 (5.s.f) y = 8452.3 + x (a) When x =1300, 50012000 y = 8452.3 + = 46923.07 = $46900 (3s.f.) 1300 Since x =1300 lies outside the data range, the linear relation may no longer hold, hence, the estimate is unreliable. (b) When x =2000, 50012000 y = 8452.3 + = 33458.30 = $33500 (3s.f.) 2000 Since x =2000, lies within the data range, and r = 0.978 is close to 1, the estimate is reliable.
(v)
No, I do not agree, as there is no causal effect between the two variables. The rise in the selling price of the car could be due to other factors like the production cost of the car.
NATIONAL JUNIOR COLLEGE PRELIMINARY EXAMINATIONS
Higher 2
MATHEMATICS
9740/01
Paper 1
13 September 2010 3 hours
Additional Materials: Answer Paper List of Formulae (MF15) Cover Sheet
0815 – 1115 hours
READ THESE INSTRUCTIONS FIRST Write your name, registration number, subject tutorial group, on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in the brackets [ ] at the end of each question or part question. This document consists of 9 printed pages.
National Junior College [Turn Over
2 1
The sum of the digits in a three-digit-number is 15. Reversing the digits in that number decreases its value by 594. Also, the sum of the tenth digit and four times the unit digit is five more than the hundredth digit. Find the number.
2
Find the value of p such that
p 0
[4]
2 . ln 3
3x dx
[2]
The graph of y 3x for 0 x 1 , is shown in the diagram below. Rectangles, each of width
1 , where n is an integer, are drawn under the curve. n
y y 3x
…
0
1 2 3 n n n
n3 n
n 2 n 1 n n
1
x
1
(i)
Show that the total area of all n rectangles, A, is given by
(ii)
State the limit of A as n .
NJC 2010
2 3n
1
n 3n 1
.
[2]
[1]
9740/01/2010
[Turn Over
3 3
The points A and B have position vectors a and b respectively, relative to the origin O such that a = b . The point P with position vector p lies on AB such that b • p = a • p . (i)
Show that AB is perpendicular to OP.
(ii)
Determine the position vector of the point D in terms of a and b, where D is the
(iii)
4
[2]
reflection of O about the line AB.
[2]
Give the geometrical meaning of a b .
[1]
The diagram shows the graph of y sin x for 0 x with coordinates
x1 ,sin x1
x ,sin x , where 1
1
π . P is a fixed point on the curve 2
and Q is a point on the curve with coordinates is measured in radians.
y
y sin x Q
P
x O sin x1 (cos 1) cos x1 sin
(i)
Prove that the gradient of PQ is
(ii)
Given that is sufficiently small for 3 and higher powers of to be neglected, express the gradient of PQ as a linear expression in terms of . [2]
.
[2]
Verify that the gradient of PQ approximates to the gradient of the tangent at P when tends to zero. [2]
NJC 2010
9740/01/2010
[Turn Over
4 5
Given that z 3 4i 5 , illustrate the locus of the point P representing the complex number z in an Argand diagram.
[1]
Hence, find the least exact value of z 1 .
[2]
The locus of point Q, representing the complex number w, satisfies the relation w 2i w ai , where a is a real number, a 2 . Find the range of values of a such that
the loci of P and Q meet more than once.
6
[3]
By using the substitution y vx , find the general solution of the differential equation
x
(i)
dy 3x y 2 . dx
[4]
State the equation of the locus where the stationary points of the solution curves lie. [1]
(ii)
Sketch, on a single diagram, the graph of the locus found in part (i) and two members of the family of solution curves where the arbitrary constant in the solution is non-zero.
[3]
7(a) By considering x 2 A(2 2 x) B where A and B are real constants, or otherwise, find
x2 x2 2x 8
[5]
dx.
1 5
(b) Show that e x cos 2 x dx e x (cos 2 x 2 sin 2 x) c , where c is an arbitrary constant. Hence, find
NJC 2010
e
x
cos 2 x dx.
[5]
9740/01/2010
[Turn Over
5 8 (a)
The sum, Sn , of the first n terms of an arithmetic progression is given by S n n 2 2n .
Write down the expression for S n Sn 1 . Hence, find the value of the common difference. [3]
(b)
A metal screw of length L (measured in millimetre) is driven into a concrete wall by an electrical screwdriver, such that its distance driven into the wall is proportional to the angle turned by the screwdriver.
L
Wall
Due to some reasons, every subsequent turn by this electrical screwdriver can only achieve an 80% of the angle turned previously. (i)
Given that the initial angle turned by the screwdriver is radians, write down the expressions for the first 3 distances driven into the concrete wall, leaving your answers in terms of and k, where k is the constant of proportionality.
(ii)
Find the total distance driven into the concrete walls after n turns, leaving your
answer in the form ak 1 b n , where a and b are constants to be determined. (iii)
[2]
[3]
Given that k 2 and assuming that the total distance driven could never exceed the length of the screw, find the minimum length of the metal screw required, giving your answer in terms of .
NJC 2010
[2]
9740/01/2010
[Turn Over
6
9
0 Relative to the origin O, the point A has position vector given by OA 1 and A lies on 0 1 the plane 1 with equation defined by r • 3 3 . Another plane 2 has equation y x . 2
The planes 1 and 2 intersect at line l.
(i)
Find the vector equation of the line l.
[1]
(ii)
Show that the cosine of the acute angle between the planes 1 and 2 is
(iii)
Find the position vector of the foot of perpendicular, OF , from point A to the line l.
7 . 7
Hence, find the exact length of projection of AF onto the plane 2 .
(iv)
[2]
[5]
Another plane 3 has equation px qy 1 , where p and q are real constants. Find the condition in which p and q must satisfy such that the planes 1 , 2 and 3 intersect at exactly one point.
[2]
10(a) Given that 2 3i is a solution to the equation z 2 (a i) z * 16 bi 0 , where z * is the conjugate of the complex number z, find the values of a and b, where a and
b are real constants
(b) (i)
[3]
Solve the equation z 5 1 0 , expressing the solutions in the form rei , where r 0 and π π . Show the roots of the equation on an Argand diagram. i
i 2 cos e 2 . 2
(ii)
For π π , show that 1 e
(iii)
Deduce the roots of w 1 1 0 in the exponential form.
NJC 2010
5
9740/01/2010
[4] [2]
[3]
[Turn Over
7 11
The curve C has equation
ax 2 bx 1 f(x) , where a, b and c are real constants. xc Given that the line y 2 x 1 is an asymptote of C, find the value of a and show that b 2c 1 .
[3]
(i) For c 1 , using algebraic method, prove that the curve C cannot lie between 2 values, which are to be determined. (ii) Sketch the graph of f(x)
[3]
2 x2 x 1 , showing clearly its asymptotes, the coordinates of x 1
the axial intercepts, and turning point(s) (if any).
[3]
Hence, state the range of x for which f(x) is concaving downwards.
[1]
(iii) Given that the line y kx k 3 ,where k is a real constant, passes through the intersection of the asymptotes of C, deduce the range of k where 2 x 2 x 1 (kx k 3)( x 1) has 2 real solutions.
NJC 2010
[1]
9740/01/2010
[Turn Over
8 12
A curve C has parametric equations x 2 2 and y 3 2 , where is a real parameter. Sketch the curve C for 0 3 .
[1]
The tangent to C at point P 2 2 , 3 2 cuts the y-axis at point Q. Show that the equation of the tangent at P may be written as
4 y 3 x 2 4 3 .
[2]
(a) (i) C cuts the y-axis at the point R. Find the area of triangle PQR, A, in terms of . [2] (ii) If x increases at a rate of 4 units per second when 2 , find the rate of change of
A at that instant. (b)
[3]
Calculate the exact area of the region bounded by the curve C, the tangent to C at point P when 2 and the y-axis.
[5]
End of Paper
NJC 2010
9740/01/2010
[Turn Over
9
NATIONAL JUNIOR COLLEGE PRELIMINARY EXAMINATIONS
Higher 2
MATHEMATICS Higher 2
9740/01 Paper 1
13 September 2010, Monday
0815 – 1115 hours
Candidate Name: ______________________
Registration No.:_____________
Subject Class: 2ma______/ 2IPma2_______
Subject Tutor: _______________ For office use
Question No.
over Page
Marks Obtained
TOTAL MARKS
1
4
2
5
3
5
4
6
5
6
6
8
7
10
8
10
9
10
10
12
11
11
12
13
INSTRUCTIONS TO CANDIDATES
Write your name, registration number, subject tutorial group, subject tutor’s name and calculator model in the spaces provided on the cover sheet and attached it on top of your answer paper. Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER. Write your calculator’s model number(s) in the box below. Scientific Calculator Model:
Graphic Calculator Model:
Presentation TOTAL
–1
/
–2 100
GRADE
NJC 2010
9740/01/2010
[Turn Over
NATIONAL JUNIOR COLLEGE PRELIMINARY EXAMINATIONS
Higher 2
MATHEMATICS
9740/02
Paper 2
17 September 2010 3 hours
Additional Materials: Answer Paper List of Formulae (MF15) Cover Sheet
0815 – 1115 hours
READ THESE INSTRUCTIONS FIRST Write your name, registration number, subject tutorial group, on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in the brackets [ ] at the end of each question or part question. This document consists of 9 printed pages.
National Junior College
[Turn Over
2 Section A: Pure Mathematics [40 marks] 1
The functions g and h are defined as follows: 1 g : x 2 x 1 , x 0, 2 2 h : x x 3x 2, x 0.
(i)
Justify why hg exists and find the range of hg.
[3]
(ii)
The function h has an inverse if its domain is restricted to x b . Find the value of b for this domain to be maximal. Sketch the graph of h and its inverse on the same
(iii)
diagram.
[2]
Solve h 1 (x) x exactly.
[2]
2 (a) A sequence of positive real numbers x1 , x2 , x3 , satisfies the recurrence relation
xn 1 3 ln xn 1 for n 1 . Given that the sequence converges to L, find the value of L. 2
[3]
(b) (i)
By expressing
r 2 7r 11 A B in the form where A and B are r 2 ! r 4 ! r 4 !
r 2 7 r 11 5 n5 . r 4 ! 4! n 4 ! r 1 n
real constants, show that
(ii)
Use the method of mathematical induction to prove your result in (i).
(iii)
Hence, find
r 2 9r 19
r 5!
.
[3]
[5] [3]
r 1
NJC 2010
9740/02/2010
[Turn Over
3 3
4
Given that y 2 esin x , show that 2
dy y cos x . dx
By further differentiation of this result, find the Maclaurin’s series for y, up to and including the term in x3 .
[5]
Deduce the Maclaurin’s series for e sin x up to and including the term in x 2 .
[3]
(i)
Use the substitution x 2 tan to show
4 x2
4 x
2 2
dx
x C , where C is an 4 x2
arbitrary constant.
[5] y
(ii) 15y 3 x 2
3 1, 5
–2
x
2
O
The diagram above shows the curve with equation y 2
4 x2
4 x
2 2
with stationary
points at x 0 . The line 15y 3 x 2 intersects the curve at 2, 0 and
3 1, . 5 (a)
y
The region bounded by the curve
4 x2
4 x2
2
and the line
15y 3 x 2 is rotated through 4 right angles about the x-axis to form a solid of revolution of volume V. Find the exact value of V, giving your answer in the form bπ. (b)
NJC 2010
[4]
Sketch the gradient graph of y
9740/02/2010
4 x2
4 x
2 2
.
[2]
[Turn Over
4 Section B: Statistics [60 marks] 5
The Head of Mathematics department of Holistics Junior College decides to take a survey of opinions of 700 graduating students regarding the quality of teaching of the subject.
6
(i)
What is the sampling frame in the context of the question?
[1]
(ii)
Describe clearly how a systematic sample of size 140 can be obtained.
[3]
The data in the table below were obtained in an experiment to estimate the relation between d, the duration of a television commercial (in seconds) and s, the average sales of a
particular brand of detergent (in thousands of bottles): Duration, d
15
18
22
25
26
29
34
39
Average Sales, s
0.43
1.12
1.75
1.98
2.11
2.26
2.40
2.44
(i)
Draw a scatter diagram to illustrate the data.
[1]
(ii)
Fit a model of the form s ln d to the data above and find the least squares estimates of and .
[1]
Hence state the product moment correlation coefficient between ln d and s, and explain whether your answer suggests that a linear model is appropriate for the transformed variables. (iii)
[2]
Using the regression model in part (ii), predict the average sales if the duration of a television commercial is 28 seconds. Comment on the reliability of your answer. [3]
NJC 2010
9740/02/2010
[Turn Over
5 7
Four married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if (i)
there are no restrictions,
[1]
(ii)
each couple must sit together.
[3]
They are to take a photo with the bride and bridegroom. The twelve people are to arrange in two rows of six and the bride and bridegroom must be together in the middle of the front row. How many ways can the photographer arrange the twelve people such that the two children must also be in the front row?
[3]
In this question, give each of your answers as an exact fraction in its lowest term.
8 (a) In a certain sample space, it is known that events A and B are independent. Given that P A B
3 2 and P A ' B , where A ' is the complement of event A, find 4 15
(i)
P B ,
[3]
(ii)
P A B A B .
[2]
(b) A teacher is to form two groups of 4 students from a class of 3 boys and 5 girls for Mathematics consultation session. Three of the girls, Ivy, Tamie and Cassy are good friends from the class. Find the probability that (i)
the boys are together in the same group,
[2]
(ii)
either Ivy or Tamie is in the same group as Cassy.
[3]
NJC 2010
9740/02/2010
[Turn Over
6 9
Suppose that the arrival and departure of aircrafts at a domestic airport follow two independent Poisson distributions. In a one-hour period, it is expected that there are 4 arrivals and 3 departures. (i)
Show that the probability that there are at least 13 arrivals in a two-hour period is 0.0638, correct to 3 significant figures.
(ii)
[1]
Find the probability that, in a randomly selected one-hour period, there are less than 2 departures given that the airport handles a total of exactly 9 arrivals and departures. [3]
A study of the domestic airport arrivals for 60 randomly selected two-hour periods is being conducted to see if there are at least 13 arrivals for each two-hour period. (iii)
Giving two reasons, in this context, explain why the binomial distribution is a suitable model for the study of the domestic airport arrivals.
(iv)
[2]
Using a suitable approximation, find the probability that there are at most 50 twohour periods with less than 13 arrivals each, explaining clearly why the approximation is appropriate.
NJC 2010
[3]
9740/02/2010
[Turn Over
7 10
The national average for monthly electricity usage measured in kilowatts hour (kWh), of Housing Development Authority (HDA) units is 380 kWh. The monthly electricity usage of 3-room units follows a normal distribution with mean of 290 kWh and variance, 2 , whereas the monthly electricity usage of 5-room units follows an independent normal distribution with mean of 450 kWh and variance 105 kWh2. (i)
Calculate the probability that the monthly electricity usage of two randomly chosen 3-room units exceeds 290 kWh each and a randomly chosen 5-room unit’s monthly electricity usage is less than 450 kWh.
(ii)
[2]
Given that the probability of the total monthly electricity usage of four randomly chosen 3-room units exceeds thrice the national average is 0.868, find 2 , correct to the nearest integer,
[3]
With effect from 1 July 2010, the monthly electricity bill is charged at 24 cents per kilowatts hour. (iii)
Determine the value of a, correct to 2 decimal places, such that the probability of the monthly electricity bill of a randomly chosen 5-room unit exceeding $a is 0.9.
(iv)
[2]
Eighty 5-room units are randomly selected. Using a suitable approximation, find the probability that there are not less than seventy 5-room units with monthly electricity bill exceeding $a.
NJC 2010
[3]
9740/02/2010
[Turn Over
8 11
The drying time, X minutes, of Noppin brand paint under specified test conditions is known to have mean value 75 minutes.
(a) On one occasion, in the manufacture of a large batch of Noppin paint, it was suspected that an accidental chemical contamination had resulted in a change in the drying time of the paint.
Due to the high costs involved in discarding the entire batch of paint, the
manufacturer decided he will only do so if there was strong evidence from a test at 5% level of significance to suggest that the drying time has changed. 50 random and independent specimens of paint samples were taken from the batch and the drying time is summarised as follows
x 3791 (i)
and
x
2
287959 .
Determine whether the manufacturer will discard the entire affected batch of paint. [6]
(ii)
Assuming that the unbiased estimate of the population variance is the same as the one found in the above sample, find the probability such that the total drying time of another 60 randomly selected specimens of paint samples obtained from the same batch is between 72 hours and 76 hours.
[3]
(b) On another occasion, chemists proposed a new additive designed to decrease the drying time. The mean and standard deviation from 20 random and independent specimens of paint samples with the new additive are x minutes and 7.5 minutes respectively. A test is to be carried out at the 5% level of significance to determine whether the new additive had been effective. (i)
(ii)
Determine the largest value of x for which the chemists can claim that their new additive had been effective in decreasing the drying time of the paint.
[3]
State a necessary assumption for validity of the test.
[1]
End of Paper NJC 2010
9740/02/2010
[Turn Over
9
NATIONAL JUNIOR COLLEGE PRELIMINARY EXAMINATIONS
Higher 2
MATHEMATICS Higher 2
9740/02 Paper 2
17 September 2010, Friday
0815 – 1115 hours
Name: _______________________________
Registration No. :_____________
Subject Class: 2ma______/ 2IPma2_______
Subject Tutor: _______________
For office use
over Page
Question No.
INSTRUCTIONS TO CANDIDATES
1
7
Write your name, registration number, subject tutorial group, subject tutor’s name and calculator model in the spaces provided on the cover sheet and attached it on top of your answer paper.
2
14
3
8
4
11
5
4
6
7
7
7
8
10
9
9
10
10
11
13
Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER. Write your calculator’s model number(s) in the box below. Scientific Calculator Model:
Marks Obtained
TOTAL MARKS
Graphic Calculator Model:
Presentation TOTAL
–1
/
–2 100
GRADE NJC 2010
9740/02/2010
[Turn Over
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
1
Let the unit, tenth and hundredth digits be z, y, x respectively. x + y + z = 15 − (1)
(100 x + 10 y + z ) − (100 z + 10 y + x) = 594 ⇒ 99 x − 99 z = 594 y + 4z = x + 5 ⇒ −x + y + 4z = 5
− (2)
− (3)
Using GC to solve the equations simultaneously, x = 8, y = 5, z = 2 . Thus the number is 852.
2
∫
p 0
3x dx =
2 ln 3
p
⎡ 3x ⎤ 2 ⎢ ln 3 ⎥ = ln 3 ⎣ ⎦0 1 2 3 p − 30 ) = ( ln 3 ln 3 p 3 −1 = 2 3p = 3 ∴ p =1 2(i)
Total area of all the n rectangles, A =
1 ⎛ 1n ⎞ 1 ⎛ n2 ⎞ 1 ⎛ n3 ⎞ 1 ⎛ nn ⎞ L 3 3 3 + + + + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜3 ⎟ n⎝ ⎠ n⎝ ⎠ n⎝ ⎠ n⎝ ⎠
2 3 n ⎞ 1 ⎛ 1n n n n = ⎜ 3 + 3 + 3 +L + 3 ⎟ n⎝ ⎠
⎛ ⎛ 1 ⎞n ⎞ ⎜ ⎜ 3n ⎟ − 1 ⎟ ⎟ 3 ⎜⎝ ⎠ = ⎜ 1 ⎟ n ⎜ n 3 −1 ⎟ ⎜ ⎟ ⎝ ⎠ 1 ⎛ ⎞ 2 ⎜ 3n ⎟ ⎝ ⎠ = ⎛ 1 ⎞ n ⎜ 3 n − 1⎟ ⎝ ⎠ 1 n
Page 1 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
2(ii)
As n → ∞ , limit of A = area under the curve y = 3x for 0 ≤ x ≤ 1 1
= ∫ 3x dx 0
= 3(i)
2 ln 3
uuuv uuuv AB ⋅ OP = ( b − a ) • p
=b•p–a•p = a • p – a • p (since b • p = a • p) =0 Hence, AB is perpendicular to OP. 3(ii)
Since a = b , then P must be the midpoint of AB.
uuur 1 Using ratio theorem, OP = ( a + b ) 2 uuuv uuuv Thus, OD = 2OP ⎛1 ⎞ = 2 ⎜ (a + b ) ⎟ 2 ⎝ ⎠ =a+b
3(iii)
a × b represents the
(1) area of rhombus OADB or OBDA. (or) (2) magnitude of a vector which is perpendicular to a and b.
4(i)
Gradient of PQ =
4(ii)
sin( x1 + θ ) − sin x1 ( x1 + θ ) − x1
=
sin x1 cos θ + cos x1 sin θ − sin x1
=
sin x1 (cos θ − 1) + cos x1 sin θ
θ
θ
(shown)
When θ is small, Gradient of PQ
2
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
=
sin x1 (cos θ − 1) + cos x1 sin θ
θ
⎡⎛ 1 ⎞ ⎤ sin x1 ⎢⎜ 1 − θ 2 ⎟ − 1⎥ + θ ( cos x1 ) ⎣⎝ 2 ⎠ ⎦ ≈
θ
1 = cos x1 − θ sin x1 2
4(iii)
As θ tends to zero,
1 ⎛ ⎞ lim ⎜ cos x1 − θ sin x1 ⎟ gradient of PQ = θ →0 ⎝ 2 ⎠ = cos x1 gradient of tangent at point P =
d ( sin x ) dx x = x1
= cos x1
5
Im (z)
9
× ( −3, 4 )
O
Re (z)
–1
Least exact value of z − 1 = (−3 − 1) 2 + (4 − 0) 2 − 5 = 32 − 5
or
4 2 −5
3
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
w − 2i = w − ai represents the perpendicular bisector y =
a+2 2
Im (z)
9
y=
a+2 2
× ( −3, 4 )
O
Re (z)
–1
For the line to meet the circle more than once, a+2 −1 < <9 2 −2 < a + 2 < 18
∴−4 < a < 16 (ans)
6
y = vx ⇒
dy dv =v+ x dx dx
dy = 3x + y − 2 dx dv ⇒ x(v + x ) = 3 x + vx − 2 dx dv 3 x − 2 ⇒ = x2 dx x
3 2 − dx x x2 2 ⇒ v = 3 ln | x | + + C x ⇒ y = 3 x ln | x | +2 + Cx
⇒ ∫ dv = ∫
4
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
6 (i)
dy = 0 ⇒ y = −3 x + 2 dx
6 (ii)
(ii) C > 0
C <0
2
y = −3 x + 2
Note: (0,2) satisfies the differential equation. (0,2) is a singular solution.
7(a)
x − 2 = A(2 − 2 x) + B By comparing coefficient of x :1 = −2 A ⇒ A = −
1 2
constant: − 2 = 2 A + B ⇒ B = −1
∫
x−2
dx −x + 2x + 8 1 − (−2 x + 2) − 1 dx =∫ 2 − x2 + 2x + 8 1 1 −2 x + 2 dx − ∫ dx =− ∫ 2 2 2 3 − ( x − 1) 2 −x + 2x + 8 =−
2
1 ⎛ x −1 ⎞ (2 − x 2 + 2 x + 8) − sin −1 ⎜ ⎟+C 2 ⎝ 3 ⎠
⎛ x −1 ⎞ = − − x 2 + 2 x + 8 − sin −1 ⎜ ⎟+C ⎝ 3 ⎠ In absence of C, deduct from presentation marks. 5
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
7 (b)
∫e
x
cos 2 x dx = e x cos 2 x − ∫ e x ( −2 sin 2 x ) dx = e x cos 2 x + 2 ⎡⎣ e x sin 2 x − 2 ∫ e x cos 2 x dx ⎤⎦ = e x cos 2 x + 2e x sin 2 x − 4 ∫ e x cos 2 x dx
∴ 5∫ e x cos 2 x dx = e x cos 2 x + 2e x sin 2 x
∫e
x
cos 2 x dx =
∫e
x
cos 2 x dx
1 x e (cos 2 x + 2 sin 2 x) + c (shown) 5
1 x e (1 + cos 2 x) dx 2∫ 1 1 = ∫ e x dx + ∫ e x cos 2 x dx 2 2 1 x 1 x = e + e (cos 2 x + 2 sin 2 x) + c 2 10 =
8(a)
Given Sn = n 2 − 2n .
(
Sn − Sn−1 = n 2 − 2n − ( n − 1) − 2 ( n − 1) 2
(ans)
)
= n 2 − 2n − ( n − 1) + 2n − 2 2
= 2n − 3
Since the progression is AP, common difference, d = Tn − Tn−1 = 2n − 3 − ( 2 ( n − 1) − 3) = 2.
OR d = T2 − T1
= 1 − ( −1) =2
8(b)
Let d be the distance driven for every turn.
(i)
n = 1 , d = kθ ⎛8⎞ n = 2 , d = ⎜ ⎟ kθ ⎝ 10 ⎠
or 0.8kθ
2
⎛ 8⎞ n = 3 , d = ⎜ ⎟ kθ ⎝ 10 ⎠
or 0.64kθ
6
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
(ii)
Total distance driven into the wall after n turns 2
⎛ 8⎞ ⎛ 8⎞ ⎛ 8⎞ = kθ + ⎜ ⎟ kθ + ⎜ ⎟ kθ + ... + ⎜ ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠
⎛ ⎛ 8⎞ ⎛ 8⎞ ⎛ 8⎞ = kθ ⎜ 1 + ⎜ ⎟ + ⎜ ⎟ + ... + ⎜ ⎟ ⎜ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 2
⎛ ⎛ 8 ⎞n ⎜ 1− ⎜ ⎟ 10 = kθ ⎜ ⎝ ⎠ ⎜ ⎛ 8⎞ ⎜ 1 − ⎜ 10 ⎟ ⎝ ⎠ ⎝
(iii)
n −1
n −1
kθ
⎞ ⎟⎟ ⎠
⎞ ⎟ ⎟ ⎟ ⎟ ⎠
⎛ ⎛ 4 ⎞n ⎞ = 5kθ ⎜1 − ⎜ ⎟ ⎟ ⎜ ⎝5⎠ ⎟ ⎝ ⎠ 4 a =5, b = 5 Distance driven in the long run ⎛ ⎛ 4 ⎞n ⎞ = lim 5kθ ⎜1 − ⎜ ⎟ ⎟ ⎜ ⎝5⎠ ⎟ n →∞ ⎝ ⎠ = 5kθ Given k = 2 , minimum length of the metal screw is 10θ unit.
9 (i)
⎛1⎞ ⎜ ⎟ r • ⎜ 3 ⎟ = 3 ⇒ x + 3y + 2z = 3 ⎜ 2⎟ ⎝ ⎠ Π2 : −x + y + 0z = 0 ⎛3⎞ ⎛ 1⎞ ⎜4⎟ ⎜− 2 ⎟ ⎜ ⎟ ⎜ ⎟ 3⎟ 1⎟ ⎜ ⎜ Using GC: l: r = + λ − , λ ∈R ⎜4⎟ ⎜ 2⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎜ 0 ⎟⎟ ⎜⎜ 1 ⎟⎟ ⎝ ⎠ ⎝ ⎠ OR ⎛3⎞ ⎜4⎟ ⎛1⎞ ⎜ ⎟ 3⎟ ⎜ ⎟ ⎜ r= + λ ⎜ 1 ⎟, λ ∈ ⎜4⎟ ⎜ −2 ⎟ ⎜ ⎟ ⎝ ⎠ ⎜⎜ 0 ⎟⎟ ⎝ ⎠
7
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
9 (ii)
For the plane Π 2 :
y = x ⇒ −x + y + 0z = 0 ⎛ −1⎞ ⎜ ⎟ Let n 2 be normal vector to plane Π 2 , then n 2 = ⎜ 1 ⎟ ⎜0⎟ ⎝ ⎠ ⎛ 1 ⎞ ⎛ −1⎞ ⎜ ⎟⎜ ⎟ ⎜ 3⎟ ⎜ 1 ⎟ ⎜ 2⎟ ⎜ 0 ⎟ 2 1 7 (shown) ∴ cos θ = ⎝ ⎠ ⎝ ⎠ = = = 7 14 2 28 7
9 (iii)
Since point F lies on line l, ⎛3⎞ ⎛3 ⎞ ⎜4⎟ ⎜ 4 +λ⎟ ⎛1⎞ ⎜ ⎟ uuur ⎜⎜ 3 ⎟⎟ ⎜ ⎟ ⎜3 Let OF = +λ⎜ 1 ⎟ = + λ ⎟ for some λ . ⎜4⎟ ⎜4 ⎟ ⎜ −2 ⎟ ⎜ ⎜ ⎟ ⎟ ⎝ ⎠ ⎜⎜ 0 ⎟⎟ ⎜⎜ −2λ ⎟⎟ ⎝ ⎠ ⎝ ⎠
⎛3 ⎞ ⎛ 3 ⎞ ⎜ 4 +λ⎟ ⎜ 4 +λ ⎟ ⎟ ⎛0⎞ ⎜ ⎟ uuur ⎜⎜ 3 ⎜ ⎟ ⎜ 1 ⎟ Then AF = + λ −⎜1⎟ = − + λ ⎟ ⎜4 ⎟ ⎜ 4 ⎟ ⎜ ⎟ ⎜⎝ 0 ⎟⎠ ⎜ ⎟ ⎜⎜ −2λ ⎟⎟ ⎜⎜ −2λ ⎟⎟ ⎝ ⎠ ⎝ ⎠ ⎛1⎞ uuur uuur ⎜ ⎟ Now, AF ⊥ l ⇒ AF • ⎜ 1 ⎟ = 0 ⎜ −2 ⎟ ⎝ ⎠
⎛ 3 ⎞ ⎜ 4 +λ ⎟ ⎜ ⎟ ⎛1⎞ ⎜− 1 + λ⎟ • ⎜ 1 ⎟ = 0 ⎜ 4 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ −2 ⎟⎠ ⎜⎜ −2λ ⎟⎟ ⎝ ⎠ 1 λ=− 12
8
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
⎛3 1 ⎞ ⎛2⎞ ⎜ 4 − 12 ⎟ ⎜ 3 ⎟ uuur ⎜⎜ 3 1 ⎟⎟ ⎜⎜ 2 ⎟⎟ − = OF = ⎜ 4 12 ⎟ ⎜ 3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎜ 1 ⎟⎟ ⎜⎜ 1 ⎟⎟ ⎝ 6 ⎠ ⎝6⎠ ⎛ 3 1 ⎞ ⎛ 2 ⎞ ⎜ 4 − 12 ⎟ ⎜ 3 ⎟ uuur ⎜⎜ 1 1 ⎟⎟ ⎜⎜ 1 ⎟⎟ = − AF = − − ⎜ 4 12 ⎟ ⎜ 3 ⎟ ⎟ ⎜ ⎟ ⎜ 1 ⎜⎜ ⎟⎟ ⎜⎜ 1 ⎟⎟ 6 ⎝ ⎠ ⎝ 6 ⎠ 2 2 2 uuur 7 ⎛ 2⎞ ⎛ 1⎞ ⎛1⎞ AF = ⎜ ⎟ + ⎜ − ⎟ + ⎜ ⎟ = 12 ⎝ 3⎠ ⎝ 3⎠ ⎝ 6⎠
Π1
A
l
θ
F
Π2
uuur Hence, exact length of projection from AF to the plane Π 2 uuur = AF cos θ
=
7 ⎛ 7⎞ ⎜ ⎟ 12 ⎜⎝ 7 ⎟⎠
=
1 1 or = 12 2 3
3 6
OR uuur exact length of projection from AF to the plane Π 2
uuur n = AF × 2 n2
9
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
9 (iv)
⎛3⎞ ⎜4⎟ ⎛1⎞ uuur ⎜⎜ 3 ⎟⎟ ⎜ ⎟ OP = + λ ⎜ 1 ⎟ , for some λ ∈ ⎜4⎟ ⎜ −2 ⎟ ⎜ ⎟ ⎝ ⎠ 0 ⎜⎜ ⎟⎟ ⎝ ⎠
Π 3 has equation px + qy = 1 . ⎛ p⎞ ⎜ ⎟ ∏3 : r • ⎜ q ⎟ = 1 ⎜0⎟ ⎝ ⎠ For the three planes to intersect exactly a point, l is not parallel to Π 3 , then: ⎛ 1 ⎞ ⎛ p⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟•⎜ q ⎟ ≠ 0 ⎜ −2 ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠
⇒ p+q ≠ 0 ∴ p ≠ − q (ans) 10(a)
z 2 + (a − i) z * + 16 + bi = 0
( 2 + 3i )
2
+ (a − i) ( 2 − 3i ) + 16 + bi = 0
⇒ −5 + 12i + 2a − 3ai − 2i − 3 + 16 + bi = 0 ⇒ 8 + 2a + (10 − 3a + b ) i = 0 By comparing real and imaginary coefficient,
Real: 8 + 2a = 0 ⇒ a = −4 (ans) Im :10 − 3a + b = 0 ⇒ b = −22 (ans) 10(b)
z5 +1 = 0 z 5 = −1 z 5 = eiπ
z 5 = ei( 2 k +1) π z=e
i ( 2 k +1) π 5
, k = 0, ±1, ± 2
10
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1 iπ
z = −1, e 5 , e
−
iπ 5
i3π
,e 5 ,e
−
i3π 5
Im( z )
1
×
× × –1
O
×
1
Re( z )
× –1
(i)
1+ e
iθ
iθ
iθ ⎞ ⎛ −iθ2 = e ⎜e + e 2 ⎟ ⎝ ⎠ 2
iθ
⎛ ⎛ θ⎞ ⎛ θ⎞ ⎛θ ⎞ ⎛θ ⎞⎞ = e 2 ⎜ cos ⎜ − ⎟ + i sin ⎜ − ⎟ + cos ⎜ ⎟ + i sin ⎜ ⎟ ⎟ ⎝ 2⎠ ⎝ 2⎠ ⎝2⎠ ⎝ 2 ⎠⎠ ⎝ θ
⎛θ ⎞ i = 2 cos ⎜ ⎟ e 2 (shown) ⎝2⎠
(ii)
Replace complex number z with w − 1 ,
z=e
i ( 2 k +1) π 5
⇒ w −1 = e ∴ w = 1+ e
i( 2 k +1) π 5
i ( 2 k +1) π 5
From (i), ∴ w = 1+ e
i( 2 k +1) π 5
⎛ ( 2k + 1) π ⎞ i = 2 cos ⎜ ⎟e 10 ⎝ ⎠
( 2k +1) π 10
for k = 0, ±1, ± 2 .
11
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
11
ax 2 + bx + 1 1 − bc + ac 2 = (ax + b − ac) + x+c x+c ax + b − ac = 2 x − 1
f(x) =
∴ a = 2 (ans) and b − ac = −1 ⇒ b = 2c − 1 (shown)
(i)
Given c = 1, f(x ) = 2 x − 1 +
2 x +1
2 x +1 ( x + 1) y = (2 x − 1)( x + 1) + 2
Let y = 2 x − 1 +
2 x 2 + (1 − y ) x + (1 − y ) = 0
For all real values of x, D ≥ 0 ⇒ (1 − y ) − 4(2)(1 − y ) ≥ 0 ⇒ ( y + 7)( y − 1) ≥ 0 ⇒∴ y ≤ −7 or y ≥ 1 Hence, y cannot lie between -7 and 1. 2
(ii) y y = 2x −1
( 0,1) x
( −2, −7 )
x = −1
f is concaving downwards for x < −1 .
12
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
2 x 2 + x + 1 = (kx + k − 3)( x + 1)
(iii) ⇒
2x2 + x + 1 = kx + k − 3 x +1
The line kx + (k − 3) passes through the point (−1, −3) , which is the intersection of the asymptotes. Since the oblique asymptote passes through the point (−1, −3) and using the graph in (ii), the gradient of the line kx + k − 3 has to be more than 2 for the above equation to have 2 real solutions. Hence, k > 2 .
12
(18, 29 )
(0 , 2 )
When λ = 0,
x = 2 ( 0 ) = 0 ; y = ( 03 ) + 2 = 2
When λ = 3,
x = 2 ( 3) = 18 ; y = ( 3) + 2 = 29
2
2
3
dx dy = 4λ , = 3λ 2 dλ dλ
dy 3λ 2 3λ = = dx 4λ 4 Equation of tangent:
13
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
3λ ( x − 2λ 2 ) 4 4 y − 4 ( λ 3 + 2 ) = 3λ x − 6λ 3 y − (λ 3 + 2) =
4 y = 3λ x − 6λ 3 + 4λ 3 + 8 4 y = 3λ x + 8 − 2λ 3 4 y = 3λ x + 2 ( 4 − λ 3 ) (shown)
(a) (i)
P( 2λ2,λ3 +2)
R ( 0, 2 ) ⎛ 4 − λ3 ⎞ Q ⎜ 0, ⎟ 2 ⎝ ⎠
When x = 0, 4 y = 3λ ( 0 ) + 2 ( 4 − λ 3 ) = 2 ( 4 − λ 3 ) ⇒ y =
4 − λ3 2
⎛ 4 − λ3 ⎞ Coordinates of Q is ⎜ 0 , ⎟. 2 ⎠ ⎝
Area of triangle PQR, A=
(ii)
Given:
A=
λ5 2
3 1⎛ 4 − λ3 ⎞ λ5 2 2 ⎛ 4−4+λ ⎞ 2 2 λ λ = − = ) ⎜ 2 ⎟ 2 units2 ⎜ ⎟( 2⎝ 2 ⎠ ⎝ ⎠
dx = 4 when λ = 2 . dt ⇒
dA 5λ 4 = dλ 2
14
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
By chain rule, (Find ∴
dA dA dλ . = ⋅ dt dλ dt
dλ dx dx dλ dλ ) For λ = 2 , = ⋅ ⇒ 4 = 4 ( 2) ⋅ dt dt dλ dt dt
dλ 1 = dt 2
dA 5 ( 2 ) = ( 0.5) = 20 units2 /sec . dt 2 4
(b)
R ( 0, 2 )
P ( 8 , 10 )
Q ( 0, −2 )
For λ = 2 ，equation of tangent at P is 2 y = 3x − 4. Coordinates of P is ( 8,10 ) . Coordinates of Q is ( 0, −2 ) .
Method 1 Area 8
8
0
0
= ∫ y dx − ∫
1 ( 3x − 4 ) dx 2 8
⎤ 1 ⎡ 3x 2 = ∫ ( λ + 2 ) ( 4λ ) dλ − ⎢ − 4x⎥ 0 2⎣ 2 ⎦0 2
2
3
= ∫ 4λ 4 +8λ dλ − 0
1 [96 − 32 − 0] 2
15
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1 2
⎡ 4λ 5 ⎤ =⎢ + 4λ 2 ⎥ − 32 ⎣ 5 ⎦0 ⎡ 208 ⎤ =⎢ − 0 ⎥ − 32 ⎣ 5 ⎦ 48 3 or 9 or 9.6 (ans) = 5 5
OR Method 2 10 1 2 y + 4 ) dy − ∫ x d y ( −2 3 2
Area of region = ∫
10
Area bounded by the tangent at P and y-axis 1 ( 2 y + 4 ) dy −2 3 10 1 = ⎡⎣ y 2 + 4 y ⎤⎦ −2 3 1 = (140 − ( −4 ) ) 3 = 48 =∫
10
Area bounded by the curve and y-axis 10
= ∫ x dy 2
2
2 ⎡6 ⎤ = ∫ 2λ 2 ( 3λ 2 ) dλ = ⎢ λ 5 ⎥ 0 ⎣ 5 ⎦0 2 4 = ∫ 6λ dλ 0
6 ( 32 − 0 ) 5 192 = 5 =
Area of region = 48 −
192 48 3 = or 9 or 9.6 (ans) 5 5 5
16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
1(i)
⎛ 1 5⎤ Rg = ⎜ , ⎥ and Dh = [ 0, ∞ ) . ⎝ 2 2⎦ Since Rg ⊆ Dh , thus hg exists.
⎛ 1 5⎤ ⎡ 1 3⎤ Dh g = Dg = ( −∞, 0] → ⎜ , ⎥ → ⎢ − , ⎥ = Rhg ⎝ 2 2⎦ ⎣ 4 4⎦ OR any appropriate method such as graphical method (ii)
For inverse of function h to exist, b = 1.5
(ii)
y=x
y y=h
−1
( x) y = h (x )
( –0.25,1.5)
x 0
(iii)
(1.5, –0.25)
To solve for exact value of h −1 ( x) = x is same as solving h( x) = x , x 2 − 3x + 2 = x x2 − 4x + 2 = 0
∴x =
4±
( −4 )
2
− 4 (1)( 2 )
2 4± 8 2 = 2 − 2 (rejected Q x ≥ 1.5) or =
2+ 2
From sketch above, for h −1 (x) ≤ x , x ≥ 2 + 2 .
2(a)
As n → ∞ , xn → L , xn +1 → L . xn +1 = 3 − ( ln ( xn + 1) )
2
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 ⇒ L = 3 − ( ln ( L + 1) )
2
⇒ 3 − ( ln ( L + 1) ) − L = 0 2
Using GC to solve 3 − ( ln ( L + 1) ) − L = 0 , 2
L = −0.860 (rejected as xn > 0 )
or L = 1.8806225 ≈ 1.88 (3sf) 2 (b)
r 2 + 7 r + 11 A B = + ( r + 4 )! ( r + 2 )! ( r + 4 )! r 2 + 7 r + 11 = A ( r + 3)( r + 4 ) + B When r = −3, B = −1 When r = 0, A = 1 Hence,
r 2 + 7r + 11 1 1 . = − ( r + 4 )! ( r + 2 )! ( r + 4 )!
⎛ r 2 + 7 r + 11 ⎞ n ⎛ 1 1 ⎞ − ⎟⎟ = ∑ ⎜⎜ ⎟⎟ r =1 ⎠ r =1 ⎝ ( r + 2 ) ! ( r + 4 ) ! ⎠ 1 1 = − 3! 5! 1 1 + − 4! 6! 1 1 + − 5! 7! 1 1 + − 6! 8! n
∑ ⎜⎜ ( r + 4 )! ⎝
+
M 1 1 + − ( n − 1)! ( n + 1)! +
1 1 − n ! ( n + 2 )!
+
1 1 − ( n + 1)! ( n + 3)!
+
1 1 − ( n + 2 )! ( n + 4 )! Page 2 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 =
(ii)
1 1 1 1 + − − 3! 4! ( n + 3) ! ( n + 4 ) !
=
4 +1 n + 4 +1 − 4! ( n + 4 ) !
=
5 n+5 − 4! ( n + 4 ) !
Let Pn be the proposition that
⎛ r 2 + 7 r + 11 ⎞ 5 n+5 for all n ∈ ⎜ ⎟⎟ = − ∑ ⎜ ( r + 4 )! ⎠ 4! ( n + 4 )! r =1 ⎝ n
Consider P1: RHS of P1 =
5 1+ 5 5 6 25 − 6 19 − = − = = . 4! (1 + 4 ) ! 4! 5! 5! 5!
LHS of P1 =
⎛ r 2 + 7 r + 11 ⎞ 12 + 7 + 11 19 = = RHS of P1 ⎜⎜ ⎟= ∑ ( r + 4 )! ⎟⎠ (1 + 4 )! 5! r =1 ⎝ 1
Hence, P1 is true. Assume Pk is true for some k ∈ ie.
+
.
⎛ r 2 + 7 r + 11 ⎞ 5 k +5 ⎜ ⎟⎟ = − ∑ ⎜ ( r + 4 )! ⎠ 4! ( k + 4 )! r =1 ⎝ k
Consider Pk+1: RHS of Pk+1
=
5 k +1+ 5 5 k +6 . − = − 4! ( k + 1 + 4 ) ! 4! ( k + 5 ) !
LHS of Pk+1
=
⎛ r 2 + 7 r + 11 ⎞ ⎜⎜ ⎟ ∑ ( r + 4 )! ⎟⎠ r =1 ⎝ k +1
⎛ r 2 + 7 r + 11 ⎞ ( k + 1) + 7 ( k + 1) + 11 = ∑ ⎜⎜ ⎟+ ( r + 4 )! ⎟⎠ ( k + 1 + 4 )! r =1 ⎝ k
2
5 k +5 k 2 + 2k + 1 + 7 k + 7 + 11 = − + 4! ( k + 4 ) ! ( k + 5)! =
5 k 2 + 10k + 25 − k 2 − 2k − 1 − 7 k − 7 − 11 − 4! ( k + 5)!
=
5 k +6 = RHS of Pk+1 − 4! ( k + 5 ) ! Page 3 of 16
+
.
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 Hence, Pk is true implies Pk+1 is also true. Since P1 is true and Pk is true implies Pk+1 is also true, by Mathematical Induction, Pn is true for all n ∈ + . (iii)
Let r = j – 1. Hence, we have ⎛ ( j − 1)2 + 9 ( j − 1) + 19 ⎞ ⎟ ∑ ⎜⎜ ⎟ j 1 5 ! − + ( ) j −1=1 ⎝ ⎠ ∞ ⎛ 2 j − 2 j + 1 + 9 j − 9 + 19 ⎞ = ∑ ⎜⎜ ⎟⎟ ( j + 4 )! j =2 ⎝ ⎠ ∞ ⎛ 2 j + 7 j + 11 ⎞ = ∑ ⎜⎜ ⎟ ( j + 4 )! ⎟⎠ j =2 ⎝ ∞ ⎛ 2 r + 7 r + 11 ⎞ = ∑ ⎜⎜ ⎟, since j is a dummy variable ( r + 4 )! ⎟⎠ r =2 ⎝
⎛ r 2 + 9r + 19 ⎞ ⎜⎜ ⎟= ∑ ( r + 5)! ⎟⎠ r =1 ⎝ ∞
∞
∞ ⎛ 2 r + 7 r + 11 ⎞ 12 + 7 + 11 = ∑ ⎜⎜ ⎟− ( r + 4 )! ⎟⎠ (1 + 4 )! r =1 ⎝ ⎛5 n + 5 ⎞ 19 = lim ⎜⎜ − ⎟− ( n + 4 )! ⎟⎠ 5! n →∞ ⎝ 4! 5 19 = − 4! 5! 5 19 = − 4! 4!( 5 )
=
3
6 5!
or
1 20
Let y = 2 esin x
y 2 = 4esin x dy = 4esin x cos x = y 2 cos x 2y dx dy = y cos x (shown) 2 dx
Page 4 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 d 2 y dy cos x − y sin x = dx 2 dx d3 y d 2 y dy dy ⇒ 2 3 = 2 cos x − sin x − sin x − y cos x dx dx dx dx 3 2 d y d y dy ∴ 2 3 = 2 cos x − 2 sin x − y cos x dx dx dx ⇒2
When x = 0, y = 2 dy =1 dx d2 y 1 = dx 2 2 d3 y 3 =− 3 dx 4 ⎛1⎞ 2 ⎛ 3⎞ 3 ⎜ ⎟x ⎜− ⎟x 2 4⎠ ∴ y ≈ 2+ x+ ⎝ ⎠ + ⎝ 2! 3! x 2 x3 y = 2+ x+ − (ans) 4 8
Method 1
esin x =
y2 1 x 2 x3 ≈ (2 + x + − ) 2 4 4 4 8 1 x 2 x3 e− sin x = esin( − x ) ≈ [2 − x + + ]2 (Replace x by − x) 4 4 8 1 x 2 x3 = [2 − ( x − − )]2 4 4 8 2 ⎛ ⎞ ⎛ 1⎡ x2 x 2 x3 ⎞ ⎤ = ⎢ 4 − 4 ⎜ x − + ... ⎟ + ⎜ x − − ⎟ ⎥ 4⎢ 4 4 8 ⎠ ⎥ ⎝ ⎠ ⎝ ⎣ ⎦ 1 ≈ ⎡⎣ 4 − 4 x + x 2 + x 2 ⎤⎦ 4 1 = ( 4 − 4x + 2x2 ) 4 x2 = 1− x + (ans) 2
OR Method 2
esin x =
y2 1 x 2 x3 ≈ (2 + x + − ) 2 4 4 4 8 Page 5 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 e − sin x = ( esin x ) ⎛ y2 ⎞ =⎜ ⎟ ⎝ 4 ⎠
−1
−1
x 2 x3 −2 − ) 4 8 x x 2 x3 = (1 + + − ) −2 2 8 16 = 4(2 + x +
⎛ x x 2 x3 ⎞ ( −2 ) ( −3 ) ⎛ x x 2 x3 ⎞ = 1− 2 ⎜ + − ⎟ + ⎜ + − ⎟ +L 2! ⎝ 2 8 16 ⎠ ⎝ 2 8 16 ⎠ 2
⎛ x2 ⎞ x2 + 3 ⎜ ⎟ +L 4 ⎝ 4 ⎠ 2 x (ans) = 1− x + 2 = 1− x −
4(i)
x = 2 tan θ ⇒
∫
4 − x2
(4 + x )
2 2
dx = 2 sec 2 θ dθ
dx = ∫ =∫
4 − 4 tan 2 θ
( 4 + 4 tan θ ) 2
4 − 4 tan 2 θ
( 4sec θ ) 2
2
( 2sec θ ) dθ 2
2
( 2sec θ ) dθ 2
4 − 4 tan 2 θ dθ 8sec 2 θ 1 1 tan 2 θ = ∫ − dθ 2 sec 2 θ sec 2 θ 1 = ∫ cos 2 θ − sin 2 θ dθ 2 =∫
1 cos 2θ dθ 2∫ 1⎛1 ⎞ = ⎜ sin 2θ ⎟ + C 2⎝2 ⎠ 1 = ( sin θ cos θ ) + C 2 1⎛ x ⎞⎛ 2 ⎞ = ⎜ ⎟⎜ ⎟+C 2 2 ⎝ 4 + x ⎠ ⎝ 4 + x2 ⎠ x = +C 2 4 + x =
Page 6 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 (ii) (a)
Volume of region rotated about x-axis, V 2
⎛ 1 =π∫ ⎜ −2 ⎜ ⎝ = π∫
4− x
(4 + x )
2 2
4 − x2
1
−2
2
(4 + x )
2 2
2 ⎞ ⎡ ⎟ − 3 ( x + 2 ) ⎤ dx ⎥ ⎟ ⎢⎣ 15 ⎦ ⎠ 2
⎛ 3⎞ 2 − ⎜⎜ ⎟⎟ ( x + 2 ) dx 15 ⎝ ⎠ 1
3 ⎡ x ⎛ 3 ⎞ ( x + 2) ⎤ = π⎢ −⎜ ⎥ ⎟ 2 ⎢⎣ 4 + x ⎝ 225 ⎠ 3 ⎥⎦ −2
⎡ 1 ⎛ 3 ⎞ ⎛ 27 ⎞ ⎛ −2 ⎞ ⎤ = π⎢ −⎜ ⎟ ⎜ ⎟ − ⎜ ⎟⎥ ⎣ 5 ⎝ 225 ⎠ ⎝ 3 ⎠ ⎝ 8 ⎠ ⎦ 33 = π (ans) 100
∴b =
33 100
(ii) (b)
y=
y
–2
O
dy dx
2
x
Section B: Statistics [60 marks] 5 (i) (ii)
It is the list of all the 700 graduating students. Arrange the list of all 700 students in some order (can be by surname, class, other reasonable category.)
Page 7 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 Calculate the interval to take samples from to obtain the 140 samples. ie.
700 =5. 140
From the first group of 5 students, select the first student using random sampling. Then select every 5th student after that. Note: Students may say they use random sampling to pick anyone in the list as the first student, but they will then need to qualify that they will need to cycle back to the names at the front of the list if it reaches the end before getting all the 140 students.
6(i)
s
(ii)
Using GC,
α = −4.975 ≈ −4.98 , β = 2.111 ≈ 2.11 r = 0.951 (to 3 sf)
Since the value of r is close to 1, it suggests an almost linear relationship. Hence a linear model is appropriate.
(iii)
When d = 28 ,
s = −4.97519 + 2.110569 ( ln 28) s = 2.05766 ≈ 2.06 or 2.058 Thus average sales is 2060 or 2058 bottles. The answer is reliable since based on the new model, the value of r is close to 1 and that suggests a linear relationship. Furthermore we are predicting s based on d which is within the range.
Page 8 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 7(i)
Number of ways = 9! = 362880 (ans)
(ii)
Number of ways to arrange the couple among themselves = 2! Number of ways to arrange 4 couples and 2 children = Number of ways =
6! 6
6! (2!)4 = 1920 (ans) 6
⎛8⎞ Number of ways to select 2 adults to in the front row = ⎜ ⎟ ⎝ 2⎠ Number of ways to arrange the bride and groom =2! Number of ways to arrange the children and two adults in the front row =4! Number of ways to arrange the 6 adults in the back row =6!
⎛8⎞ Number of arrangements = ⎜ ⎟ ( 4!)( 2!)( 6!) = 967680 (ans) ⎝ 2⎠ OR (iii)
Alternative Solution
⎛ 4⎞ Number of ways to arrange the children front row = ⎜ ⎟ ( 2!) ⎝ 2⎠ Number of ways to arrange the bride and groom =2! Number of ways to arrange the 8 adults = 8! ⎛ 4⎞ Number of arrangements = ⎜ ⎟ ( 2!)( 2!)( 8!) = 967680 (ans) ⎝ 2⎠
8(a)
A
(i)
B 2 15
P ( A) = P ( A ∪ B ) − P ( A ' ∩ B )
3 2 − 4 15 37 = 60 =
Page 9 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 Since events A and B are independent, P ( A '∩ B ) = P ( A ') P ( B ) =
2 15
(1 − P ( A) ) P ( B ) = 152 2 ⎛ 37 ⎞ ⎜1 − ⎟ P ( B ) = 15 ⎝ 60 ⎠ P ( B) = OR
8 23
Alternative solution P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) 3 37 37 = + P ( B) − P ( B ) 4 60 60 23 2 ⇒ P ( B) = 60 15 8 (ans) ∴P ( B) = 23 ⇒
(ii)
P ( A ∩ B A ∪ B) =
=
=
P (( A ∩ B ) ∩ ( A ∪ B )) P ( A ∪ B)
P ( A ∩ B) P ( A ∪ B) P ( A) × P ( B ) P ( A ∪ B)
(OR: =
P ( B ) − P ( A' ∩ B ) P ( A ∪ B)
⎛ 37 ⎞⎛ 8 ⎞ ⎜ ⎟⎜ ⎟ 60 23 = ⎝ ⎠⎝ ⎠ 3 4 =
296 1035
Page 10 of 16
)
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 8(b)
P(all 3 boys in the same group)
(i)
⎛ 3 ⎞⎛ 5 ⎞⎛ 4 ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 3 ⎠⎝ 1 ⎠⎝ 4 ⎠ 2! = ⎛ 8 ⎞⎛ 4 ⎞ ⎜ ⎟⎜ ⎟ ⎝ 4 ⎠⎝ 4 ⎠ 2! =
(ii)
5 1 = (ans) 70 14
P(either Ivy or Tamie is in the same group as Cassy) = P(Ivy together with Cassy but not Tamie) + P(Tammy together with Cassy but not Ivy) + P(Ivy and Tamie together with Cassy) ⎛ 5 ⎞⎛ 3 ⎞ ⎛ 5 ⎞⎛ 3 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 1 ⎠ ⎝ 4 ⎠ 2! + 2! + 2! = ⎛ 8 ⎞⎛ 4 ⎞ ⎛ 8 ⎞⎛ 4 ⎞ ⎛ 8 ⎞⎛ 4 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ 4 ⎠⎝ 4 ⎠ ⎝ 4 ⎠⎝ 4 ⎠ ⎝ 4 ⎠⎝ 4 ⎠ 2! 2! 2! =
25 5 = 70 14
(ans)
OR (either Ivy or Tamie is in the same group as Cassy) = P(Ivy together with Cassy) + P(Tammy together with Cassy) – P(Ivy and Tamie together with Cassy) ⎛ 6 ⎞⎛ 4 ⎞ ⎛ 6 ⎞⎛ 4 ⎞ ⎛ 5 ⎞⎛ 4 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 4 ⎠ ⎝ 2 ⎠⎝ 4 ⎠ ⎝ 1 ⎠⎝ 4 ⎠ 2! + 2! − 2! = 25 = 5 = 8 4 8 ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ 4 ⎞ ⎛ 8 ⎞⎛ 4 ⎞ 70 14 ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ 4 ⎠⎝ 4 ⎠ ⎝ 4 ⎠⎝ 4 ⎠ ⎝ 4 ⎠⎝ 4 ⎠ 2! 2! 2!
(ans)
Page 11 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 9(i)
Let X be the number of arrivals at the airport in a two-hour period. X ~ Po ( 8 ) P ( X ≥ 13) = 1 − P ( X ≤ 12 ) = 0.063797 ≈ 0.0638
(ii)
Let W be the number of arrivals in a one-hour period. Let Y be the number of departures in a one-hour period. W ~ Po ( 4 ) ; Y ~ Po ( 3) ; W + Y ~ Po ( 7 )
P (Y < 2 | W + Y = 9 ) =
P (Y < 2 ∩ W + Y = 9 ) P (W + Y = 9 )
=
P (Y = 0, W = 9 ) + P (Y = 1, W = 8 ) P (W + Y = 9 )
=
P (Y = 0 ) P (W = 9 ) + P (Y = 1) P (W = 8 ) P (W + Y = 9 )
0.0051052524 0.1014046695 = 0.0503453384 ≈ 0.0503 (ans) =
(iii)
(1)There are two mutually exclusive outcomes – either there are at least 13 arrivals in each two-hour period or there isn’t. (2)The probability of having at least 13 arrivals for each two hour period remains constant for each of the 60 two-hour periods. (3)There is a fixed number of 60 two-hour periods independently selected under consideration.
(iv)
Notice that we are unable to define the random variable as the number of two-hour periods, out of 60, with less than 13 arrivals each as it will not be possible to do any approximations. np = 56.172 ( > 5) and nq = 3.828 ( < 5)
Let V be the number of two-hour periods, out of 60, with at least 13 arrivals each. V ~ B ( 60, P ( X ≥ 13) )
Since n = 60 ( > 50, large) and p = P ( X ≥ 13) = 0.0638 ( < 0.1, small), such that np = 3.828
Page 12 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 ( < 5), we have V ~ Po ( 3.828 ) approximately.
At most 50 two-hour periods with less than 13 arrivals each means the same as at least 10 two-hour periods with at least 13 arrivals each.
P (V ≥ 10 ) = 1 − P (V ≤ 9 ) = 0.0060899731 = 0.00609
Note:
Students may choose to use the more accurate value for P ( X ≥ 13) = 0.0637971966 . If they do so, the following values will be obtained:
np = 3.827831796 V ~ Po ( 3.827831796 ) P (V ≥ 10 ) = 0.0060881936 = 0.00609.
This will still be considered as correct and will be awarded the necessary marks.
10
Let X denote the monthly electricity usage of a 3-room unit. X
N ( 290, σ 2 )
Let Y denote the monthly electricity usage of a 5-room unit. Y
(i)
N ( 450,105 )
P ( X > 290 ) × P ( X > 290 ) × P (Y < 450 ) = 0.5 × 0.5 × 0.5 = 0.125 (ans)
(ii)
X1 + X 2 + X 3 + X 4
N (1160, 4σ 2 )
P ( X 1 + X 2 + X 3 + X 4 > 3 ( 380 ) ) = 0.868 ⇒ P ( X 1 + X 2 + X 3 + X 4 < 1140 ) = 0.132 ⎛ 1140 − 1160 ⎞ ⇒ P⎜Z < ⎟ = 0.132 4σ 2 ⎝ ⎠ Page 13 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 ⇒
1140 − 1160 4σ 2
= −1.1169867
20 1.1169867 ⇒ σ = 8.9526579 ⇒ 2σ =
∴σ 2 = 80.1500827 ≈ 80 (ans) (iii)
Let C denote the monthly electricity charge of a 5-room unit. C = 0.24Y ∴C
N ( 0.24(450), 0.242 (105) )
N (108, 6.048 )
P ( C > a ) = 0.9 ⇒ 1 − P ( C ≤ a ) = 0.9 ⇒ P ( C ≤ a ) = 0.1
∴ a = 104.848321 ≈ 104.85 (ans)
(iv)
Let W denote the number of 5-room units with monthly electricity bill exceeding $a, out of eighty 5-room units. W
B ( 80, 0.9 )
Since n = 80 is large and np = 80(0.9) = 72( > 5), nq = 80(0.1) = 8( > 5) , ∴W
N ( 72, 7.2 ) approx
P (W ≥ 70 ) = P (W > 69.5 ) = 0.8242529 ≈ 0.824 (ans)
11(a)
x=
3791 = 75.82 50
s2 =
1 ⎡ 37912 ⎤ 525.38 26269 287959 − = = = 10.72204 ⎢ 49 ⎣ 50 ⎥⎦ 49 2450
Let μ be the mean drying times of Noppin paint. Page 14 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 Test
H0:
μ = 75 against H1:
μ ≠ 75
Level of Significance: 5% (2-tailed Z-test) Test Statistic: Z =
X −μ s n
~ N ( 0 ,1) approximately.
Method 1: Using critical region and observed test statistic, zcalculated
Critical region: z > 1.95996
zcalculated =
75.82 − 75 10.72204 50
= 1.77076
0.05
−1.95996
1.95996
0
1.77076
Since zcalculated = 1.77076 < 1.95996 , we do not reject H0.
Method 2: Using p-value
p-value = 0.07660
Since p-value = 0.07660 > 0.05, we do not reject H0.
There is insufficient evidence at 5% level of significance to claim that the mean drying times of the Noppin paint has changed. Hence, the manufacturers will not discard the entire affected batch of paint. 11(a) Since X is unknown and n = 60 is large, (ii)
by Central Limit Theorem, X1 + X 2 + X 3 + L + X 60
N ( 60 × 75, 60 ×10.72204082) approx .
Page 15 of 16
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2 P ( 60 ( 72 ) < X 1 + X 2 + X 3 + L + X 60 < 60 ( 76 ) ) P ( 4320 < X 1 + X 2 + X 3 + L + X 60 < 4560 ) = 0.990999 ≈ 0.991(ans) (b)(i) Test H0:
μ = 75 against
μ < 75
H1:
Level of Significance: 5% (lower-tailed) s2 =
n 20 20 2 σ n 2 = ( 7.5 ) ⇒ s = 7.5 = 7.69484 19 19 n −1
Test Statistic: T =
X −μ s n
~ t (19 )
0.05 tcalculated =
x − 75 7.69484 20
−1.72913
0
For the chemist to claim that the new additive had been effective, we need to reject H0 in favour of H1. Hence, we need tcalculated < −1.72913 . x − 75 7.69484 20
< −1.72913
x < 72.02482 Largest value of x = 72.0 (to 3 sig. fig.)
(ii)
It is necessary to assume that the drying time of Noppin paint, X, follows a normal distribution.
Page 16 of 16
NANYANG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Higher 2
MATHEMATICS
9740/01 16th September 2010
Paper 1
3 Hours Additional Materials:
Answer Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages. NANYANG JUNIOR COLLEGE Internal Examinations © NYJC 2010
[Turn Over
2
1
Find the equation of the quadratic function which has a line of symmetry at x the points (1,9) and (1,5) .
2
3
[4]
The complex numbers z and w are 1 + ai and b – 2i respectively where a and b are real and a is negative. Given that zw* = 8i, find the exact values of a and b.
[3]
Find the smallest positive integer value of n such that wn is purely imaginary.
[2]
Find the expansion of
1 4x
in ascending powers of x, up to and including the term in x 2 . Hence
find an approximate value of the integral
2 0
1 dx , expressing your answer in the form 4 2x
a b 2 , where a and b are constants to be determined.
4
1 and passes through 3
[6]
d2 x dx The differential equation 2 9 x 3t is such that x = 0, = 1 when t = 0. Find the Maclaurin’s dt dt
series for x up to and including the t 3 term.
[3]
It is known that this differential equation has a general solution of the form x = Asin3t +Bt, where A and B are constants. Assuming that t is sufficiently small for terms with powers of 4 and above to be neglected, find the values of A and B.
NYJC 2010 JC2 Preliminary Examination
[4]
9740/01
5
3 With respect to the origin O, the points A, B and C have position vectors a, b and c respectively, and
are such that OACB is a parallelogram in an anti-clockwise sense. (i)
Express c in terms of a and b.
[1]
(ii)
Show that the area of parallelogram OACB is given by a b .
[2]
(iii)
Show that the maximum area of parallelogram OACB is a b .
[2]
(iv)
b1 a1 By considering a·b where a a2 and b b2 , show that b a 3 3
(a1b1 a2b2 a3b3 ) 2 a12 a22 a32 b12 b22 b32 .
6
The diagrams below show the graphs of y = |f(x)| and y =
f ( x) for x
[3]
. The point A, B and A’ has
the coordinates of (1, 2) , (1,1.5) and (1, 2) respectively.
y f ( x)
y
A (1, 2)
y
y f ( x)
B (1,1.5) A ' (1, 2) x
x
On separate diagrams, sketch the graphs of, (i)
y f ( x) ,
[2]
(ii)
y
1 , f ( x)
[3]
(iii)
y f '( x) ,
[3]
showing clearly any asymptotes and the coordinates of any stationary point(s).
NYJC 2010 JC2 Preliminary Examination
9740/01
[Turn Over
7
4 Functions f and g are respectively defined on the domain of real numbers by
f : x x2 2x 2
, x > 1,
g: x x3
, x –3.
(i)
By considering the derivative of f(x), prove that f is a one-one function.
[2]
(ii)
Solve the equation f(x) = f 1 x where f 1 x denotes the inverse function of f(x).
[2]
(iii)
Deduce the solution set to the inequality f x f 1 x .
[2]
(iv)
Show that the composite function gf exists and define it in a similar form. State also its range.
8
[3]
A curve C is defined by the parametric equations x 2t ,
y 3et , where t . 2
(i)
Sketch the curve C.
(ii)
Given that the point P lies on the curve with coordinates (2, 3e), show that the equation of the
[1]
tangent to the curve at point P is y 3e( x 1) . (iii)
[2]
The normal to the curve at point P cuts the x-axis at point Q. Given also that the tangent to the curve at point P cuts the y-axis at point R, find the coordinates of Q and R. Hence, show that the area of PQR is 3e(1 9e 2 ) units2.
(iv)
State the range of values of m for which the line y mx 3e intersects the curve at 2 distinct points.
9
[5]
[1]
Given that f ( x) ax 1
a 1 , x , a 0. 4x 1 4
(i)
Find the equation of the asymptote(s) of y f ( x) .
[2]
(ii)
Find the coordinates of the turning point(s) for y f ( x) .
[4]
(iii)
(a)
Sketch the graph of y f ( x) , for 0 < a < 1, indicating clearly the equations of the asymptote(s), and the coordinates of any turning point(s) and axial intercept(s).
(b)
[3]
Determine the range of values of m such that there will be intersection between the 1 1 line y a 1 = m x and y = f(x). 4 4
NYJC 2010 JC2 Preliminary Examination
9740/01
[1]
5 10 An innovation is introduced into a community of 100 farmers at time t 0 . Let x denote the number
of farmers who have adopted the innovation at time t. Assume that x is a continuous function of time. The rate at which the number of farmers in that community who adopted the innovation at a particular instant is proportional to the product of the number of farmers who have already adopted and the number of farmers who have not adopted the innovation.
Initially, one farmer adopted the innovation and the rate at which the number of farmers who adopted the innovation is one farmer per unit time.
dx k(100x x 2 ) , where k is to be determined. dt
(i)
Show that
(ii)
Find the particular solution of x , in terms of t.
[5]
(iii)
Sketch the graph of x versus t, for t 0 .
[2]
(iv)
Using the graph in (iii) or otherwise, find the time taken for 75% of the population of farmers
(v)
[1]
to adopt the innovation, leaving your answer to 2 decimal places.
[1]
Give a reason why the model may not be suitable.
[1]
11 Solve the equation ( 2)5 32 0 , leaving your answers in the form rei , where r 0 and .
(i)
[4]
State the equation of the circle in the form | z a | b that passes through all the points represented by the roots. Sketch this circle, showing clearly the relationship between the roots and the locus clearly. [3]
(ii)
The roots 1 and 2 are such that
2
arg(1 2) 0 arg(2 2)
2
. On the same
Argand diagram, sketch the locus of the points representing z given that arg(z 1 )
3 . [2] 5
Find the complex number represented by the point of intersection between the two loci. (iii)
Find the least value of z 2 if arg(z 1 )
NYJC 2010 JC2 Preliminary Examination
9740/01
3 . 5
[1] [2]
[Turn Over
6 12 (a)
(b)
(i)
Write down the derivative of (1 x 2 )n .
[1]
(ii)
Find x3 (1 x 2 )n dx where n 1, 2 .
[3]
Region A is bounded by the curve y sin x cos x , the lines y 2 and x
3 . 4
(i)
Find the exact area of region A.
(ii)
Deduce the exact area of the region in the first quadrant bounded by the curve
[4]
y sin x 4 cos x 4 and the two axes.
[2]
(iii) Find the volume generated when region A is rotated through four right angles about the
x-axis.
[2]
-----END OF PAPER-----
NYJC 2010 JC2 Preliminary Examination
9740/01
NANYANG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Higher 2
MATHEMATICS
9740/02 17th September 2010
Paper 2
3 Hours Additional Materials:
Answer Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages. NANYANG JUNIOR COLLEGE Internal Examinations © NYJC 2010
[Turn Over
2 Section A: (40 Marks) 1
Two scientists are studying the growth of a certain species on an island. (a)
The first scientist proposes the following model: “Let xn be the population of the species in the nth year and d n xn xn1 , n 2 with d1 x1 100 . It is believed that d n is a geometric sequence with common ratio e r .” (i)
Find an expression for xn in terms of r and n.
(ii)
Obtain an inequality for r in order for the population to stabilise at K. Find the value of K in terms of r.
(b)
[2]
[2]
The second scientist proposes the following model: “Let yn be the population of the species (in thousands) in the nth year. It can be hypothesised that y1 0.1 and yn 1 2 yn (1 yn ) .”
2
(i)
Find the limiting population of this model.
[2]
(ii)
Find the value of r in order for both models to have the same limiting population.
[2]
A circular cylinder is inscribed in a right circular cone of base radius 2c and vertical height c. One of the circular ends of the cylinder lies on the base of the cone and the other end is in contact with the inner surface of the cone. If the cylinder has base radius x, find expressions for the volume V and the total surface area S of the cylinder in terms of c and x. If c is fixed and x varies, find the maximum
3
value of V.
[8]
Prove that S cannot exceed 8πc2.
[2]
The sequence of numbers {u r } where r = 1, 2, 3, …, is such that it satisfies the recurrence relation
rur 1 ur r 2 and u1 = 1. r 1 (i)
By dividing the above recurrence relation by r and using the method of difference, show that un
(ii)
n 2 n n 2 2
for n = 1, 2, 3, ….
Prove the result in (i) using mathematical induction.
(iii) Find the exact value of
NYJC 2010 JC2 Preliminary Examination
un as n . n3
9740/02
[5] [4] [1]
4
(a)
3 The line l passes through the point A with coordinates (1,-1,1) and is parallel to the vector
2i + j – 2k. The plane p has equation r (i + j + k) = 3. Find, in exact form,
(b)
(i)
the position vector of B, the point of intersection between l and p,
[3]
(ii)
the sine of the acute angle between l and p,
[2]
(iii)
the shortest distance from A to p,
[1]
(iv)
the length of the projection of AB onto p.
[2]
Given that the system of linear equations
x + y + 3z = (, ) 3x + y + 4z = 9
x+y = 3 has infinite solutions, obtain the numerical values of and .
[4]
Section B: (60 Marks)
5
(a)
Out of the 29 basketball teams, 14 teams have 12 players and 15 teams have 13 players. A sample of 58 players is to be chosen as follows. Each team will be asked to place cards with its players’ names (1 card for 1 name) into a hat and randomly draw out two names. The two names from each team will be combined to make up the sample. Explain why this procedure will not result in a simple random sample of the 363 basketball players. Describe a procedure that will result in a simple random sample in this situation.
(b)
[2]
A cable company plans to survey potential customers in a small city currently served by satellite dishes. It intends to select a sample of families from each of the five non-overlapping neighbourhoods that make up the city. Suggest, with justification, a sampling technique that could be used in this case.
NYJC 2010 JC2 Preliminary Examination
[2]
9740/02
[Turn Over
6
4 A coffee production factory claims that the average amount of coffee in a packet is at least 10 grams.
A consumer suspects the factory has overestimated the mean. To check the claim, a random sample of 8 packets is weighed and the values are given as follows. 9.5 10.1 9.3 10.4 9.6 9.1 9.9 9.0 Making necessary assumption(s), test the factory’s claim at 3% level of significance.
[4]
A second sample of 64 packets is obtained and the data are summarised as follows:
(x 10) 25.6, (x 10)
2
151.99
Test the factory’s claim at 5% level of significance.
[2]
If an observation of mass 9.6 grams is added to the second sample, without conducting the test, decide whether the conclusion of the second test will still remain the same?
7
[1]
An experiment was conducted to investigate the relationship between the amount of unreacted chemical, x, and the time that elapsed since the start of the experiment, t.
(i)
x
25.5
28
31
33.5
41
43.5
45.5
51
57.5
58
73
t
46
44
35.5
30
20
15.6
17
12.3
11
8.3
6
Obtain the scatter diagram for x against t and comment on any relationship between x and t. Calculate the linear product moment correlation coefficient, r, between x and t.
(ii)
[4]
State with a reason (without any calculations) which of the following models is more appropriate for the data:
(iii)
(a)
x = atb, where a > 0 and b < 0
(b)
x = a + bt2, where a > 0 and b < 0.
[1]
If there was an error in recording the t values and all the t-values must be increased by 3, what would be the effect on (a)
t ,
[1]
(b)
standard deviation of t,
[1]
(c)
the correlation coefficient, r?
[1]
NYJC 2010 JC2 Preliminary Examination
9740/02
8
5 Four soccer players, three tennis players, two badminton players and one swimmer sit at a round
table with 10 seats. Find the number of possible seating arrangements (i)
if the two badminton players sit together,
[2]
(ii)
if the two badminton players sit directly opposite each other,
[2]
(iii)
if the two badminton players sit together and none of the four soccer players sit next to either
(iv)
9
of the two badminton players,
[2]
if the two people sitting beside the swimmer are from different sports.
[2]
A vehicle rental company has 7 cars and 4 vans available for rental per day. It is known that the request for cars has a mean of 4 per day; and independently, the request for vans has a mean of 2 per day. (i)
Find the probability that the number of request for a vehicle exceeds 11 on a particular day.[2]
(ii)
Find the probability that some requests for a vehicle have to be refused on a particular day. [2]
(iii)
Explain why the value found in (ii) is larger than the value found in (i).
(iv)
Using a suitable approximation, find the least number of days such that the average number
[1]
of requests for vehicles exceeding 7 is less than 0.001.
[3]
10 Chris takes a bus to school every day in a 5-day week. The bus journey consists of two intermediate
stops. The probability of delay at stop A is
4 2 while the probability of delay at stop B is . The 5 5
delays are independent of each other. If the bus is delayed at either stop, Chris will be late for school. (i)
Given that Chris was late on Monday, find the probability he was late exactly three times in a week.
[3]
(ii)
Given that Chris was late one day, find the probability that he was delayed at stop A.
[2]
(iii)
Given that Chris was delayed at exactly one stop, find the probability that he was delayed at stop B.
NYJC 2010 JC2 Preliminary Examination
[3]
9740/02
[Turn Over
6 11 On the tropical island of Stabletree, records show that the number of floods occurring each month
may be modelled using a Poisson distribution with mean 2. A “bad” month is a month where there are at least 4 floods occurring, and a “bad” year is one where there are more than 2 bad months in the year. Regarding a month as a twelfth part of a year, and assuming independence of flood occurrences, (i)
show that the probability that there are at most 2 bad months in a year is 0.760.
(ii)
use a suitable approximation to find the probability that out of fifty years, there are less than 5 bad years.
(iii)
[3]
[4]
Comment on whether it is suitable to use the model to estimate the probability that the year 2060 (which is 50 years from now) will be a bad year.
[1]
12 The mass of a Munchi pear is normally distributed with mean mass 120g and standard deviation 10g. (i)
If two Munchi pears are chosen at random, find the probability that one of the pears will have a mass between 100g and 126g while the other will have a mass of less than 115g. [2]
(ii)
If ten Munchi pears are chosen at random, find the probability that exactly three of the pears will have a mass more than 122g each. [2]
The mass of a Fuchi apple is normally distributed with mean mass 115g and standard deviation 8g. (iii)
Find the probability that the total mass of three randomly chosen Fuchi apples will be more than three times the mass of a randomly chosen Munchi pear. [2]
(iv)
A random sample consisting of n Fuchi apples is chosen. Find the least value of n such that there is a probability of not more than 0.3 that the sample mean differs from its mean mass by more than 4g. [3]
-----END OF PAPER-----
NYJC 2010 JC2 Preliminary Examination
9740/02
2010 NYJC JC2 Prelim 9740/1 Solutions Qn 1 Let the equation of parabola be y = ax 2 + bx + c At points (-1, 9) and (1, 5) 9 = a− b+ c 5= a + b+ c dy = 2ax + b dx dy 1 = 0, x = At line of symmetry, dx 3 2 0 = a+b 3 using GC, a = 3, b = −2, c = 4
2010 NYJC JC2 Prelim 9740/1 Solutions Qn 3
Hence, arg(wn) = − ∴ Least n = 3
(then argument is
5nπ 6
1
( )( )
∫
2 0
2 1 1 2x 3 2 dx ≈ ∫ + ( 2 x ) dx 1 + 0 2 8 128 4 − 2x 1 2 x 3 2 = ∫ 1 + + x dx 2 0 4 32
(1+ai)(b+2i) = 8i => (b-2a)+(ab+2)i = 8i Comparing real/imaginary parts, b-2a = 0 and ab+2 = 8 i.e. b = 2a and a(2a) + 2 = 8 i.e. a 2 = 3 i.e. a = - 3 (since a < 0) and b = - 2 3 w = - 2 3 - 2i => arg(w) = −
−
1
− x 2 2 = (4) 1 − 4 −1 −3 2 1 1 x 2 2 − x + ... = 1 + − − + 2 2 4 (1)(2) 4 1 x 3 2 ≈ 1 + + x 2 8 128
Equation of parabola is y = 3x 2 − 2 x + 4
2
1 − 1 = (4 − x) 2 4− x
=
x2 1 3 1 x + 8 + 32 x 2 0
=
1 2+ 2
∴b =
5π 6
4
( 2) 8
2
2
+
1 32
( 2 ) = 17322 + 81 3
17 1 , a= 32 8
dx d 3x d2 x + 9 x = 3t , 3 + 9 = 3 . 2 dt dt dt d2x d 3x dx = 1, 2 = 0 , 3 = −6 . Thus the Maclaurin’s series is When t = 0, x=0, dt dt dt x = t − t 3 +⋯ x = Asin 3t + Bt Since
−5π , so that wn is of form – ki with k > 0). 2
(3t)3 + Bt ≈ A 3t − 6 9A 3 t 2 Equating coefficients: 3A + B = 1 1 2 ⇒A= ,B= 9A =1 3 9 2 = (3A + B)t −
Page 1 of 9
Page 2 of 9
2010 NYJC JC2 Prelim 9740/1 Solutions Qn (i) 5 (ii)
c=a+b
ˆ = OA OB sin AOB ˆ = OA × OB = a × b Area = OA × OB × sin AOB
2010 NYJC JC2 Prelim 9740/1 Solutions 6
i
y
ˆ has maximum value of 1, thus maximum area of OACB is (iii) Since sin AOB a b. a b 1 1 (iv) Let a = a2 and b = b2 . We have aib = a b cos θ ≤ a b . a3 b3 2
2
x
2
Thus aib ≤ a b . Since aib = a1b1 + a2b2 + a3b3 , we have (a1b1 + a2b2 + a3b3 ) 2 ≤ ( a12 + a22 + a32 )( b12 + b22 + b32 ) .
ii
iii
7
7(i) f ′( x ) = 2 x − 2 = 2( x − 1) > 0 since x > 1 So f is strictly increasing on its domain and therefore one-one. (ii)
Page 3 of 9
f(x) = f −1 ( x ) ⇒ f(x) = x since the graphs of y = f(x), y = f −1 ( x ) and y = x all meet at the same point(s). Thus x 2 − 2 x + 2 = x (x − 1)(x − 2) = 0 x = 2 or x = 1 (reject since x > 1) Hence x = 2. Page 4 of 9
2010 NYJC JC2 Prelim 9740/1 Solutions (iii)
2010 NYJC JC2 Prelim 9740/1 Solutions Coordinates of point Q: ( 2 + 9e 2 , 0)
y y = f-1(x) y = f(x) Soln set = (1, 2]
O (iv)
1
=
(x (x
2 2
− 2x + 2 + 3 − 2x + 5
)(
= 3e (1 + 9e 2 )
x
9
(i)
2
(ii)
2
2 dy dy dt 6tet = × = = 3tet dx dt dx 2
At point P, by observation, t = 1 .
(iii)
y
P
(
(iv)
y = 3et
2
1 = ( PR )( PQ ) 2 1 = 2 1 + 9e2 3e 1 + 9e 2 2
) (x > 1) ( x − 2 x + 5) , x > 1
3 O
− 2 ) + (0 − 3e) 2
2
Area of ∆PQR
)
(i)
x = 2t ,
( 2 + 9e
+ (6e) 2 = 4 + 36e 2 = 2 1 + 9e 2
= 3e 1 + 9e 2
Rgf = (2,∞).
(ii)
PQ =
2
= 81e4 + 9e 2
2
Therefore, gf : x ֏
8
( 2)
x
2
Since Rf = (1,∞) ⊂ [–3, ∞) = Dg, gf exists. gf(x) = g( x 2 − 2 x + 2 ) =
PR =
3
O
)
(shown)
For the line y = mx − 3e to intersect the curve at 2 distinct points, the absolute value of m must be greater than the tangent to the curve at point P : ∴ m > 3e 1 and 4 y = ax – 1. a 1 y = ax − 1 + , x≠ , a>0 4x −1 4 dy 4a =a − 2 dx ( 4 x − 1)
Asymptotes needed: x =
dy = 0 when a − dx
y − y1 = m ( x − x1 )
i.e. ( 4 x − 1) = 4
y − 3e = 3e( x − 2) ∴ y = 3e( x − 1) (shown) Coordinates of point R: (0, -3e) Equation of normal at point P: 1 y − y1 = − ( x − x1 ) m 1 y − 3e = − ( x − 2) 3e 1 2 ∴ y = − x + + 3e 3e 3e Let y = 0: x = 2 + 9e2
4a
3 1 or − 4 4 Coordinates of turning points: 3 3 5 1 , a − 1 or − , − a − 1 4 4 4 4 (a)
( 4 x − 1)
2
=0
2
x=
Page 5 of 9
x
R
Equation of tangent at point P:
(iii)
Q
Page 6 of 9
2010 NYJC JC2 Prelim 9740/1 Solutions
2010 NYJC JC2 Prelim 9740/1 Solutions
x = A100e x=
100 t 99
A100e 1 + Ae
(iii)
100 t 99
100 t 99
− xAe
100 t 99 100
100e 99
=
99 + e
t
100 t 99
x 100
1 0 1 1 (b) The line y − a − 1 = m x − passes through the point of 4 4 intersection between the 2 asymptotes. For the 2 graphs to intersect, m > a. 10
(i)
(ii)
dx = k ( x)(100 − x) dt dx At t = 0 , x = 1, = 1 dt 1 1 = k (1)(99) ⇒ k = 99 dx 1 therefore = (100 x − x 2 ) dt 99 dx 1 = (100 x − x 2 ) dt 99 1 1 dx = dt x(100 − x) 99 1 1 1 1 + dx = dt 100 x 100 − x 99
∫
(iv)
Using GC, 5.64 years.
(v)
The farmers may be influenced by adoption of innovation from other sources, eg mass media, besides farmers. Or any other reasonable answer (ω − 2)5 − 32 = 0 ⇒ (ω − 2)5 = 25 e 2 kπ i
11
ω = 2 + 2e
(i)
2 kπ i 5
(
kπ i
− k5π i
)
W2 4
C B
∫
A O
100 t +C ln x − ln(100 − x) = 99 x 100 ln t +C = 100 − x 99 100
kπ i
= 2e 5 e 5 + e
kπ kπ5 i = 4 cos e , k = 0, ± 1, ± 2 5 z−2 = 2 Im
∫
∫
t
2π 5
2 4
4
Re
3π 5
4 W1 4
100
t +C t x = e 99 = Ae 99 100 − x
1 When t = 0 , x = 1 , A = 99
Page 7 of 9
3π i
(ii)
c = 2 + 2e 5 .
(iii)
By symmetry, ∡BAW2 =
π 5
π . Thus BW2 = 2sin . 5
Page 8 of 9
12
2010 NYJC JC2 Prelim 9740/1 Solutions d 2 n (a)(i) (1 − x ) = −2xn(1 − x 2 ) n−1 dx (ii) 1 3 2 n 2 2 n ∫ x (1 − x ) dx = 2 ∫ − x −2 x 1 − x dx
( )
(
( ) n +1 1 − x2 ) ( du = −2 x , v = dv = −2 x 1 − x 2 dx
Let u = - x 2 ,
dx
3
(
∴∫ x 1− x
)
2 n
n
n +1 x 2 (1 − x 2 )n +1 1 − ∫ −2 x ⋅ dx = − (1 − x 2 )n +1 dx 2(n + 1) 2(n + 1)
2 n +1
2
)
=−
x (1 − x ) 2(n + 1)
=−
x 2 (1 − x 2 )n+1 1 1 (1 − x 2 ) n+ 2 + C − 2(n + 1) 2 (n + 1)(n + 2)
+
1 2 n +1 ∫ x ⋅(1 − x ) dx n +1
(b)(i) Area of A = 3π 3π π 4 2 − − ∫π sin x + cos x dx 4 4 4 = =
π 2 2
π 2 2
3π
− [sin x − cos x ]π4 4
− 2
(ii) By translation, req’d area =
(iii) Volume = π ( 2 )2 = 4.93
π 2
π 2 2
− Area of A = 2
3π 4
− π ∫π (sin x + cos x)2 dx 4
Page 9 of 9
2010 NYJC JC2 Prelim 9740/2 Solutions
2010 NYJC JC2 Prelim 9740/2 Solutions Qn
Qn 1
d1 (1 − (er )n ) 100(1 − ern ) = . 1 − er 1 − er k =1 In order for population to stabilise at K, the sum of the geometric sequence must converge. Thus er < 1 . Hence r < 0 . n
(a)(i) (a)(ii)
Since xn = ∑ d k =
100 . 1 − er Let y = lim yn . Taking limits K = lim xn = n →∞
(b)(i)
n →∞
lim yn +1 = lim 2 yn (1 − yn ) n →∞
dV = 0: dx
dV π x 2 = ( −1) + π x ( 2c − x ) dx 2 x = π x 2c − x − 2 πx = ( 4c − 3 x ) 2
n →∞
⇒ y = 2 y (1 − y ) 1 ⇒ y = 0, 2
∴
Using GC, yn is an increasing sequence, thus y = (b)(ii)
At max/min V,
Thus, the limiting population is 500. Need K = 500. Thus 100 4 = 500 ⇒ r = ln . 1 − er 5
2
1 . 2
πx
2 4c x= 3
( 4c − 3 x ) = 0 or
d 2V π x π = ( − 3 ) + ( 4c − 3 x ) dx 2 2 2 =
π 2
( − 3 x + 4c − 3 x )
= π ( 2c − 3 x ) when x =
x c
h
x = 0 (rejected)
4c d 2V 4c , = π 2c − 3 < 0 3 dx 2 3
Therefore, volume V is maximum when x =
4c . 3
2c 2
1 4c 4c max volume = π c − 2 3 3 16π c3 max V = 27
Using similar triangle, c c−h = 2c x x h=c− 2 x Volume, V = π x 2 c − 2 π x2 V= ( 2c − x ) 2
Surface area, S = π x ( x + 2c ) S is a strictly increasing function for x > 0 ∴ S is maximum when x is maximum: Since x < 2c S = π x ( x + 2c ) < π ( 2c )( 2c + 2c ) = 8π c 2
∴ S < 8π c 2
x Surface area, S = 2π x 2 + 2π x c − 2 = 2π x 2 + 2π xc − π x 2 S = π x ( x + 2c ) Page 1 of 9
Page 2 of 9
2010 NYJC JC2 Prelim 9740/2 Solutions Qn 3(i) Dividing the reurrence relation by r gives 3 ur +1 ur − =r. r +1 r Summing both sides from r = 1 to r = n – 1, we have n −1
(iii)
n −1
ur
r =1
4
r =1
u2 u1 − 2 1 u u + 3− 2 3 2 ⋮ u u + n − n−1 n n −1
=
n −1 (1 + n − 1) 2
[Note that LHS is a telescopic series and RHS is an arithmetic series] n ( n − 1) un u1 − = which gives n 1 2 n ( n − 1) n 2 un = n 1 + = n − n + 2 as desired. 2 2
Thus
(
)
(ii) Let Pn denotes the proposition in (i) For n = 1, LHS = u1 = 1 1 RHS = 12 − 1 + 2 = 1 = LHS. 2 So P1 is true. Assume Pk is true for some k = 1, 2, 3, …. k That is, uk = ( k 2 − k + 2 ) ------ (IH) 2 k +1 2 To prove uk +1 = ( k + 1) − ( k + 1) + 2 . 2 For n = k + 1, k +1 LHS = uk +1 = k 2 + uk by the recurrence reln k k k +1 by (IH) = k 2 + k 2 − k + 2 2 k k +1 2 = k +k +2 2 k +1 2 = ( k + 1) − ( k + 1) + 2 = RHS. 2 So Pk+1 is true. Since P1 is true and Pk true ⇒ Pk+1 is true, by the principle of MI, Pn is true.
(
(
un 1 n2 − n + 2 = 3 n 2 n2
1 1 2 1 as n → ∞ . 1 − + → 2 n n2 2 4(a)(i) Equation of l is r = i – j + k + λ(2i + j – 2k). 1 + 2λ Substitute r = − 1 + λ into equation of p gives 1 − 2λ 1 + 2 λ 1 − 1 + λ • 1 = 3 1 − 2λ 1 ⇒ 1 + 2λ – 1 + λ + 1 – 2λ = 3 ⇒ λ = 2. 5 Therefore OB = 1 . − 3 =
∑ r + 1 − r = ∑ r ur +1
2010 NYJC JC2 Prelim 9740/2 Solutions Qn
φ θ B
)
(
)
)
Page 3 of 9
p (side view)
Let φ be the angle between l and the normal vector of p. Taking scalar product of the direction vector of l and the normal vector of p gives 2 1 1 . 1 • 1 = 3 3 cos φ ⇒ cos φ = 3 3 − 2 1 [Note that φ is acute since cos φ > 0] Let θ be the acute angle between l and p. 1 sin θ = sin(90° – φ) = cos φ = . 3 3
)
(
A
(ii)
(iii)
shortest distance from A to p 4 6 2 1 = . = AB sin θ = 2 = 3 3 3 3 3 −4
Page 4 of 9
2010 NYJC JC2 Prelim 9740/2 Solutions Qn
2010 NYJC JC2 Prelim 9740/2 Solutions Qn
(iv)
X − 10 ~ t (7) s/ n Reject H0 if p-value < 0.03 Since x = 9.6125 , s = 0.49117 , p-value = 0.0304
4 2 AB = 2 = 42 + 22 + ( −4 ) = 6 −4 length of the projection of AB onto p
Under H0, T =
2
=
(b)
Since p-value > 0.03, there is insufficient evidence at 3% level of significance level to reject H0, and we conclude that the factory’s claim could be correct. To test H 0 : µ = 10 vs H1 : µ < 10 at 5% level of significance
26 2 62 − = 2 3 using Pythagoras’ Theorem 3
By GC, the equation of the line of intersection of the planes given by the last 2 equations is 3 −2 r = 0 + λ 2 . 0 1 For the system to have infinite solutions, the above line must lie on the plane given by the first equation. So the line is perpendicular to the normal vector of the plane. That is, 1 −2 1 α • 2 = 0 ⇒ -2 + 2α + 3 = 0 ⇒ α = − 2 3 1 Furthermore, any point of the line is also a point of the plane. Take (3,0,0) and substitute into the first equation gives 3 + α(0) + 3(0) = β ⇒ β=3
Under H0, using Central Limit Theorem, Z =
X − 10 ~ N (0,1) approx s/ n
Reject H0 if p-value < 0.05 25.6 1 (614.4 − 10 × 64) 2 Since x = 10 − = 9.6 , s 2 = 151.99 − = 2.25 64 63 64 Thus, p-value = 0.0164 Since p-value < 0.05, there is sufficient evidence at 5% level of significance level to reject H0, and we conclude that the factory’s claim may not be correct. Since sample mean remains unchanged and s2 decreases, the p-value will decrease. Thus we will still reject H0.
7
1(i) 80 70 60 50
5
(a) - The first method does not give a simple random sample as each player does not have the same chance of being selected. - A simple random sample can be obtained by numbering all the 363 from 001 to 363 and then picking three digits at a time from a random number table, ignoring numbers over 363 and ignoring repeats, until a group of 58 numbers is obtained. The players corresponding to these 58 numbers will be a simple random sample.
40 20 10 0 0
6
To test H 0 : µ = 10 vs H1 : µ < 10 at 3% level of significance Assume that the distribution of the amount of coffee in a packet is normally distributed. Page 5 of 9
10
20
30
40
50
There appears to be a curvilinear/non-linear relationship between x and t. r ≈ 0.920. (ii) Model (a) is more appropriate because the graph of x = atb, where a > 0 and b < 0 is concave upwards whereas the graph of x = a + bt2, where a > 0 and b < 0 is concave downwards. The points on the scatter diagram fall on a graph that is concave upwards. (iii)(a) t will increase by 3 (b)standard deviation of t remains the same (a) r remains unchanged
(b) Stratified sampling, where the population is divided into homogeneous groups called strata and random individuals from each stratum are chosen. Stratified samples can give useful information about each stratum (in this case, about each of the five neighbourhoods) in addition to information about the whole population (the city population) Quota sampling as the sampling frame is not available. The population is divided into non-overlapping groups and the sample size is determined without any basis.
x
30
8
(i) 8! x 2! = 80640 (ii) 8C4 x 4! x 4! or 7! x 8 = 40320 (iii) 4C2 x 2! x 2! x 6! = 17280 (iv) [4C1. 3C1 +3C1. 2C1 + 4C1 .2C1] x 2! x 7! = 262080 Page 6 of 9
2010 NYJC JC2 Prelim 9740/2 Solutions Qn 9 Let X be the number of requests for cars on a particular day. X ~ Po(4) Let Y be the number of requests for vans on a particular day. Y ~ Po(2) (i) Let T be the number of requests for vehicles on a particular day. Thus T ~ Po(6) prob.req′d = P (T > 11) = 1 − P (T ≤ 11) (ii)
= 0.0201 Either demand for a car or a van is not met. Thus prob.req′d = P ( X > 7 or Y > 4)
2010 NYJC JC2 Prelim 9740/2 Solutions Qn (iii) P(Chris was delayed at B | he was delayed at exactly one stop) = P(Chris was delayed at stop B only)/P(delayed at exactly one stop) 1 2 ( )( ) 1 5 5 = = 4 3 1 2 ( )( ) + ( )( ) 7 5 5 5 5
11
i)
Let X be the no. of floods in a month. Then, X ~ Po(2), and P(bad month) = P(X ≥ 4) = 0.142877 Let Y be the no. of bad months in 12 months (note: 1 year = 12 months) Then, Y ~ B(12, 0.142877) Hence, P(Y ≤ 2) = 0.76006
ii)
Let W be the no. of bad years, out of fifty years. Then, W ~ B(50, 1 – 0.76006) ~ B(50, 0.23994)
= 1 − P ( X ≤ 7 and Y ≤ 4)
(iii) (iv)
= 1 − P ( X ≤ 7) P (Y ≤ 4) = 0.101 The event in (i) is a subset of the event in (ii). Thus the value obtained in (i) will be smaller. Let n be the number of days needed. Assume that n is large. By Central 6 Limit Theorem, T ~ N 6, approx. n P (T > 7) < 0.001
Since n = 50 is large, np = 11.997 > 5 and nq = 38.003 > 5, W may be approximated using N(11.997, 9.11844). P(W < 5) → (c.c.) P(W < 4.5) = 0.00652 or 0.00650
7−6 ⇒ PZ > < 0.001 6 n
iii)
n ⇒ P Z ≤ > 0.999 6
n > 3.09023 ⇒ n > 57.3 6 Thus least number of days required is 58. 4 2 4 2 22 (i) P(Chris is late in a day) = + − ( )( ) = 5 5 5 5 25 Thus
10
22 ). 25 P(Chris was late exactly thrice in a week | Chris was late on Mon) = P(Chris was late twice in the remaining four days of the week) = P(X = 2) ≈ 0.0669 (ii) P(Chris was delayed at A | Chris was late) 4 10 = P(Chris was delayed at A)/P(Chris was late) 5 = 22 11 25
12
(i)
Let M be the mass of a randomly chosen Munchi pear. M ∼ N (120,10 2 ) Using GC, P (100 < M < 126 ) ≈ 0.7030
P ( M < 115) ≈ 0.3085
Let X be the number of days Chris is late out of 4 days. X~B(4,
Page 7 of 9
It is not suitable since the weather conditions may have changed over such a long time period and the model may no longer be applicable.
Required probability = ( 0.7030 )( 0.3085 ) × 2! ≈ 0.434 (3 s.f.) (ii)
Using GC, P ( M > 122 ) ≈ 0.4207 Let S be the number of pears of mass that is more than 126g. S ∼ B (10, 0.4207 )
P ( S = 3 ) ≈ 0.196 (3 s.f.) (iii)
Let F be the mass of a randomly chosen Fuchi apple. F ∼ N (115,82 )
Page 8 of 9
2010 NYJC JC2 Prelim 9740/2 Solutions Qn
(
F1 + F2 + F3 − 3M ∼ N ( 3 × 115 − 3 × 120 ) , ( 3 × 82 + 9 × 102 ) ⇒ F1 + F2 + F3 − 3M ∼ N ( −15,1092 ) Using GC, Required probability = P ( F1 + F2 + F3 > 3M
)
= P ( F1 + F2 + F3 − 3M > 0 ) ≈ 0.325 (iv)
(3 s.f.)
F ∼ N (115,82 )
82 ⇒ F ∼ N 115, n
(
)
P F − 115 > 4 ≤ 0.3
⇒ P( F − 115 < −4) + P( F − 115 > 4) ≤ 0.3 4 P Z < − 2 8 n
4 P + Z > 2 8 n
≤ 0.3
n n ∴P −
Page 9 of 9
)
3 1
It is known that the number of diagonals that can be drawn in a polygon of n sides can be expressed as a quadratic polynomial in n. By considering the number of diagonals in a triangle, a quadrilateral and a pentagon, find the number of diagonals that can be drawn in a polygon of 200 sides. [4]
2
Given that f ( x) ln(cos x) , show that f ''( x) sec 2 x .
By further differentiation, find the first two non-zero terms in the Maclaurin’s series of f ( x) . Hence obtain an approximate value for 0.4
0 3
ln cos x dx .
[6]
Solve the inequality ln( x 1) 9 x 2 . Hence, solve the inequality ln( x 1) 9 x 2 .
[5]
A
4
3m C
10 ms 1 xm
2m
B An athlete at point C is running towards the finishing line AB, at a constant speed of 10 ms 1 in a direction perpendicular to AB, as shown in the diagram above. (i)
Given that ACB is and x is the distance of the athlete from AB, show that
3 x
2 x
tan 1 tan 1 . (ii)
[2]
Find the exact rate of change of when the athlete is 10 m from the finishing line. [3] [Turn over
4 5
(i)
Sketch the graph of y x
1 , stating the equations of any asymptotes and x
the coordinates of any points of intersection with the axes.
(ii)
[2]
The curve C has equation
y
p 2 x 2 pqx 1 , px q
where p and q are positive constants. State a sequence of transformations which transform the graph in (i) to the graph of C.
6
1 , of 3 x x expressing the coefficients in their simplest form. State the set of values of x for which the expansion is valid. Find the first three terms in the expansion, in ascending powers of
Hence, by substituting x 25 , find an approximate value for
7
[3]
7 as a fraction. [6]
Find the exact value of
6 2x
e
0
sin x dx .
[5]
5
y 8
C (1, 4) (3, 2) D
A 1
B 0
5 2
y f ( x)
1 y 2 x
x2
The diagram shows the graph of y f ( x) . The curve crosses the x-axis at the 5 points A (1, 0) and B , 0 , and has a minimum point at C (1, 4) and a 2 maximum point at D (3, 2) . The lines y
1 , x 0 and x 2 are asymptotes to 2
the curve. Sketch, on separate diagrams, the graphs of
(i)
y f ( x) ,
[3]
(ii)
y f '( x) ,
[3]
labeling each graph clearly and showing the asymptotes and coordinates of the points corresponding to A, B, C and D.
[Turn over
6 9
Find the roots of the quartic equation z 4 2 i2 3 0 . Give your answers exactly, in the form rei , where r 0 and . [3] The polynomial P(z) has degree eight and real coefficients. All the roots of the equation z 4 2 i2 3 0 are also the roots of the equation P(z) = 0. By considering P(z) as a product of two quartic factors, find P(z), expressing all the coefficients in real and non-trigonometrical form. [3]
10
A curve is defined by the parametric equations x et ,
Find an expression for
11
y t2 t .
dy in terms of t. dx
[1]
(a)
Given that the tangent to the curve at the point with parameter p passes [3] through the origin, find the exact values of p.
(b)
Given that
d 2 y d dy dx , deduce the exact range of t for which the dx 2 dt dx dt curve is concave upwards. [3]
A seed from a type of tree was planted and the change in the height of the tree is noted at the end of each year. It is observed that the change in height of the tree follows an arithmetic progression in the first 10 years of growth, and subsequently, follows a geometric progression from the end of the 10th year. Given that the tree’s change in height was 40 cm and 70 cm at the end of the 3rd year and 5th year of its growth respectively, find (i)
the change in height of the tree at the end of the 4th year of its growth, [1]
(ii)
the height of the tree at the end of the 10th year of its growth.
[3]
Given that the tallest the tree can ever grow is 20 m, find the year at which [6] the tree is at least twice as tall as at the end of 10th year of its growth.
7 12
By using substitution u 4 e x or otherwise, show that
ln p 0
1 1 5p x dx 4 ln 4 p . 4e
[4]
Hence, (i)
find the exact volume of the solid of revolution formed when the region 1 bounded the curve y 2 , the y-axis and the line x ln 6 is rotated 4 ex through 2 right angles about the x-axis, [3]
(ii)
by considering the area under the curve y
1 , find the exact value of 4 ex
1 5
1 4 y ln dy . 1 y
[3]
8
13
x jogging direction treadmill
O
A woman starts to jog at a rate proportional to the distance x from O, along a treadmill which is running at a constant rate of a ms 1 , as shown in the diagram above. At time t seconds, the distance she has moved is x m. When her distance from O is 1 m, her position remains stationary. Show that dx a x 1 . dt [3] 1 Given that at t = 2s, she is moving at 2 ms1 , find the speed of the treadmill. e [7] Hence find the exact distance she has further moved after another 2 seconds. [2]
[Turn over
8
14
1 2 The planes p1 and p2 have equations r. 1 4 and r. 2 respectively, 1 where and are negative constants. p1 and p2 are inclined at 60 to each
other. The point A with position vector i − 2j, where is a constant, is in both p1 and p2 .
(i)
Find the position vector of A.
[1]
(ii)
The point B has position vector 3i − 3j + 2k. Find the shortest distance [3] from B to p1 .
(iii)
Find the equation of p2 in the form r.n d .
[4]
The plane p3 has equation x y z , where and are constants. (iv)
Find the values of and if p3 passes through the origin and p1 , p2 and p3 have only one common point, which is also the foot of perpendicular from A to p3 .
(v)
[3]
What can be said about the values of and if p1 , p2 and p3 have no common point and all points on p3 are equidistant from p1 ? [2]
3 Section A: Pure Mathematics [40 marks]
x cm
1
A 5 cm
B l cm D
C
10 cm
A student wants to draw a straight line of length l cm that begins at the side AB, ends at the side BC and passes through the point D, as shown in the diagram above. (AB and BC may be assumed to be infinitely long.) (i)
By using similar triangles, show that l 2 x2
(ii)
2
25 x 2
x 10
2
.
[3]
Hence find the length of the shortest straight line he can possibly draw. [3]
The function f is defined by b f :x a 2 , x
1 x , 2
where a and b are positive constants. (i)
Sketch the graph of f and state range of f.
[2]
(ii)
Find f 1 .
[3]
(iii)
Given that b2 gf : x 2 , x
find g( x) in terms of a and b.
1 x , 2
[2]
[Turn over
4
3
D
C
5m 6m 10 m
A H 3m
6m
4m
E
k
B j
G
3m
O i
10 m
F
The diagram above shows a partial design of the roof of the gallery of an amphitheatre. ABCD is an inclined rectangular roof, where AB = 10 m, and BC 5 m . EFGH is a rectangle on a horizontal ground, where EF = 10 m and EH 4 m . The points A and B are 3 m directly above E and F respectively. The points C and D are 6 m directly above G and H respectively. Point O is the centre of EFGH andis taken as the origin. Perpendicular unit vectors i, j, k are in the direction of EF , EH and EA respectively.
(i)
10 Show that AC 4 and find a vector equation of the line passing 3
through A and C. (ii)
A spot light is fixed at the top of a vertical pillar erected at point Q with coordinates 15, 6, 0 . Find the height of the pillar if the top of the pillar is collinear with A and C.
(iii)
[3]
[3]
The main electrical supply to the gallery runs along the diagonal AC. Find the position vector of point X on AC such that the length of the electric cable that is required to provide electricity from the main electrical supply to an amplifier mounted at D, will be the least. [3]
5 4
A sequence {xn } of negative numbers is defined by x0 2 and 1 xn xn1 2
(i)
n 1
for n 1 .
4 1 Prove by mathematical induction that xn 3 2
n 1
1 for n 0 . [4]
(ii)
The sequence {xn } converges to l as n tends to infinity. State the exact value of l. [1]
(iii)
Determine, with a reason, whether
(iv)
By considering xn xn1 , deduce that xn xn1 if n is odd and xn xn1 if n is even. [2]
N
5
xn converges.
n1
[1]
Sketch on an Argand diagram the set of points representing all complex numbers z satisfying both the inequalities iz 2i 2 2 and Re z 1
3i .
[4]
Find
(i)
the range of arg z 2 2i ,
[2]
(ii)
the complex number z where arg z 2 2i is a maximum.
[2]
The locus of the complex number w is defined by w 5 2i k , where k is a real and positive constant. Find the range of values of k such that the loci of w and z will intersect. [2]
[Turn over
6 Section B: Statistics [60 marks] 6
7
(a)
The Royal Club has membership of 1600 people, of which 1000 are males and 600 are females. The boss wants to find out what members think of the new set meals menu offered by the Oiishi restaurant of the Club. A sample of 80 members is to be chosen. State a suitable sampling method and describe in detail how this can be carried out. [3]
(b)
The amount of tips a member gives after a meal in Oiishi restaurant of Royal Club has mean $5 and standard deviation $1. Estimate the probability that the amount of tips collected from 80 randomly chosen customers is between $350 and $410. [2]
In an experiment, a new computer game and a new mathematics quiz are given to a group of teenagers. It may be assumed that the teenagers are playing the computer game and attempting the mathematics quiz for the first time. The computer game score, u, and the mathematics quiz score, v, of 8 teenagers are given in the table below. Teenager 1 2 3 4 5 6 7 8 Computer game score, u 52 90 64 74 80 82 76 74 Mathematics quiz score, v 60 90 68 74 74 90 82 70
(i)
(ii)
Find the linear product moment correlation between v and u and the equation of the regression line of v on u.
[2]
Comment on the suitability of using the regression line in (i) to predict u given v. Use the appropriate regression line to predict u given that v is 85, giving your answer to the nearest whole number. [3]
The scores are actually recorded for 9 teenagers. However, the scores for the last teenager are lost. Teenager 1 2 3 4 5 6 7 8 9 Computer game score, u 52 90 64 74 80 82 76 74 p Mathematics quiz score, v 60 90 68 74 74 90 82 70 q It is known that the inclusion of the scores of the last teenager does not alter the mean computer game score and one of the regression lines is given by u 45.556 0.395v .
(iii)
Find p and q, giving your answers to the nearest whole number.
[3]
7 8
9
10
A taxi company wants to track the number of morning, afternoon and night trips made in a day. Of the total trips in a day, p% are morning trips, q% are afternoon trips and the remaining are night trips. Past records indicated that the probability that a randomly chosen morning, afternoon and night trip is paid by NETS is 0.85, 0.65 and 0.25 respectively.
(a)
If p 30 and q 50 , find the probability that a randomly chosen trip is paid by NETS. [3]
(b)
Given that the probability of a randomly chosen trip paid by NETS is taken in the morning is 0.5, express q in terms of p.
[4]
Defective spots are found randomly on a roll of ribbon. On average, 5 defective spots are found per 10 m of ribbon. The roll of ribbon is cut into short ribbons, each of length 1 m. A short ribbon is not discarded if it contains at most 1 defective spot.
(i)
State a condition under which a Poisson distribution would be a suitable probability model and find the percentage of short ribbons that do not need to be discarded. [3]
(ii)
Given 52 randomly chosen short ribbons, find an approximate probability that there are more than forty-five short ribbons that do not need to be discarded. [4]
(iii)
Determine the maximum number of short ribbons required for which the probability of at least 3 short ribbons discarded does not exceed 0.1. [2]
A six-digit number is to be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. For each of the following cases, find how many different ways the six-digit number can be formed.
(i)
The even and odd digits of the number must alternate and any digit may appear more than once. [2]
(ii)
The number must be odd and is less than 600 000 and no digit may appear more than once. [3]
(iii)
The number is formed from four different digits, eg. 621313, 255567. [4]
[Turn over
8 11
A star fruit plantation claims that it has developed a method of producing star fruits that are larger in size with a mean weight of 200 grams. A wholesaler suspects that the plantation is overstating the weight of the star fruits produced and decides to test the plantation’s claim. The wholesaler obtains a random sample of 10 star fruits. The weights (to the nearest gram) of the 10 star fruits are as follows: 189
192
200
205
206
198
188
190
200
192
Find an inequality satisfied by the significance level that will confirm the wholesaler’s suspicion. State an assumption made in conducting the test. [5] It is later known that the standard deviation of the weight of the star fruits produced by the plantation is 6 grams. Another random sample of 10 star fruits is obtained. A second test is conducted at 2% significance level and the wholesaler is unable to confirm his suspicion. Find, to the nearest gram, the range of values for the mean weight of these 10 star fruits in order to arrive at this conclusion. [4] Explain, in the context of the second test, the meaning of ‘at 2% significance level’? [1]
12
The weight, Y, of a Yummy cereal bar is normally distributed with mean (120 – k) g and standard deviation 10 g. The weight, F, of a Fullness cereal bar is normally distributed with mean 180 g and standard deviation 20 g.
(i)
Given that P(Y < 80) = P(Y > 150), show that the value of k is 5.
(ii)
5 Yummy cereal bars are randomly chosen. Find the probability that exactly one bar weighs lesser than the lower quartile weight and exactly one bar weighs more than the median weight. [2]
(iii)
Find the probability that the weight of 2 randomly chosen Yummy cereal bars differs from one and a half times the weight of a randomly chosen Fullness cereal bar by at most 5g. State the assumption needed for your working. [4]
[1]
The Fullness cereal bars are sold at $2 per 100g.
(iv)
Find the probability that a randomly chosen Fullness cereal bar costs more than $3.50. [2]
(v)
A random sample of 100 Fullness cereal bars is to be taken. Using a suitable approximation, find the probability that there are less than 65 bars with each costing more than $3.50. [3]
1
RI 2010
9740/01/10
[Turn over
2
1
By writing z x iy, x, y
, solve the simultaneous equations
z 2 zw 2 0 and z *
2
where z * is the conjugate of z .
[3]
Sketch the graph of y e x x . By adding a suitable graph to the sketch, find the set of values of x that satisfies 2 x 5 2e x
3
w , 1 i
2 . x2
[4]
Four friends returned from a trip to Europe and converted their foreign currencies back to Singapore Dollars. The amounts of foreign currencies converted and the total amounts received in Singapore Dollars are shown in the following table. Donald 36 77 42 269.90
Sterling Pound Euro Dollar Swiss Franc Total amount in Singapore Dollars
Leonard 55 18 63 233.45
Michael 40 31 26 175.50
Raphael k 59 24 313.00
Assuming that, for each foreign currency, the exchange rate quoted for each of the friends is the same, calculate the value of k. [4] 4
Find (a)
(b)
RI 2010
1 1 2 x x2
[2]
dx,
x2 x 3 ( x 2 2)(1 x)
dx .
[3]
9740/01/10
3
5
A sequence u0 , u1 , u2 ,
is defined by
u0 1 and un 1 run d for n 0 where r and d are non-zero constants. (i)
If r = d = 1, write u n in terms of n.
[1]
d d If r 1, prove by induction that un r n 1 for all non-negative integral r 1 r 1 values of n. [4] (ii)
6
If r 1, state the limit of the sequence as n .
[1]
The diagram shows the graph of y f ( x) . The curve has a maximum point at (0,2) and it cuts the x-axis at the points (a,0) and (a,0) where a is a positive constant. The lines
x 2 , x 2 and y 2 are asymptotes to the curve. y
y f(x) 2
y2 a O
x 2
a
x
x2
Sketch, on separate diagrams, the graphs of (i)
y 2 f ( x),
(ii)
y
[3]
1 . f (2 x)
[3]
Your sketches should show clearly the equations of asymptotes, stationary points and intersection with the axes, where applicable.
RI 2010
9740/01/10
[Turn over
4
7
The curve C has equation y f (x) 2 x 1 (i) (ii)
a where a and b are positive constants. bx 1
Show that C has exactly two stationary points. Given that C passes through the point (0,3) , find the value of a.
[3] [1]
Hence sketch the graph of y f '(x) , indicating clearly any intersection with the axes and the equation(s) of asymptote(s) of the curve in terms of b. [3]
8
A cylindrical water tank has a horizontal base with a fixed area of A m 2 and is initially empty. Water is poured into the tank at a constant rate of 5 m3s1 , and leaks out through a small hole in the base at a rate which is proportional to the depth of water in the tank. The depth of water in the tank is x metres at time t seconds. Show that
A
dx 5 kx , where k is a positive constant. dt
Solve the differential equation, expressing x in terms of A, k and t .
9
(a)
[5]
Solve the equation z 6 4 2(1 i) , expressing the solutions in the form rei , where
r 0 and .
(b)
[2]
(i)
[3]
The two complex numbers 2 3 + 2i and 2 3 + 2i satisfy the cubic equation a3 z 3 a2 z 2 a1 z a0 0 . Explain clearly whether it is possible for all the coefficients a3 , a2 , a1 , a0 to be real numbers.
(ii)
RI 2010
[1]
In an Argand diagram, the point A represents the complex number 2 3 i . If A, B, and C are the vertices of an equilateral triangle taken in clockwise order, and these three points lie on a circle with centre at the origin, find the complex number represented by B in the form p iq where p, q , giving the exact values of p and q. [3]
9740/01/10
5
10
The curve C has equation y xe x for x 0 and P(a, ae a ) is a point on C . (i)
Sketch the curve C .
[1]
(ii)
Find, in terms of a , the equation of the tangent to the curve at P .
[2]
This tangent cuts the y axis at the point Q(0, h) . Using differentiation, find, as a varies, the exact maximum value of h .
11
[6]
The equations of two planes p1 , p2 are 2x 5y + 3z = 3, 3x + y + 6z = , respectively, where and are constants. (i)
Given that the two planes intersect in a line l, with a vector equation given by 4 2 r = 2 s 1 , s , 5 3
show that the value of is 12 and find the value of .
[3]
A third plane p3 has equation given by 5x + 8y + tz = 12, where t is a constant. (ii)
With the values of and found in (i), find the exact value of t if the three planes have no point in common. [2]
(iii) The plane p4 contains the line l and the point (1, 1, 2). Find the cartesian equation of p4 and the acute angle between p1 and p4. [5]
RI 2010
9740/01/10
[Turn over
6
12
Relative to the origin O, three points A, B and P have position vectors given by a = 14i + 8j + 6k, b = 11i + tj + 12k and p = 12i 4j + 10k respectively, where t is a constant. (a)
If P divides the line segment AB in the ratio : 2, find the values of and t. [3]
(b)
It is given that t = 2. (i)
Find the exact length of projection of PB onto AB . Deduce the exact shortest distance of P from line AB.
(ii)
13
[4]
BAPQ forms a parallelogram. Find the position vector of the point Q and the area of the parallelogram correct to 2 decimal places. [4]
Given that f ( x)
4
1 ,x . 3 (1 3x) x 2 4
(i)
Express f ( x) as a series expansion in ascending powers of x, up to and including the term in x3 . [4]
(ii)
Find the range of values of x for which the expansion in (i) is valid.
(iii)
A curve is given by the equation
RI 2010
[2]
2tan y f (x). By differentiation, find the series expansion for y in ascending powers of x, up to an including the term in x 2 . [5]
9740/01/10
7
14
(a)
pn The sum of the first n terms of a series, Sn , is given by n1 5 , where p is a non5 zero constant and p 5 . Obtain an expression for Tn , the nth term of the series and prove that this is a geometric series. Find the range of values of p for the sum to infinity to exist.
(b)
[5]
An arithmetic progression is grouped into sets of numbers as follows: {2} , {6 , 10} , {14 , 18 , 22} , {26 , 30 , 34 , 38} , … where the number of terms is 1 for the first set, 2 for the second set, 3 for the third set, 4 for the fourth set and so forth. (i)
Show that the first term in the nth set is 2n2 2n 2 .
(ii) Find, in terms of n, a simplified expression for the last term in the nth set. (iii) Hence, find the sum of all the terms in the nth set.
RI 2010
9740/01/10
[5]
[Turn over
1
RI 2010
9740/02/10
[Turn over
2 Section A : Pure Mathematics [40 marks]
1
1 1 1 na , show that a . 2 2 2 2 2 2(n 1) 2n n (n 1)
(i)
Given that
(ii)
Given that S N
N
2n 1
2n (n 1)
nM
2
2
[1]
, state the smallest possible value of M, where M
and M N , such that S N can be defined.
[1]
(iii)
If M = 3, find S N in terms of N.
[3]
(iv)
Deduce that the sum to infinity of the series
1 1 1 ... 2 2 (2)(3) (3)(4) (4)(5)2
is less than
2
1 . 8
[3]
The point A represents a fixed complex number a where
arg(a)
4 2 numbers ia and a are represented by the points B and C respectively.
. The complex
On a single clearly labelled Argand diagram, show the points A, B, C and the set of points representing all complex numbers z satisfying both relations
z ia and arg( z a) arg(a) . 4
[4]
Find (i)
the minimum value of | z | in terms of | a | ,
[2]
(ii)
the range of values of arg( z ) in terms of arg(a) .
[2]
RI 2010
9740/02/10
3 3
(a)
Functions f and g are defined by
f:x
( x 2)( x 4) ,
for x , x 4 ,
g:x
x , x3
for x \ 3 .
Determine if f is one-one, justifying your answer.
[1]
Only one of the composite functions fg and gf exists. Give a definition (including the domain) of the composite that exists, and explain why the other composite does not exist. Find the range of the composite that exists.
(b)
The function h is defined by h : x a(1 e x ) ,
[6]
for x
0
, where a is a positive constant.
Find h 1 (x) and state the domain of h 1 .
[3]
Given that the graphs of y h( x) and y h 1 ( x) intersect at the point where x = b, and that
b 0
h( x) dx I , find the area bounded by the curves y h( x), y h 1 ( x) and the
axes, giving your answer in terms of I and b.
4
The curve C has equation y f ( x) , where f ( x)
1
(i)
Find the exact value of
(ii)
Using the substitution u 3x e2 3 , find
1
[2]
4x , x . x 1 2
f (| x |) dx .
[2]
ln(3x e2 3) dx .
Hence find the exact area of region R bounded by C, the curve y ln(3 x e 2 3) and the line y 0 .
[7]
(iii) Find the volume of solid formed when R is rotated completely about the x-axis, giving your answer correct to 2 decimal places. [3]
RI 2010
9740/02/10
[Turn over
4
Section B : Statistics [60 marks]
5
(a) Find the number of three-letter code-words that can be formed from the letters of the word WHYOGEE. [3] (b) A country is invited to send a delegation of six youths selected from six badminton players, six tennis players and five football players to participate in the opening ceremony of the Singapore 2010 Youth Olympic Games. No youth plays more than one game. The delegation is to consist of at least one, and not more than three players selected from each sport. (i)
Find the number of ways in which the delegation can be selected.
[3]
During the ceremony, the youths from the delegation are to be seated in six out of ten chairs which are arranged in a row. (ii)
6
Find the number of ways this can be done if no two empty chairs are adjacent. [2]
1 , and the 15 probabilities of getting ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’ are consecutive terms of an arithmetic progression with common difference p . Show the probability of getting ‘6’ is 1 15p . [2] A six-sided die is thrown. The probability of getting a ‘1’ is p , where 0 p
The die is thrown twice and events A and B are defined as follows: A : the sum of the scores of the two throws is at least 10; B : the score of the first throw is 5. Find (i) (ii)
P(A) , P(A|B).
[3] [2]
Hence, show that there is no value of p for which A and B are independent.
RI 2010
9740/02/10
[2]
5 7
A researcher investigates the relationship between the stress level of a person and his job performance. The table below shows the findings for 8 different workers for the same job. Stress level, x
2
2.5
3
3.5
4
5
6
7
Performance, y
3.30
3.10
3.00
2.86
2.72
2.60
2.55
2.50
(i)
Draw a scatter diagram for the data and calculate the product moment correlation coefficient between x and y. [2]
(ii)
Comment on whether a linear model is appropriate.
[2]
(iii) The researcher proposes that a model of the form b y a where a and b are positive constants x is more appropriate. Explain why he thinks this is so. [1] (iv) Calculate the least square estimates of a and b for the model in (iii). [2] (v) Estimate the performance when the stress level is 4.5. Comment on the reliability of the estimate. [2]
8
The mass of a male student in Aishan Secondary School is denoted by X kilograms. The masses of a random sample of 150 male students are summarized by
x 8400 (i)
and
2 ( x56) 5555.
Calculate unbiased estimates of the mean and variance of X.
[2]
Aishan Secondary School claimed that the mean mass of a male student in the school is 55 kilograms. (ii)
Test, at the 3% significance level, whether the school is understating the mean mass of a male student. Does the Central Limit Theorem apply in this context? [5] (iii) State what you understand by the expression ‘at the 3% significance level’ in the context of the question in (ii). [1] In a separate study, the opinions of female students in Aishan Secondary School are to be collected to determine if they are satisfied with their own weights. Describe how a quota sample of size 100 might be obtained. [2]
RI 2010
9740/02/10
[Turn over
6 9
Every weekday, the last train from Bedok Station leaves the station at 11.29 pm. Its travel time X (in minutes) from Bedok Station to Redbridge Station may be taken to follow a normal distribution with mean 30 and variance 9. (i)
Find the travel time exceeded by 90% of the trips made by the last train.
[2]
At 11.40 pm every weekday, Abel walks from his workplace to Redbridge Station to catch the last train. The time Y (in minutes) he takes follows a normal distribution with mean 16 and variance 4. (ii)
Find the probability that the train arrives at Redbridge Station after 12 am and Abel arrives before 12 am. [2]
(iii) Find the probability that Abel will not miss the train.
[3]
(iv) Explain why the answer to part (iii) is greater than the answer to part (ii).
[1]
The time W (in minutes) taken by Dina to walk from her workplace to Redbridge Station has a mean of 4 and a variance of 6. (v)
Explain why W is unlikely to be normally distributed.
[1]
(vi) Find the probability that the mean time taken by Dina to walk from her workplace to Redbridge Station, in a random sample of 40 trips, is less than 3.5 minutes. [2] 10
Incoming telephone calls to the management office of a shopping mall are received randomly and independently, at an average rate of 6.75 per hour. The mall (including its management office) is open from 10 am to 10 pm. (i)
State, in the context of the question, a condition required for a Poisson distribution to be a suitable model for the number of incoming calls from 10 am to 10 pm. [1]
Assume that the condition in (i) is satisfied. (ii)
Find the probability of receiving no fewer than 8 calls in a particular hour.
[2]
(iii) In 4 non-overlapping one-hour periods, find the probability of receiving at most 6 calls in a one-hour period, exactly 7 calls in another one-hour period, and no fewer than 8 calls in each of the 2 remaining one-hour periods. [3] (iv) The number of incoming calls from 10 am to 10 pm, Y , has mean and standard deviation . Use a suitable approximation to find P Y , giving your answer correct to 4 decimal places. [4] (v)
RI 2010
A day is considered busy if there are more than 90 incoming calls received from 10 am to 10 pm. Find the probability that the 3rd busy day in a month occurs on the 15th day of the month. [3]
9740/02/10
RAFFLES INSTITUTIION matics 974 40 H2 Mathem 2010 Year 6
__ _________ ________ ________ _____ 20 010 H2 Math hs Prelimin nary Exam Paper P 1 Solu utions Q1 [3]
Let z = x + iy. Su ubstitute the second s equaation into thee first. 2 z + z ( z * (1 + i)) − 2 = 0 ( x + iy ) 2 + ( x 2 + y 2 )(1 + i) − 2 = 0 2 x 2 + 2ixy + ( x 2 + y 2 )i − 2 = 0 On com mparing reall and imaginnary parts, 2 2x − 2 = 0, 2xy + x 2 + y 2 = ( x + y ) 2 = 0 x = ±1, 1 y = m1 . When z = 1 − i , w = 2i When z = −1 + i , w = − 2i
Q2 [4]
y
y = ex − x
y=
α
x
βO
2x + 5 − 2e x ≤
5 1 − 2 x2
2 x2
2 ≤ 2e x − 2 x x2 5 1 − 2 ≤ ex − x 2 x
⇒ 5− ⇒
Q3 [4]
t diagram,, x ≤ α or β ≤ x < 0 or From the o x>0 Using G.C., x ≤ −2.18 or − 0.920 ≤ x < 0 or x > 0 (3s.f.) Thereffore solution n set is ( −∞, −2.18] ∪ [ −00.920, 0) ∪ (00, ∞ ) . Let x, y, y z be the exxchange ratee quoted for Sterling Pouund, Euro Dollar and Sw wiss Franc, respectively (i.e. 1 Sterling Poound = x Sinngapore Dolllars, 1 Euro Dollar = y Singapore Dollars D S Franc = z Singappore Dollars)). and 1 Swiss 36x + 77 y + 42 z = 269.9 55x + 18 y + 63z = 233.45 40x + 31y + 26 z = 175.5 Using the GC, x = 2.15, y = 1.7 78, z = 1.32 kx + 599 y + 24 z = 313 3 313 − 59(1.778) − 24(1.32) ∴k = = 82 2..15 ____ ____________ _____________ ________
Q4 [5] (a)[2]
∫
1 1 + 2 x − x2
dx =
∫
1 2 − ( x − 1)2
dx
⎛ x −1 ⎞ = sin −1 ⎜ ⎟+c ⎝ 2 ⎠ (b)[3]
∫
x2 − x + 3 2
( x + 2)(1 − x)
dx = =
∫
1
1 dx x + 2 1− x 2
+
⎛ x ⎞ tan −1 ⎜ ⎟ − ln |1 − x | + c 2 ⎝ 2⎠
1
Q5 [6] (i) [1]
un = n + 1
(ii) [4]
d ⎞ d ⎛ for n ∈ Z0+ . Let Pn be the statement un = r n ⎜1 + ⎟− ⎝ r −1 ⎠ r −1 When n = 0, LHS: u0 = 1 (given) d ⎞ d ⎛ =1 RHS: r 0 ⎜1 + ⎟− ⎝ r −1 ⎠ r −1 ∴ P0 is true. Assume Pk is true for some k ∈ Z + ,
d ⎞ d ⎛ i.e. uk = r k ⎜1 + ⎟− ⎝ r −1 ⎠ r −1 Prove that Pk +1 is true, d ⎞ d ⎛ i.e. to prove uk +1 = r k +1 ⎜1 + ⎟− ⎝ r −1 ⎠ r −1 uk +1 = ruk + d ⎛ ⎛ d ⎞ d ⎞ =r ⎜ r k ⎜1 + ⎟− ⎟+d ⎝ ⎝ r −1 ⎠ r −1 ⎠ d ⎞ rd ⎛ =r k +1 ⎜1 + +d ⎟− ⎝ r −1 ⎠ r −1 d ⎞ −rd + rd − d ⎛ =r k +1 ⎜1 + ⎟+ r −1 ⎝ r −1 ⎠ d ⎞ d ⎛ =r k +1 ⎜1 + ⎟− ⎝ r −1 ⎠ r −1 Since Pk is true implies Pk +1 is true, and P0 is true, by mathematical induction, Pn is true for all
n ∈Z0+ . (iii) [1]
As n → ∞, un →
d 1− r
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1 Page 2 of 8
y
Q6 [6] (i) [2]
y 2 = f(x) 2 −a O
y= 2 x
a
− 2
y=− 2
x = −2
x=2
y
(ii)[2]
y=
1 f ( 2x)
1 2
|
y= |
O
−1
x = − a2 Q7[7] (i)[3]
y = 2x + 1 +
x=
1 2
x
1
a 2
2 bx + 1 2b
dy = 2− 2 dx ( bx + 1)
When
dy = 0, dx
2b
=2
( bx + 1) 2 ( bx + 1) = b 2
x=
(
1 −1 ± b b
)
Since b is a positive constant, there are 2 distinct real solutions for x. Hence, C has exactly two stationary points. (shown) y (II)[3]
y = f ' (x) y=2
2
−
( b 1
|
b +1
)
|
O
2 − 2b
__
x=−
( b 1
b −1
)
x
1 b __________________________________ 2010 H2 Maths Preliminary Exam Paper 1 Page 3 of 8
dV dx =A dt dt
8 [7]
V = Ax
[2]
dV = 5 − kx , where k is a positive constant. dt dx ⇒ A = 5 − kx (shown) dt
⇒
dx = 5 − kx dt 1 ⇒ A∫ dx = ∫ dt 5 − kx A ⇒ − ln 5 − kx = t + c1 , where c1 is an arbitrary constant k k ⇒ ln 5 − kx = − t + c, where c is an arbitrary constant A A
[5]
⇒ 5 − kx = e
k − t +c A
⇒ 5 − kx = Be
k − t A
, where B is an arbitrary constant
k − t 1 (5 − Be A ) k When t = 0, x = 0 ⇒ B = 5
⇒ x=
∴ x= Q9 [7]
k − t 5 (1 − e A ) k
z = 4 2(1 + i) = 8e 6
(a)[3]
z = 2e (b)(i)[1]
(ii)[3]
iπ (8 k +1) 24
,
i
π 4
= 8e
⎛π ⎞ i ⎜ + 2 kπ ⎟ ⎝4 ⎠
k = −3, −2, −1, 0,1, 2
Since the two complex roots are non-real, and not conjugate, for all the coefficients a3 , a2 , a1 , a0 to be real, the degree of the polynomial has to be at least four. Since the degree is only three, it is not possible. Since the points A, B and C are on a circle centered at the origin, and they form an equilateral 2π 2π triangle, angle AOB is . OA rotated by clockwise about O will give OB. Let b be the 3 3 complex number represented by B. b=e
−
2π i 3
(2 − 3 + i)
⎛ 1 3⎞ b = ⎜⎜ − − i ⎟⎟ (2 − 3 + i) 2 2 ⎝ ⎠ b = 3 − 1 + i(1 − 3) (i)[1]
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1 Page 4 of 8
(ii)[8] [2]
dy = x (−e − x ) + e − x = e − x (1 − x) dx dy At P , x = a, y = ae − a , = e − a (1 − a ) dx Equation of tangent to the curve at P is y − ae − a = e − a (1 − a )( x − a )
y = xe − x ⇒
⇒ y = e − a (1 − a )( x − a ) + ae − a At Q , x = 0, y = h
h = e − a (1 − a )(0 − a ) + ae − a = e − a (a − 1)(a ) + ae − a = a 2 e − a dh ⇒ = 2ae − a + a 2 (−e − a ) = ae − a (2 − a ) da dh For stationary values of h , =0 da ⇒ a = 0 (N.A. since a > 0 ) or a = 2 ⇒
[6]
Q11[10] (i)[3]
dh da Tangent
2−
2
2+
>0
0
<0
Maximum value of h = 4e −2 Since the two planes intersect in a line l, with a vector equation given by ⎛ 4 ⎞ ⎛ −2 ⎞ r = ⎜⎜ −2 ⎟⎟ + s ⎜⎜ 1 ⎟⎟ , s ∈ R, ⎜ −5 ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠
The line l lies in p2.
⎛ −2 ⎞ ⎛ 3 ⎞ So ⎜⎜ 1 ⎟⎟ ⋅ ⎜⎜ λ ⎟⎟ = 0 ⇒ −6 + λ + 18 = 0 ⇒ λ = −12 ⎜ 3 ⎟ ⎜6⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 4⎞ ⎜ ⎟ ⎜ −2 ⎟ satisfies equation of p2. ⎜ −5 ⎟ ⎝ ⎠ ⎛ 4 ⎞ ⎛ 3⎞ So ⎜⎜ −2 ⎟⎟ ⋅ ⎜⎜ λ ⎟⎟ = μ ⇒ μ = 12 − 2λ − 30 = 12 + 24 − 30 = 6 ⎜ −5 ⎟ ⎜ 6 ⎟ ⎝ ⎠ ⎝ ⎠ (ii)[2]
Since p1 and p2 intersect in line l, and the 3 planes have no common point of intersection, l does not intersect p3. Hence l is parallel to p3, but l does not lie in p3. ⎛ −2 ⎞ ⎛ 5 ⎞ 2 ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⋅ ⎜ 8 ⎟ = 0 ⇒ −10 + 8 + 3t = 0 ⇒ t = 3 ⎜ 3 ⎟ ⎜t⎟ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ 4 ⎞ ⎜5⎟ ⎜ ⎟ 2 Also ⎜⎜ −2 ⎟⎟ . ⎜ 8 ⎟ = ≠ 12 which means (4, −2, −5) is not in p3 and so l does not lie in p3. ⎜ −5 ⎟ ⎜ 2 ⎟ 3 ⎝ ⎠⎜ ⎟ ⎝3⎠
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1 Page 5 of 8
(iii)[5]
⎛ −2 ⎞ Since p4 contains l, ⎜⎜ 1 ⎟⎟ is parallel to p4. ⎜ 3⎟ ⎝ ⎠
Since p4 contains the points (1, −1, 2) and (4, −2, −5). ⎛ 4⎞ ⎛1⎞ ⎛ 3⎞ Hence ⎜⎜ −2 ⎟⎟ − ⎜⎜ −1⎟⎟ = ⎜⎜ −1 ⎟⎟ is parallel to p4. ⎜ −5 ⎟ ⎜ 2 ⎟ ⎜ −7 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 3 ⎞ ⎛ −2 ⎞ ⎛ 4 ⎞ A vector normal to p4 is ⎜⎜ −1 ⎟⎟ × ⎜⎜ 1 ⎟⎟ = ⎜⎜ 5 ⎟⎟ . ⎜ −7 ⎟ ⎜ 3 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 4⎞ ⎛ 1 ⎞ ⎛ 4⎞ Equation of p4 is given by : r⋅ ⎜⎜ 5 ⎟⎟ = ⎜⎜ −1⎟⎟ ⋅ ⎜⎜ 5 ⎟⎟ = 1 ⇒ 4 x + 5 y + z = 1 ⎜1⎟ ⎜ 2 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Let θ be the angle between p1 and p4.
cos θ =
⎛ 2 ⎞ ⎛ 4⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −5 ⎟ ⋅ ⎜ 5 ⎟ ⎜ 3 ⎟ ⎜1⎟ −14 ⎝ ⎠ ⎝ ⎠ ⇒ θ = 110.514 = 2 2 2 2 2 2 38 42 2 + 5 + 3 4 + 5 +1
Acute angle between p1 and p4 = 180° − θ = 69.5° (1 d.p.) Q12[11] (a) [3]
By Ratio Theorem, uur uuur uuur 2OA + λ OB OP = λ+2
⎛ 12 ⎞ ⎛ 14 ⎞ ⎛ 11 ⎞ 2 ⎜ ⎟ λ ⎜ ⎟ ⎜ ⎟ ⇒ ⎜ −4 ⎟ = ⎜ 8 ⎟+ ⎜t ⎟ ⎜ 10 ⎟ λ + 2 ⎜ 6 ⎟ λ + 2 ⎜ 12 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
So 12 =
28 11λ + ⇒ 12λ + 24 = 28 + 11λ ⇒ λ = 4 λ +2 λ + 2
Also, − 4 =
(b) [8] (i)[4]
16 λt 16 4t + ⇒ −4 = + ⇒ t = −10 λ +2 λ + 2 6 6
⎛ 11 ⎞ ⎛ 12 ⎞ ⎛ −1⎞ uur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ PB = OB − OP = ⎜ 2 ⎟ − ⎜ −4 ⎟ = ⎜ 6 ⎟ ⎜12 ⎟ ⎜ 10 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 11 ⎞ ⎛ 14 ⎞ ⎛ −3 ⎞ uuur uuur uur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = OB − OA = ⎜ 2 ⎟ − ⎜ 8 ⎟ = ⎜ −6 ⎟ ⎜12 ⎟ ⎜ 6 ⎟ ⎜ 6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ uur uuur Length of projection of PB onto AB ⎛ −1⎞ ⎛ −3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 6 ⎟ • ⎜ −6 ⎟ ⎜2⎟ ⎜ 6⎟ 3 − 36 + 12 21 7 ⎝ ⎠ ⎝ ⎠ = = = = 9 3 81 32 + 62 + 62 2
7 49 320 8 5 Shortest distance of P from AB = BP 2 − ⎛⎜ ⎞⎟ = 41 − = = ⎝3⎠
9
3
3
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1 Page 6 of 8
(ii)[4]
BAPQ forms a parallelogram. uuur uuur Hence AB = PQ
⎛ −3 ⎞ ⎛ 12 ⎞ ⎛ 9 ⎞ uuur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⇒ OQ = AB + OP = ⎜ −6 ⎟ + ⎜ −4 ⎟ = ⎜ −10 ⎟ ⎜ 6 ⎟ ⎜ 10 ⎟ ⎜ 16 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Area of parallelogram BAPQ uuur uuur = AB × AP
⎛ −3 ⎞ ⎛ −2 ⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ −6 ⎟ × ⎜ −12 ⎟ ⎜6⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 48 ⎞ ⎜ ⎟ =⎜0⎟ ⎜ 24 ⎟ ⎝ ⎠ = 482 + 02 + 242
= 53.67 (to 2 dec places) 13[11] (i)[4]
f ( x) =
4 (1 − 3x) x 2 + 4
⎡ ⎛ x2 ⎞⎤ = 4(1 − 3x) −1 ⎢ 4 ⎜1 + ⎟ ⎥ 4 ⎠⎦ ⎣ ⎝ ⎛ x2 ⎞ 4 = (1 − 3x) −1 ⎜1 + ⎟ 4⎠ 4 ⎝
−
−
1 2
1 2
⎡ 1 ⎛ x2 ⎞ ⎤ = 2 ⎡⎣1 + 3x + (3x) 2 + (3x)3 + ...⎤⎦ ⎢1 − ⎜ ⎟ + ...⎥ ⎣ 2⎝ 4 ⎠ ⎦ ⎡ ⎤ ⎛ 1 ⎞ ⎛ 3 ⎞ = 2 ⎢1 + 3x + ⎜ − + 9 ⎟ x 2 + ⎜ − + 27 ⎟ x3 + ...⎥ ⎝ 8 ⎠ ⎝ 8 ⎠ ⎣ ⎦ 71 213 3 x + ... = 2 + 6 x + x2 + 4 4 (ii)[2]
For expansion in (i) to be valid, x2 <1 3 x < 1 and 4 1 and x 2 < 4 3 1 ⇒ x< and x < 2 3 ⎛ 1 1⎞ Range of values of x = ⎜ − , ⎟ ⎝ 3 3⎠ ⇒
x<
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1 Page 7 of 8
(iii)[5]
2 tan y = f ( x) ----- (1) Differentiate (1) with respect to x, dy = f '( x) ----- (2) dx Differentiate (2) with respect to x, 2sec2 y
2 ⎡ d2 y ⎤ ⎛ dy ⎞ 2 2 ⎢ 2sec y tan y ⎜ ⎟ + sec 2 y 2 ⎥ = f "( x) ----- (3) dx ⎦⎥ ⎝ dx ⎠ ⎣⎢ When x = 0, 71 from (i), f (0) = 2, f '(0) = 6 and f ''(0) = 2 π (1) gives 2 tan y = 2 ⇒ y = 4 dy dy 3 = (2) gives 2(2) = 6 ⇒ dx dx 2 2 ⎡ d 2 y ⎤ 71 d 2 y 35 ⎛3⎞ ⇒ = (3) gives 2 ⎢ 2(2)(1) ⎜ ⎟ + 2 2 ⎥ = dx ⎥⎦ 2 dx 2 8 ⎝2⎠ ⎢⎣ 2 π 3 π 3 35 ⎛ 35 ⎞ x Hence y = + x + ⎜ ⎟ + ... = + x + x 2 + ... 4 2 4 2 16 ⎝ 8 ⎠ 2!
Q14 [5]
pn − 5. 5n −1 p n −1 S n−1 = n − 2 − 5. 5
Given Sn =
⇒
Tn = S n − S n −1
∴
⎛ p n −1 ⎞ pn 5 − − ⎜ n−2 − 5 ⎟ n −1 5 ⎝5 ⎠ n−1 ⎛ p⎞ ⎡ ⎛ 1 ⎞⎤ = ⎜⎜ ⎟⎟⎟ ⎢ p − ⎜⎜ −1 ⎟⎟⎟⎥ ⎝⎜ 5 ⎠ ⎢⎣ ⎝⎜ 5 ⎠⎥⎦
=
n−1
⎛ p⎞ = ( p − 5)⎜⎜ ⎟⎟⎟ ⎝⎜ 5 ⎠
n−2
⎛ p⎞ Now, Tn−1 = ( p − 5)⎜⎜ ⎟⎟⎟ ⎜⎝ 5 ⎠
n−1
∴
Tn Tn−1
⎛ p⎞ ( p − 5)⎜⎜⎜ ⎟⎟⎟ ⎝5⎠ = n−2 ⎛ p⎞ ( p − 5)⎜⎜⎜ ⎟⎟⎟ ⎝5⎠
=
p 5
(constant)
⇒ series is a geometric series. (shown) For the sum to infinity to exist,
p <1 5
∴ −5 < p < 5 , p ≠ 0
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1 Page 8 of 8
RAFFLES INSTITUTION H2 Mathematics 9740 2010 Year 6
____________________________ 2010 H2 Maths Preliminary Exam Paper 2 Solutions Q1 [8] (i)[1]
1 1 − 2 2 2(n − 1) 2n
n 2 − n 2 + 2n − 1 2n 2 (n 2 − 1) n − 12 1 = 2 ∴ a = − (shown) 2 n (n − 1) 2 Smallest value of M = 2 N 2n − 1 SN = ∑ 2 2 n =3 2n ( n − 1) =
(ii) [1] (iii)[3]
n − 12 2 2 n =3 n ( n − 1) N
=∑ N
1 1 − 2) 2 2n n =3 2( n − 1) 1 1 = − 2 2(2) 2(3)2 1 1 − + 2 2(3) 2(4) 2 M 1 1 − + 2 2( N − 2) 2( N − 1) 2 1 1 − + 2 2( N − 1) 2 N 2 1 1 = − 8 2N 2 1 1 1 + + + ... 2 2 (2)(3) (3)(4) (4)(5) 2 ∞ 1 =∑ 2 n = 3 n ( n − 1) = ∑(
n −1 2 n = 3 n ( n − 1) ∞
=∑
2
n − 12 2 2 n = 3 n ( n − 1) ∞
<∑
n − 12
N
= lim
N →∞
∑ n (n − 1) n =3
2
2
1 ⎤ ⎡1 = lim ⎢ − N →∞ 8 2 N 2 ⎥⎦ ⎣ __________________________________ 2010 H2 Maths Preliminary Exam Paper 2 Page 1 of 8
=
1 (shown) 8
Q2 [8] [4]
(i)[2]
(ii)[2]
Q3 [12] (a)[7]
Minimum value will be the perpendicular distance from origin to the line segment CB. Let the point of intersection of this perpendicular and CB be D. By observing the right angle triangle π |a| . formed by O, C, and D, OD = | a | sin = 4 2 It is clear that the locus of z satisfying both relations is the line segment CB, not including C. The π argument of the point represented by B is arg(a) − , while the argument of the point represented 2 by C is arg(a ) − π . Hence the range of values is π arg(a ) − π < arg( z ) ≤ arg(a) − . 2
Since f (2.5) = f (3.5) = −0.75 , f is not one-one. OR
y = f ( x)
O
2 y = −0.75
°4
(3, −1)
D f = (−∞, 4)
R f = [ −1, ∞ )
Since the horizontal line y = −0.75 does not cut the graph of f at one and only one point, f is not one-one.
y = g( x) y =1 O x = −3
D g = R \ {−3}
R g = R \ {1}
fg does not exist because R g = R \ {1} ⊄ ( −∞, 4) = D f . Since R f = [ −1, ∞ ) ⊆ R \ {−3} = D g , gf exists. gf ( x ) =
( x − 2)( x − 4) x2 − 6x + 8 3 D gf = D f = ( −∞, 4) . = 2 = 1− ( x − 2)( x − 4) + 3 x − 6 x + 11 ( x − 3) 2 + 2 __________________________________ 2010 H2 Maths Preliminary Exam Paper 2 Page 2 of 8
R gf = R g whose domain is restricted to R f = R g whose domain is restricted to [−1, ∞) = [ − 12 ,1)
OR sketch y = gf(x) y = gf ( x)
2
O
y =1
°4
(3, − ) 1 2
(b)[5]
D gf = (−∞, 4)
R gf = [− 12 ,1)
Let y = a (1 + e − x ) y − 1 = e− x a x = − ln ( ay − 1)
h −1 (x) = − ln ( ax − 1) D h −1 = R h = ( a , 2 a ]
y = h −1 (x)
A(0, 2a) B(b, b)
y=x
y = h(x) y=a
C (2a,0) x=a
O
Dh =
+ 0
R h = ( a, 2a ]
Since y = h −1 ( x) can be obtained by reflecting y = h( x ) about the line y = x, therefore at the point of intersection of y = h( x ) and y = h −1 ( x) , y = x = b.
∫
b 0
b
a (1 + e − x ) d x = ∫ a (1 + e − y ) d y = I . 0
∴ Area bounded by y = h( x), y = h −1 ( x) and the axes b
b
0
0
= ∫ a (1 + e − x ) dx + ∫ a (1 + e − y ) d y − (b × b) = 2I − b2
OR b
Area OAB =
∫
b 0
h( x ) d x − ∫
b 0
⎡ x2 ⎤ b2 x dx = I − ⎢ ⎥ = I − 2 ⎣ 2 ⎦0
Since y = h( x ) and y = h −1 ( x) are symmetrical about y = x , Area OAB = Area OBC = I −
b2 . 2
∴ Area bounded by y = h( x), y = h −1 ( x) and the axes ⎛ b2 ⎞ = 2 ⎜ I − ⎟ = 2I − b2 2⎠ ⎝
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2 Page 3 of 8
Q4 [12] (i)[2]
∫
1
f (| x |)dx = 2 −1
∫
1 0
−4 x x2 + 1
= 2(−2)
∫
1
dx 2x
x +1 2
0
dx
1
= −4 ⎡⎣ln( x 2 + 1) ⎤⎦
0
= −4 ln 2 (ii)[7]
∫
ln(3 x + e 2 + 3)dx
=
∫
=
1 3
=
1 1 [u ln u ] − 3 3
1 ln(u ) du 3
∫
ln(u )du
∫
1du
1 [u ln u − u ] + c 3 1 = ⎡⎣ (3 x + e2 + 3) ln(3 x + e 2 + 3) − (3 x + e 2 + 3) ⎤⎦ + c 3 =
4
y
3
(−1, 2)
y = ln(3 x + e 2 + 3) 2
1
R –6
–4
x
–2
−4 x
2
− ( e + 2) 3 2
y= –1
4
6
x +1 2
–2
–3
–4
∫
−1 − ( e2 + 2) 3
ln(3 x + e 2 + 3)dx +
∫
0
−4 x
−1
x2 + 1
dx
−1 1 = ⎡⎣(3 x + e 2 + 3) ln(3 x + e 2 + 3) − (3 x + e 2 + 3) ⎤⎦ ( e2 + 2) + 2 ln 2 − 3 3
1 = [e2 ln e2 − e2 + 1] + 2 ln 2 3 1 = (e 2 + 1) + 2 ln 2 3 (iii)[3]
π
∫
−1
− ( e2 + 2) 3
(
)
2
ln(3 x + e2 + 3) dx + π
∫
0
2
⎛ −4 x ⎞ ⎜ 2 ⎟ dx −1 ⎝ x + 1 ⎠
= 20.55 (2d.p.) __________________________________ 2010 H2 Maths Preliminary Exam Paper 2 Page 4 of 8
Q5[8] (a)[3]
(b) [5] (i) [3]
(iii) [2]
Q6 [7] [2]
W,Y,O,G – 1 letter each E – 2 letters Case 1: All 3 letters are different, i.e. W, Y, O, G, H, E. Number of ways = 6C3 × 3! = 120 . Case 2: 2 letters are identical, i.e. E, E and one of the remaining 5 letters, W, Y, O, G, H. 3! Number of ways = 5C1 × = 15 . 2! Total number of ways = 120 + 15 = 135 Ways to select the players Badminton Tennis Football No. of ways (6) (6) (5) 6 1 2 3 C1 × 6C2 × 5C3 =900 6 1 3 2 C1 × 6C3 × 5C2 =1200
C2 × 6C1 × 5C3 =900
2
1
3
6
2
3
1
6
2
2
2
6
3
1
2
6
3
2
1
6
C2 × 6C3 × 5C1 =1500
C2 × 6C2 × 5C2 =2250 C3 × 6C1 × 5C2 =1200 C3 × 6C2 × 5C1 =1500
Total number of ways = 9450. Arrange the 6 youths in 6! ways and slot the 4 empty chairs into the row in 7C4 ways. Total number of ways = 6! × 7C4 = 25200 . Probability of getting ‘6’ = 1 − (p + 2 p + 3 p + 4 p + 5 p ) = 1 − 15 p
(i)[3]
First die
Second die +
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
P(A) = P(sum = 10)+P(sum = 11)+P(sum = 12) = 2(4p)(1 − 15 p)+(5 p )(5p )+2(5p)(1 − 15 p )+(1 − 15 p )2 = 1 − 12 p − 20 p 2
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2 Page 5 of 8
(ii)[2]
P(A B ) = P(2nd die is 5 1st die is 5)+P(2nd die is 6 1st die is 5) = 5 p + 1 − 15 p
= 1 − 10 p Alt :
P(A B ) P(A ∩ B) P(B ) P(1st die is 5, 2nd die is 5)+P(1st die is 5, 2nd die is 6) = P(1st die is 5) (5 p )(5 p )+(5 p )(1 − 15 p ) = 5p = 1 − 10 p =
For A and B to be independent, P(A) = P(A|B) 1 − 12 p − 20 p 2 = 1 − 10 p 20 p 2 + 2 p = 0 ∴ p = 0 or −
1 10
Since p > 0, there are no values of such that A and B are independent. Q7[9] (i)[2]
The product moment correlation coefficient between x and y,
r is -0.943(3 s.f.). (ii)[2]
Since r is close to 1, it suggests a linear model is appropriate. However the scatter diagram
(iii)[1]
The scatter diagram shows that as x increases, the rate of decrease in y becomes smaller and y appears to approach a value. For a linear model, the rate of decrease is constant. OR The product moment correlation coefficient between 1/x and y, r2 is 0.993 where r2 is almost 1 as
shows the relationship is non-linear.
compared to r =0.943.
b where a > 0 ,b > 0 is better. x a = 2.16 (3s.f.), b = 2.33 (3s.f.)
Therefore the model y = a + (iv) [2]
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2 Page 6 of 8
(v) [2]
When x = 4.5, y = 2.68 (3 s.f..) Since x = 4.5 is within the data range of the x values and value of r2 is close to 1, the estimate is reliable.
8 [10]
Unbiased estimate of the mean =
(i)[2]
8400 = 56 150 5555 . 149 H1 : μ > 55
Unbiased estimate of the variance = (ii)[5]
(i)
H0 : μ = 55
vs
Perform a 1-tail test at 3% level of significance. ⎛ s2 ⎞ Under H0 , X ~ N ⎜ μ0 , ⎟ approximately, by Central Limit Theorem, where n⎠ ⎝
μ0 = 55, s =
5555 , x = 56, n = 150 149
Using a z-test, p-value = 0.0224 (3 s.f.) Since the p -value = 0.0224 < 0.03, we reject H0 and conclude that there is sufficient evidence, at 3% significance level, that the school is understating the mean mass of male students in the school.
(iii)[1] [2]
Central Limit Theorem has to be applied here as X is not stated to be normally distributed and the sample size of 150 is large. It means that the probability of concluding that the mean mass is more than 55 kg when it actually is 55 kg is 0.03. Suppose that there are 4 different levels of female students in the school. Select 25 from each level and carry out the data collection by interviewing the first 25 female students of each level who enter the school gate in the morning.
Q9 [11]
P ( X > t ) = 0.9
(i)[2]
P ( X < t ) = 0.1 t = 26.2 (in minutes)
(ii)[2]
P ( X > 31 and Y < 20 ) = P ( X > 31) × P (Y < 20 ) = 0.36944 × 0.97725
= 0.361 (3sf) (iii)[3]
(
P ( X > Y + 11) = P ( X − Y > 11) where X − Y ~ N 14, 13
2
)
= 0.797 (3sf) (iv)
Part (iii) includes the case in (ii), as well as other cases in which Abel reaches the station before the train, e.g. train arrives after 12.01am and Abel arrives before 12.01am.
(v)
If W ~ N ( 4, 6 ) , then 99.7% of the values would lie within 4 ± 3 6 , which contains a significant range of negative values. __________________________________ 2010 H2 Maths Preliminary Exam Paper 2 Page 7 of 8
(vi)
2 ⎛ 6 ⎞ ⎟ approximately. Since sample size 40 is large, by Central Limit Theorem, W ~ N ⎜ 4, ⎜ 40 ⎟ ⎝ ⎠
(
)
P W < 3.5 = 0.0984 (3sf)
Q10[13] (i)[1]
(ii)[2]
The average number of incoming calls received per hour is constant throughout the opening hours of the mall. OR The probability of 2 or more incoming calls received in a very short interval of time is negligible. Let X be the number of incoming calls received in an hour. X ~ Po ( 6.75 ) . P ( X ≥ 8) = 1 − P ( X ≤ 7 )
(iii)[3]
= 0.364 (3sf) Required probability is
P ( X ≤ 6 ) × P ( X = 7 ) × ⎡⎣ P ( X ≥ 8 ) ⎤⎦ × 2
≈ 0.48759 × 0.14832 × ( 0.36409 ) × 2
(iv)[4]
Y ~ Po ( 81) . So μ = 81 , σ = 9 .
(
4! 2!
4! = 0.115 (3sf) 2!
)
Since μ = 81 > 10 , Y ~ N 81, 92 approximately. P ( 81 − 9 < Y < 81 + 9 ) = P ( 72 < Y < 90 ) = P ( 72.5 < Y < 89.5 ) by continuity correction (v)[3]
= 0.6551 (4dp) Let W be the number of busy days in 14 days. W ~ B (14, P (Y > 90 ) ) ,that is, W ~ B (14, 0.14593) .
Required probability is P (W = 2 ) × P (Y > 90 ) ≈ 0.29191× 0.14593
= 0.0426 (3sf)
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2 Page 8 of 8
Name
(
)
Class
RIVER VALLEY HIGH SCHOOL 2010 Year 6 Preliminary Examinations Higher 2
MATHEMATICS
9740/01 14 September 2010
Paper 1
3 hours Additional Materials:
Answer Paper List of Formulae (MF15) Cover Page
READ THESE INSTRUCTIONS FIRST
Do not open this booklet until you are told to do so. Write your name, class and index number in the space at the top of this page. Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers.
At the end of the examination, place the cover page on top of your answer paper and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages.
2
1
A shampoo manufacturing company produces shampoo that comes in three different sizes of bottles (small, medium and large). The amount of plastic required to manufacture the bottles, and the profit made from the sale of each bottle of shampoo are shown in the table below: Volume of shampoo (cm3) Amount of plastic (cm2) Profit ($) per unit sold
Small 200 150 2.50
Medium 450 335 3.80
Large 600 475 4.20
On a particular round of production, the volume of shampoo and the amount of plastic used were 370700cm3 and 280400cm2 respectively. When all the bottles of shampoo from this round of production were sold, the profit made from the sale of the medium bottles was twice the total profit made from the sale of the small and large bottles. Determine the number of each size of shampoo produced and state an assumption made about the bottles manufactured. [6]
2
3
The complex number z satisfies the relations iz + 1 ≤ i and z − z* ≤ 2i . (i)
Illustrate both of these relations on a single Argand diagram.
[3]
(ii)
Find the greatest and least possible values of arg ( z + 1 + i) , giving your answers in radians correct to 3 decimal places. [3]
Given that θ is sufficiently small for θ 3 and higher powers of θ to be neglected, express 4 − cos θ as a quadratic expression in θ . [4] 2 + tan θ By letting θ =
π , show that 6
RIVER VALLEY HIGH SCHOOL
3≈
9 11 π 55π 2 . + − 7 56 2016
9740/01/2010
[2]
3
4
d sin −1 x 2 ) . ( dx
(i)
Find
(ii)
Evaluate the integral
(iii)
[1]
∫
5
1
x ln tan −1 dx numerically. 10
Given that 0 < k ≤ 1 and
∫ show that
∫
k
k
x 2 sin −1 x 2 dx =
0
∫
5 1
x ln tan −1 dx , 10
x4 = dx ak 3 sin −1 k 2 + b , where a and b are constants to be 4 1− x
0
determined.
5
[1]
[4]
The sequence of numbers u1 , u2 , u3 , is given by u1 = 2 and un +1 =
n+2 for all positive nun
integers.
6
(i)
By writing down the terms u2 and u3 , make a conjecture for un in terms of n . [2]
(ii)
Prove your conjecture by mathematical induction.
[4]
(iii)
Write down the limit of unun −1 as n tends to infinity.
[1]
(i)
Verify that r 4 + r 2 + 1= (r 2 + r + 1)(r 2 − r + 1).
[1]
(ii)
By considering
1 1 − 2 , show that r + r +1 r − r +1 2
∑r N
4
r =1
∞
(iii)
Evaluate
∑ r =2
r = + r2 +1
1 1 1 − 2 . 2 N + N +1
r . r + r2 +1
RIVER VALLEY HIGH SCHOOL
[4]
[2]
4
9740/01/2010
4
7
ax 2 + bx + c where a , b , c and x+d d are constants. The asymptotes, also shown in the diagram, are x = −2 and y= x − 4 .
The diagram below shows a sketch of the graph of y =
y
4
−2 O
x
−4
(i)
Write down the value of d and find the values of a and b .
[4]
(ii)
Given that the curve passes through the point ( 0 , −4.5 ) , find the value of c .
[1]
(iii)
By sketching the graph of ( x − 4 ) + y 2 =, k 2 determine the range of values of k 2
2
ax 2 + bx + c such that the equation ( x − 4 ) + k 2 , where a , b , c and d are = x+d the values found above, has at least one negative root. [3] 2
8
(i)
Solve the equation ( z + 2)3 = −8 , giving the roots in the form reiθ − 2 , where r > 0 and − π < θ ≤ π .
[4]
(ii)
Express ( z + 2)3 + 8 as a product of three linear factors.
[2]
(iii)
Hence, find the values of the real numbers a and b in the equation z 2 + az + b = 0 where the roots are the complex numbers found in (i). [2]
(iv)
Explain why the roots in (i) lie on a circle centre (−2,0) and radius 2.
RIVER VALLEY HIGH SCHOOL
9740/01/2010
[1]
5
9
x ) ln (1 + sin x ) , where − It is given that f (=
3π 3π . ≤x≤ 8 8
(i)
Show that (1 + sin x)f ′ ( x ) = cos x .
(ii)
By further differentiation of this result, find the Maclaurin series for f ( x ) , up to
[1]
and including the term in x3 . Hence, write down the equation of the tangent of [5] y = f ( x ) at x = 0 . Denote the Maclaurin series of f ( x ) in (ii) by g ( x ) . (iii)
On the same diagram, sketch the graphs of y = f ( x ) and y = g ( x ) for −
(iv)
10
3π 3π . ≤x≤ 8 8
[2]
3π 3π , the set of values of x for which the value of g ( x ) is ≤x≤ 8 8 within ±0.05 of the value of f ( x ) . [2]
Find, for −
(a)
Given that the first, third and fourth terms of an arithmetic progression are three consecutive terms of a geometric progression, and that the sum of the first twenty even-numbered terms of the arithmetic progression is 960, find the common difference of the arithmetic progression. [4]
(b)
Annie puts $x on 1 January 2010 into a bank account which pays compound interest at a rate of 3% per month on the last day of each month. On the first day of each subsequent month, she puts in an amount which is $x more than the amount she puts in the previous month. For example, she puts in $x on 1 January, $2x on 1 February and so on. (i)
Show that the amount of money in the bank account on the last day of 103 x 103(1.03)3 − 112 . March is $ [4] 9
(ii)
Find the least integer value of x so that the amount of interest earned for the first three months of the year 2010 exceeds $100. [2]
RIVER VALLEY HIGH SCHOOL
9740/01/2010
6
11
The lines l1 and l2 meet at P. The line l3 lies on the same plane containing l1 and l2 , and is perpendicular to l1 . Given that l1 and l2 are parallel to the vectors a and b respectively,
a ⋅b show that l3 is parallel to the vector b − 2 a . a The
lines
l1
and
have
l2
equations
[3]
r = 3i − 5 j + 2k + t ( 4i − j + 2k )
and
r = 3i − 5 j + 2k + s ( 8i − 4 j + 3k ) respectively. The line l3 lies on the plane containing l1 and l2 and is perpendicular to l1 . Given that the three lines intersect, find the equation of [3] l3 . Find also the angle between l2 and l3 . The plane Π1 contains the lines l1 , l2 and l3 , and the plane Π 2 has equation 5 kx + 2 y − 4 z = 9 . Given that Π1 is parallel to Π 2 , show that k = . Hence, find the exact 2 distance between the two planes. [6]
12
(i)
Sketch the graph of y =
2 3 2 2
(4 − x )
.
[2]
(ii)
The region R is bounded by the curve, the axes and the line x = 3 . Using the substitution x = 2 cos θ , find the exact area of R. [5]
(iii)
Find the exact volume of revolution when R is rotated completely about the y-axis.
− End of Paper −
RIVER VALLEY HIGH SCHOOL
9740/01/2010
[6]
1
Name
(
)
Class
Tutor
Calculator Model
RIVER VALLEY HIGH SCHOOL 2010 Year 6 Preliminary Examinations Higher 2
MATHEMATICS
9740/01 14 September 2010
Paper 1 Cover Page
INSTRUCTIONS TO CANDIDATES Attach this cover page on top of your answer paper. Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER.
Question
Mark
Max. Mark
Question
1
6
7
8
2
6
8
9
3
6
9
10
4
6
10
10
5
7
11
12
6
7
12
13
Total
100
Grade
RIVER VALLEY HIGH SCHOOL
9740/01/2010
Mark
Max. Mark
Name
(
)
Class
RIVER VALLEY HIGH SCHOOL 2010 Year 6 Preliminary Examinations Higher 2
MATHEMATICS
9740/02 15 September 2010
Paper 2
3 hours Additional Materials:
Answer Paper List of Formulae (MF15) Cover Page
READ THESE INSTRUCTIONS FIRST
Do not open this booklet until you are told to do so. Write your name, class and index number in the space at the top of this page. Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, place the cover page on top of your answer paper and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages.
2
Section A: Pure Mathematics [40 marks]
1
The diagram shows the graph of y = f(x). The curve passes through the origin O, the points A(–3a, 2) and B(–a, –2), where 0 < a < 1 . y
x=a
A(−3a, 2)
x
0 B (−a, −2)
Sketch on separate clearly labelled diagrams, the graphs of
2
(a)
1 y= −f x + 1 ; 2
[4]
(b)
y = f '( x) .
[2]
Points O, A and B are such that OA =i + 2 j − 3k and OB= 4i − j .
(i)
Use vector product to find the exact area of the triangle OAB.
(ii)
A point P on line AB is such that AP : AB = 2 : 3. Given that point Q lies on OP produced, show that OQ [2] = c(3i − k ) , for some constant c.
(iii)
Given also that triangle AQB has a right angle at Q, find the value of c.
RIVER VALLEY HIGH SCHOOL
9740/02/2010
[2]
[3]
3
3
The curve C has parametric equations x= 2t + t 2 ,
y= (1 − t ) 2 ,
where t ∈ .
4
(a)
Find the range of values of t for which C is decreasing.
[3]
(b)
Find the equation of the normal to the curve at t = 2 . Determine if the normal meets C again. [4]
A long cylindrical metal bar is submerged into iced water. A researcher claims that the rate at which the length, l cm, of the bar is shrinking at any time t seconds is proportional to the volume of the bar at that instant, assuming that the cross-sectional area of the bar remains constant during the shrinking process. Formulate and integrate a differential equation to show that l = Ae kt , where A and k are constants. State the range of values of A and of k. [5] Given that the initial length of the bar is L, sketch a graph to show the relation between l and t. Comment on the suitability of the claim by the researcher. [3] It is later found that l and t are related by the equation l= B + Le kt , where B is some constant. Given that l = 0.5L when t = T, show that the length of the bar when t = 3T, is given by
5
( 0.5L − B ) B+ L2
3
.
[2]
The functions f and g are defined by f : x 2 x 3 + 1, g : x ln( x − a ),
x > 0, x>a.
(i)
Give a reason why f -1 exists. Hence find f -1 ( x) and state the domain of f -1 .
(ii)
Only one of the composite functions fg and gf exists. State the greatest possible value of a such that this composite function exists and using this value of a, give a definition of this composite function. Write down the range of the composite function. Explain why the other composite function does not exist. [5]
(iii)
By sketching suitable graphs or otherwise, find the set of values of x such that [2] f -1f ( x) = ff -1 ( x) .
RIVER VALLEY HIGH SCHOOL
9740/02/2010
[3]
4
Section B: Statistics [60 marks]
6
Research has shown that 5 out of 8 students of a particular school make use of the school’s e-learning portal. Three random samples of 40 students are chosen. Using a suitable approximation, find the probability that more than 80 students make use of the school’s e-learning portal. [3]
7
The sales of movie tickets sales are known to be high during the weekend, low during the beginning of the week (i.e. Monday and Tuesday) and somewhat moderate at midweek. A sample of 50 is to be chosen from 350 days of a year of movie ticket sales to estimate the ticket sales per day. Describe how the sample could be chosen using systematic random sampling.
[2]
Describe any disadvantage of using systematic sampling and explain briefly whether it would be more appropriate to use stratified sampling instead. [2]
8
A group of 10 students consisting of 6 males and 4 females are seated at a round table with chairs of different colours for lunch. Find the number of arrangements such that all the female students are seated together. [3] After lunch, the same group of 10 students bought tickets to attend a concert. If the tickets are for a particular row of 10 adjacent seats, find the number of possible seating arrangement when
9
(i)
the first and last seats are occupied by the same gender,
[3]
(ii)
four particular students did not turn up for the concert.
[1]
The average daily sales at a small food store is estimated to be $452.80. The daily sales, $x, over a period of 12 days is summarized by
622.8 ∑ ( x − 450) =
and
78798 . ∑ ( x − 450)2 =
(i)
Test, at 1% significance level, whether the food store has underestimated its average daily sales. State an assumption you made about the data. [6]
(ii)
Suppose in a test at 5% significance level, it is found that there is significant evidence of an increase in average daily sales. Using only this information, and giving a reason, discuss whether there is significant evidence at 5% significance level that the average daily sales has changed. [2]
RIVER VALLEY HIGH SCHOOL
9740/02/2010
5
10
(a)
(b)
A student threw a fair die four times and recorded the number obtained for each throw. Find the probability that (i)
at least two of the four numbers obtained are the same,
[2]
(ii)
exactly two of the four numbers obtained are the same given that the number 6 appeared at least once.
[4]
Events A and B are such that P( B′) = the probability P( A) .
11
1 7 4 , P( A′ ∩ B′) = and P( B A) = . Find 2 20 7 [3]
An experiment was conducted to verify how the radiation intensity x from a radioactive source varies with time t. The following data were obtained from a particular source. x t
1 15.3
4 6.9
6 4.4
8 2.7
9 2.1
10 1.7
12 1.0
15 0.49
(i)
Plot a scatter diagram for the above data.
(ii)
Calculate the product moment correlation coefficient for the above data and comment on it, in relation to your scatter diagram in (i). [2]
(iii)
Calculate the equation of the least square regression line of x on t. Give a practical interpretation of the coefficient of t. [2]
(iv)
The variable y is defined by y = ln x. For the variables y and t, calculate the product moment correlation coefficient and hence compare the suitability of this logarithmic model and the linear model in (iii). [2]
(v)
Use an appropriate regression line to give the best estimate that you can of the time when x = 10. [2]
RIVER VALLEY HIGH SCHOOL
9740/02/2010
[1]
6
12
13
The two most common types of disciplinary offences in a particular boy school is keeping long hair and failure to wear the school badge. The mean number of disciplinary offences recorded per day involving long hair is 1.12. Assuming that each school week consists of five school days, the mean number of disciplinary offences recorded per school week involving failure to wear the school badge is 4.2. The number of cases for each disciplinary offence is assumed to have an independent Poisson distribution. (i)
Find the probability that at most 9 cases of disciplinary offence are recorded in a given school week. [3]
(ii)
In a school week in which there are more than 7 cases of disciplinary offence involving long hair, find the probability that at most 9 cases of disciplinary offence are recorded. [3]
(iii)
Calculate the probability that on a Thursday in a particular school week, it is the third day in the school week in which the discipline master caught at least 4 students having long hair in a day. (You may assume that Monday is the first day of school for a school week.) [3]
(iv)
Explain why the Poisson distribution may not be a good model for the number of disciplinary cases involving long hair, in a school year. [1]
A farm in Kranji rears chickens and ducks for sale to the local market. (a)
The mass of a randomly chosen chicken has mean 2.5 kg and standard deviation 0.4 kg. If the probability that the mean mass of a large sample of n chickens is greater than 2.45 kg exceeds 0.95, find the least value of n. [3]
(b)
The variable X denotes the mass, in kg, of a randomly chosen duck which is normally distributed with mean 2.7 and standard deviation σ . The ducks are sold at $4 per kg. (i)
50 P( X < 2.9) , show that the value of σ is 107 0.424, correct to 3 significant figures. [2]
Given that P( X < 2.5) =
(ii)
A housewife brings $50 to the farm. Calculate the probability that she has less than $27.50 after buying 2 ducks. [3]
(iii)
A sample of 4 ducks is randomly chosen. Calculate the probability that the most expensive duck costs less than $11. [2]
− End of Paper −
RIVER VALLEY HIGH SCHOOL
9740/02/2010
1
Name
(
)
Class
Tutor
Calculator Model
RIVER VALLEY HIGH SCHOOL 2010 Year 6 Preliminary Examinations Higher 2
MATHEMATICS
9740/02 15 September 2010
Paper 2 Cover Page
INSTRUCTIONS TO CANDIDATES Attach this cover page on top of your answer paper. Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER.
Question
Mark
Max. Mark
Question
1
6
6
3
2
7
7
4
3
7
8
7
4
10
9
8
5
10
10
9
11
9
12
10
13
10
Total RIVER VALLEY HIGH SCHOOL
100
Grade
9740/02/2010
Mark
Max. Mark
RVHS 2010 Preliminary Exam: 9740 H2 Mathematics Paper 1 Solutions Question 1 [6 marks] Let the no. of small, medium and large bottles manufactured be denoted by s, m and l respectively. So, 150 s + 335m + 475l = 280400 , 200 s + 450m + 600l = 370700 and 3.8m = 2(2.5s + 4.2l ) Using GC, solving the augmented matrix: ⎛ 150 335 475 280400 ⎞ ⎜ ⎟ ⎜ 200 450 600 370700 ⎟ ⎜ 5 −3.8 8.4 0 ⎟⎠ ⎝ Ans: s = 166, m = 550 and l = 150
Assumption: The plastic bottles are of negligible thickness. OR The plastic bottles are of the same thickness. Question 2 [6 marks] (i) iz + 1 ≤ i z+
1 ≤1 i
z −i ≤1
(ii)
z − z* ≤ 2 x + iy − ( x − iy ) ≤ 2 2iy ≤ 2 −1 ≤ y ≤ 1 Im(z)
1 1
Re(z)
0
RVHS 2010 H2 Math Prelim P1
1
arg( z + 1 + i) = arg( z − (−1 − i))
Im(z)
1
1
Re(z)
0 −1 − i Max. arg( z + 1 + i) =
π
2 Min. arg( z + 1 + i) =α.
α
.
α = β − θ = tan −1 2 − sin −1
1 = 0.644 (3 dec. pl.) 5
Question 3 [6 marks] ⎛ θ2 ⎞ 4 − ⎜1 − ⎟ 2 ⎠ 4 − cos θ ⎝ = 2 + tan θ 2 +θ
3+ =
θ2
2 2 +θ −1
⎛ θ 2 ⎞ ⎡ ⎛ θ ⎞⎤ = ⎜ 3 + ⎟ ⋅ ⎢ 2 ⎜1 + ⎟ ⎥ 2 ⎠ ⎣ ⎝ 2 ⎠⎦ ⎝ ⎛ θ2 ⎞ 1⎛ θ θ2 ⎞ = ⎜ 3 + ⎟ ⋅ ⎜1 − + + ... ⎟ 2 ⎠ 2⎝ 2 4 ⎝ ⎠ 3 3 5 ≈ − θ + θ2 2 4 8
3 4− 2 π 2 ≈ 3 − π + 5π Let θ = : 6 2+ 1 2 8 288 3 (8 − 3) 3 3 π 5π 2 ≈ − + 2 8 288 2(2 3 + 1)
RVHS 2010 H2 Math Prelim P1
2
(8 3 − 3)(2 3 − 1) 3 π 5π 2 ≈ − + 2(4(3) − 12 ) 2 8 288
51 − 14 3 3 π 5π 2 ≈ − + 22 2 8 288 9 11π 55π 2 (shown) ∴ 3≈ + − 7 56 2016
Question 4 [6 marks]
(i) (ii) (iii)
d 2x sin −1 x 2 ) = ( dx 1 − x4 5 ⎛ −1 ⎛ x ⎞ ⎞ ∫ 1 ln ⎜⎝ tan ⎜⎝ 10 ⎟⎠ ⎟⎠ dx = −5.292774105 = −5.29 (3 s.f.) k dv 2 −1 2 (letting u = sin −1 x 2 and = x2 ) ∫0 x sin x dx dx k
k 2 ⎡ x3 ⎤ x4 dx = ⎢ sin −1 x 2 ⎥ − ∫ ⋅ ⎣3 ⎦0 0 3 1 − x4 k 2 k3 x4 dx = sin −1 k 2 − ∫ ⋅ 0 3 3 1 − x4 k 2 k3 x4 sin −1 k 2 − ∫ ⋅ dx = − 5.292774105 0 3 3 1 − x4 Hence k ⎤ x4 3 ⎡ k3 −1 2 d x = ⎢ sin k + 5.292774105⎥ ∫0 1 − x4 2⎣ 3 ⎦
∴
1 = k 3 sin −1 k 2 + 7.94 , 2 where a = 0.5 and b = 7.94
Question 5 [7 marks] (i) 3 4 u2 = , u3 = 2 3 Conjecture: un = (ii)
n +1 n
Let P(n) be the statement “ un =
n +1 ”, for all positive integers n. n
When n = 1: LHS = u1 = 2 = RHS (given)
∴ P(1) is true.
RVHS 2010 H2 Math Prelim P1
3
Assume P(k) is true for some integers k ∈ Z + , i.e uk = To show: P(k+1) is true, i.e. uk +1 =
k +1 k
k+2 . k +1
Show: LHS = uk +1 =
k +2 kuk
k +2 k . k k +1 k +2 = = RHS k +1 =
Since P(1) is true and P(k) is true ⇒ P(k+1) is true,
∴ By Mathematical Induction, P(n) is true for all positive integers n. (iii)
un =
n +1 (n − 1)un −1
⇒ unun −1 =
2 n +1 = 1+ → 1 as n → ∞ (n − 1) n −1
Question 6 [7 marks] (i) RHS = (r 2 + r + 1)( r 2 − r + 1)
= r4 − r3 + r 2 + r3 − r 2 + r + r 2 − r +1 = r 4 + r 2 + 1 =LHS (ii)
1 1 − 2 r + r +1 r − r +1 r 2 − r + 1 − (r 2 + r + 1) = 2 (r + r + 1)(r 2 − r + 1) −2r = 2 (r + r + 1)(r 2 − r + 1) −2r = 4 2 r + r +1 2
RVHS 2010 H2 Math Prelim P1
4
∑ N
r =1
=−
r 1 =− 4 2 r + r +1 2 1 2
∑ ⎛⎜⎝ r N
r =1
2
∑r N
r =1
4
2r + r2 +1
1 1 ⎞ − 2 ⎟ + r +1 r − r +1⎠
1⎛1 = − ⎜ −1 2⎝3 1 1 + − 7 3 1 1 + − 13 7 +... +
1 1 ⎞ − 2 ⎟ + + − + N N 1 N N 1⎠ 2
1⎛ 1 ⎞ = − ⎜ −1 + 2 ⎟ N + N +1 ⎠ 2⎝ 1⎛ 1 ⎞ = ⎜1 − 2 ⎟ 2 ⎝ N + N +1 ⎠ (iii)
∞
∑ r =2
r = 4 r + r2 +1
∞
∑r
4
r =1
1⎛ 1 r 1 ⎞ 1 = lim ⎜1 − 2 − 4 2 ⎟− 2 N →∞ 2 ⎝ N + N +1⎠ 3 + r +1 1 +1 +1
1 Since lim 2 → 0, N →∞ N + N + 1 1⎛ 1 ⎞ 1 1 1 1 ∴ lim ⎜1 − 2 ⎟− = − = . N →∞ 2 ⎝ N + N +1⎠ 3 2 3 6 Question 7 [8 marks] d =2 (i) Method 1: Using long division: c − 2 ( b − 2a ) ax 2 + bx + c y= = ax + ( b − 2a ) + x+d x+2 ∴ a = 1 , b = −2
Method 2:
( x − 4 )( x + 2 ) + h = x 2 − 2 x − 8 + h h = x+2 x+2 x+2 ∴ a = 1 , b = −2 y = x−4+
(ii)
y=
x2 − 2x + c x+2 c 2 ∴ c = −9
At ( 0, −4.5 ) : −4.5 =
RVHS 2010 H2 Math Prelim P1
5
(iii) y
(4, k)
4
−2 O −4
( x − 4)
2
+ y 2 = k 2 is a circle centre (4, 0), radius k
k > 4.52 + 42 ⇒k<−
145 145 or k > 2 2
Question 8 [9 marks] (i) ( z + 2)3 = −8 ( z + 2)3 = 23 e(π + 2 kπ )i z + 2 = 2e
π 2 kπ ( + )i 3 3
π 2 kπ
( +
(ii)
, k = 0, ±1
)i
z = 2e 3 3 − 2, k = 0, ±1 When k = 1 , z = 2eπ i − 2 = 2(cos π + i sin π ) − 2 = −2 − 2 = −4 π
i π π⎞ ⎛ k = 0, z = 2e 3 − 2 = 2 ⎜ cos + i sin ⎟ − 2 3 3⎠ ⎝
⎛1 3 ⎞ i ⎟⎟ − 2 = −1 − 3i = 2 ⎜⎜ − ⎝2 2 ⎠ π − i ⎛ ⎛ π⎞ ⎛ π ⎞⎞ k = −1, z = 2e 3 − 2 = 2 ⎜ cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎟ − 2 ⎝ 3⎠ ⎝ 3 ⎠⎠ ⎝ ⎛1 3 ⎞ = 2 ⎜⎜ + i ⎟⎟ − 2 = −1 + 3i ⎝2 2 ⎠ So the 3 roots are z = −1 ± 3i and −4 .
( z + 2)3 + 8 = [ z − (−1 + 3i ) ][ z − (−1 − 3i ) ][ z + 4 ]
= [ z + 1 − 3i ] [ z + 1 + 3i ] [ z + 4 ] RVHS 2010 H2 Math Prelim P1
6
4+k
x
(iii)
[ z + 1 − 3i ][ z + 1 + 3i ]
(
) (
)
= z 2 + 1 − 3i z + 1 + 3i z + 4 = z2 + 2z + 4
(iv)
Hence a = 2 and b = 4 . ( z + 2)3 = −8 3
z+2 =8 z+2 =2 Thus, the roots in (i) lie on a circle centre (-2,0) and radius 2. (or deduce answer from the form in answers of (i)) Question 9 [10 marks] (i) f ( x ) = ln (1 + sin x ) cos x 1 + sin x ⇒ (1 + sin x ) f ′ ( x ) = cos x (shown)
∴f ′( x) = (ii)
Differentiate with respect to x: (1 + sin x ) f ′′ ( x ) + ( cos x ) f ′ ( x ) = − sin x Differentiate with respect to x: (1 + sin x ) f ′′′ ( x ) + ( cos x ) f ′′ ( x ) − ( sin x ) f ′ ( x ) + ( cos x ) f ′′ ( x ) = − cos x At x = 0 : f ( x ) = 0 , f ′ ( x ) = 1 , f ′′ ( x ) = −1 , f ′′′ ( x ) = 1 ∴f ( x) ≈ x −
x 2 x3 + 2 6
Equation of tangent of y = f ( x ) at x = 0 is y = x . (iii)
y ⎛ 3π ⎞ ⎜ , 0.757 ⎟ ⎝ 8 ⎠ ⎛ 3π ⎞ ⎜ , 0.654 ⎟ ⎝ 8 ⎠
x
O
g ( x) = x −
1 2 1 3 x + x 2 6
⎛ 3π ⎞ ⎜ − , − 2.14 ⎟ ⎝ 8 ⎠
f ( x ) = ln (1 + sin x )
⎛ 3π ⎞ ⎜ − , − 2.58 ⎟ ⎝ 8 ⎠
RVHS 2010 H2 Math Prelim P1
7
(iv)
To find x such that f ( x ) − g ( x ) < 0.05 Using GC, draw the graph of y = f ( x ) − g ( x ) to find the intersection with y = 0.05 .
−0.7745598 < x < 0.96923979 Range is −0.775 < x < 0.969. Question 10 [10 marks] (a) Let the first term and common difference of the AP be a and d respectively. a + 2d a + 3d = a a + 2d 2 ( a + 2d ) = a ( a + 3d )
a 2 + 4ad + 4d 2 = a 2 + 3ad a = −4d ------ (1) 20 ⎡ 2 ( a + d ) + (19 ) 2d ⎤⎦ = 960 2 ⎣ 2a + 40d = 96 ------ (2) Subs. (1) into (2): −8d + 40d = 96 ⇒d = 3 (b) (i)
Mth Jan Feb Mar
Amt at beginning of mth x 1.03x + 2x 1.032x +2(1.03)x + 3x
Amt at end of mth 1.03x 1.032x +2(1.03)x 1.033x + 2(1.03)2x + 3(1.03)x
∴ Amt on the last day of March = $[1.033x + 2(1.03)2x + 3(1.03)x] = $x[1.033 + 1.032 + 1.03 + 1.032 + 1.03 + 1.03]
⎡1.03 (1.033 − 1) 1.03 (1.032 − 1) 1.03 (1.031 − 1) ⎤ ⎥ + + 1.03 − 1 1.03 − 1 ⎥ ⎢⎣ 1.03 − 1 ⎦
=$x ⎢ = $x
1.03 (1.033 + 1.032 + 1.03 − 3) 0.03
RVHS 2010 H2 Math Prelim P1
8
(b) (ii)
3 103 x ⎛ 1.03 (1.03 − 1) ⎞ ⎜ =$ − 3⎟ ⎟ 3 ⎜ 0.03 ⎝ ⎠ 3 ⎛ ⎞ 103 x 103 (1.03 − 1) − 9 ⎜ ⎟ =$ ⎟ 3 ⎜ 3 ⎝ ⎠ 103x =$ (103(1.03)12 − 112 ) (shown) 9 103x 103(1.03)3 − 112 ) − ( x + 2 x + 3 x ) > 100 ⇒ x > 328.37 ( 9 ∴ least value of x is 329
Question 11 [12 marks] uuur Let PB be b and F be the foot of perpendicular from B to l1. l3
l1
F
a
P b
B
l2
⎛ a ⋅b ⎞ uuur Then ⎜ 2 ⎟ a = b ⋅ a$ a$ = PF ⎜ a ⎟ ⎝ ⎠ ⎛ a ⋅b ⎞ uuur uuur uuur ∴ b − ⎜ 2 ⎟ a = PB − PF = FB , ⎜ a ⎟ ⎝ ⎠ which is parallel to l3.
( )
Alternative: Use cross product. ⎛4⎞ ⎛8⎞ ⎜ ⎟ ⎜ ⎟ −1 ⋅ −4 ⎛ 8 ⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎛ 4 ⎞ ⎛ a ⋅b ⎞ 2 3 ⎜ ⎟ ⎜ ⎟ b − ⎜ 2 ⎟ a = ⎜ −4 ⎟ − ⎝ 2 ⎠ 2⎝ 2⎠ ⎜ −1⎟ ⎜ a ⎟ ⎜ 3 ⎟ (4 +1 + 2 ) ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛8⎞ ⎛4⎞ ⎛ 0⎞ = ⎜⎜ −4 ⎟⎟ − 2 ⎜⎜ −1⎟⎟ = ⎜⎜ −2 ⎟⎟ ⎜ 3 ⎟ ⎜ 2 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Since P(3, –5, 2) is a common point, ⎛ 3⎞ ⎛ 0⎞ ⎜ ⎟ ∴ l3 : r = ⎜ −5 ⎟ + μ ⎜⎜ −2 ⎟⎟ , μ ∈ . ⎜ 2⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎝ ⎠
RVHS 2010 H2 Math Prelim P1
9
⎛ ⎞ ⎛8⎞ ⎛ 0⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −4 ⎟ ⋅ ⎜ −2 ⎟ ⎜ 3 ⎟ ⎜ −1 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎟ Angle between l2 and l3 = cos −1 ⎜ ⎜ 64 + 16 + 9 0 + 4 + 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ o = 76.3 ⎛4⎞ ⎛ 0⎞ ⎛5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ n1 = ⎜ −1⎟ × ⎜ −2 ⎟ = ⎜ 4 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ −1 ⎠ ⎝ −8 ⎠ ⎛k ⎞ ⎛5⎞ ⎜ ⎟ So, ⎜ 2 ⎟ = c ⎜⎜ 4 ⎟⎟ ⎜ −4 ⎟ ⎜ −8 ⎟ ⎝ ⎠ ⎝ ⎠
1 ⎧ ⎪⎪ c = 2 ⇒⎨ ⎪k = 5 ⎪⎩ 2 ⎛5⎞ ⎜ ⎟ Π 2 : r ⎜ 4 ⎟ = 18 ⎜ −8 ⎟ ⎝ ⎠ ⎛2⎞⎛5⎞ Since ⎜⎜ 0 ⎟⎟ ⎜⎜ 4 ⎟⎟ = 18 , a point on Π 2 is Q ( 2, 0, −1) . ⎜ −1⎟ ⎜ −8 ⎟ ⎝ ⎠⎝ ⎠ ⎛ 2 ⎞ ⎛ 3 ⎞ ⎛ −1 ⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ PQ = ⎜ 0 ⎟ − ⎜ −5 ⎟ = ⎜ 5 ⎟ ⎜ −1 ⎟ ⎜ 2 ⎟ ⎜ − 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ uuur Required distance = PQ ⋅n1
⎛ −1 ⎞ ⎜ ⎟ = ⎜ 5 ⎟⋅ ⎜ −3 ⎟ ⎝ ⎠
=
⎛5⎞ 1 ⎜ ⎟ 4 25 + 16 + 64 ⎜⎜ ⎟⎟ ⎝ −8 ⎠
13 105 35
Alternative Method 1 to calculate distance:
RVHS 2010 H2 Math Prelim P1
10
∏1 : r.
⎛5⎞ ⎜ ⎟ ⎜4⎟ ⎜ −8 ⎟ ⎝ ⎠ 105
∏ 2 : r.
⎛5⎞ ⎜ ⎟ ⎜4⎟ ⎜ −8 ⎟ ⎝ ⎠ 105
=
−21 105
=
18 105
Thus, O is between the planes. 21 + 18 = 39 So, distance = 105 105 105 Alternative Method 2 to calculate distance: Equation of line through P and perpendicular to Π1 and Π 2 : ⎛ 3⎞ ⎛5 ⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜ −5 ⎟ + μ ⎜ 4 ⎟ ⎜ 2⎟ ⎜ −8 ⎟ ⎝ ⎠ ⎝ ⎠ Let N be the foot of perpendicular from P to Π 2 . ⎡⎛ 3 ⎞ ⎛ 5 ⎞⎤ ⎛ 5 ⎞ 13 ⎢⎜ ⎟ ⎜ ⎟⎥ ⎜ ⎟ ⎢⎜ −5 ⎟ + μ ⎜ 4 ⎟ ⎥ ⎜ 4 ⎟ = 18 ⇒ μ = 35 ⎜ −8 ⎟ ⎥ ⎜ −8 ⎟ ⎢⎣⎜⎝ 2 ⎟⎠ ⎝ ⎠⎦ ⎝ ⎠ ⎛ 3⎞ ⎛5 ⎞ uuur ⎜ ⎟ 13 ⎜ ⎟ ∴ ON = ⎜ −5 ⎟ + ⎜ 4 ⎟ ⎜ 2 ⎟ 35 ⎜ −8 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛5 ⎞ uuur 13 ⎜ ⎟ 13 105 Required distance = PN = 4 = 35 ⎜⎜ ⎟⎟ 35 ⎝ −8 ⎠
Question 12 [13 marks] (i)
y=
y
2 3 2 2
(4 − x )
2 0.25 –2
(ii)
0
Required area =
32
∫
3
0
RVHS 2010 H2 Math Prelim P1
2 3 2 2
(4 − x )
x
dx
11
=
=
∫
∫
π 6 π 2
π 2 π 6
1 = 2
2
( 4 − 4 cos θ ) 2
2 8 ( sin θ ) 2
∫
π 2 π 6
3 2
3 2
⋅ −2sin θ dθ
⋅ 2sin θ dθ
cos ec 2θ dθ
π 1 [ − cot θ ] π2 2 6 1⎡ = ⎣0 + 3 ⎤⎦ 2 3 = units2 2
=
(iii) y=
2 3 2 2
(4 − x )
⇒
Required Vol = π
3 2 2
(4 − x )
2
⎛ 2 ⎞3 ⇒ x =4 −⎜ ⎟ ⎝ y⎠
2 = y
2
⎛
( 3 ) ( 2) − π ∫ ⎜⎝ 4 − 2 2
2
1 4
(or π ∫
2
0
2
3 ⋅ (2) dy ) 2
2 1 ⎡ ⎤ 3 = 6π − π ⎢ 4 y − 2 ⋅ 3 y 3 ⎥ ⎣ ⎦1
4
= 6π − π [ (8 − 6) − (1 − 3)] = 2π units3
RVHS 2010 H2 Math Prelim P1
12
2 3
⋅y
−
2 3
⎞ ⎟ dy ⎠
RVHS 2010 Preliminary Exam: 9740 H2 Mathematics Paper 2 Solutions Question 1 [6 Marks] (a)
x = 2(a − 1)
y
B '(−2a − 2, 2)
x
−2
0
A '(−6a − 2, −2)
dy dx
(b)
−3a
−a
x a
0
Question 2 [7 Marks] (i) 1 uuur uuur Area of triangle OAB = OA × OB 2 ⎛1⎞ ⎛4⎞ 1⎜ ⎟ ⎜ ⎟ = ⎜ 2 ⎟ × ⎜ −1⎟ 2⎜ ⎟ ⎜ ⎟ ⎝ −3 ⎠ ⎝ 0 ⎠ 1 = 2 =
⎛ −3 ⎞ ⎜ ⎟ ⎜ −12 ⎟ ⎜ −9 ⎟ ⎝ ⎠
3 26 units2 2
RVHS 2010 H2 Math Prelim P2 Solutions
1
(ii)
uuur 1 uuur uuur OP = OA + 2OB by ratio thm 3 ⎡⎛ 1 ⎞ ⎛ 4 ⎞ ⎤ ⎛ 3 ⎞ 1 ⎢⎜ ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ = ⎢⎜ 2 ⎟ + 2 ⎜ −1⎟ ⎥ = ⎜ 0 ⎟ 3 ⎜ ⎟ ⎜ ⎟ ⎢⎣⎝ −3 ⎠ ⎝ 0 ⎠ ⎥⎦ ⎝⎜ −1⎠⎟
(
)
⎛3⎞ uuur uuur uuur ⎜ ⎟ Since OQ // OP , OQ = c ⎜ 0 ⎟ for some constant c. ⎜ ⎟ ⎝ −1⎠
(iii)
uuur uuur AQ BQ = 0 ⎛ 3c − 1⎞ ⎛ 3c − 4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −2 ⎟ ⎜ 1 ⎟ = 0 ⎜ 3 − c ⎟ ⎜ −c ⎟ ⎝ ⎠ ⎝ ⎠
9c 2 − 12c − 3c + 4 − 2 − 3c + c 2 = 0 5c 2 − 9c + 1 = 0 Using GC, c = 1.68 or c = 0.119 (rej. since OQ > OP) Question 3 [7 Marks] (a) x = 2t + t 2 , y = (1 − t ) 2 dx dy = 2 + 2t , = 2(1 − t ) 2 (−1) = −2(1 − t ) dt dt dy −2(1 − t ) −2(1 − t ) (t − 1) = = = dx 2 + 2t 2(1 + t ) t +1 dy For decreasing C, < 0 . dx (t − 1) <0 t +1 Method 1: −
+
−1
+
1
So −1 < t < 1 .
Method 2: Plot the graph of y =
(t − 1) and find the range of values of t such that y < 0 . t +1
RVHS 2010 H2 Math Prelim P2 Solutions
2
So −1 < t < 1 . (b)
At t = 2 , x = 2(2) 2 + 22 = 8, y = (1 − 2) 2 = 1 . At ( 8, 1) , the equation of normal is
gradient = −
t +1 3 = − = −3 t − 1 t =2 1
y − 1 = −3 ( x − 8 ) y = −3 x + 25
Substitute the parametric equations into the equation of the normal. (1 − t ) 2 = −3(2t + t 2 ) + 25 1 − 2t + t 2 = −6t − 3t 2 + 25 4t 2 + 4t − 24 = 0 t2 + t − 6 = 0 1 − 4(1)(−6) = 25 > 0 Thus, the normal will meet the C again. [Alternative: Factorise and solve the equation.] Question 4 [10 Marks] dl = kl dt 1 ∫ l dl = ∫ k dt ln l = kt + C (note: l > 0) l = e kt + C l = Ae kt (shown) A > 0 and k < 0
When t = 0, l = L: A = L ∴ l = Le kt l L
0
RVHS 2010 H2 Math Prelim P2 Solutions
t
3
The model suggested by the researcher is not suitable as l → 0 when t → ∞ (i.e. the bar vanished). t = T , l = 0.5L : 0.5 L = B + Le kT 0.5 L − B ⇒ = e kT L When t = 3T, l = B + Le
3 kT
=
( 0.5L − B ) B+
3
L2
Question 5 [10 Marks] (i)
Any horizontal line cuts the graph of f at most 1 point. Therefore, f is a one to one function and f -1 exists. f : x a 2 x3 + 1,
x > 0.
Let y = 2 x + 1 3
3
x=
y −1 2 3
f -1 : x → (ii)
x −1 , x >1 2
For gf to exist, (1, ∞) = R f ⊆ Dg = (a, ∞) . Hence largest possible a = 1 . For a = 1 , gf ( x) = g (2 x3 + 1) = ln(2 x3 + 1 − 1) = ln(2 x3 ), x > 0 Rgf = (−∞, ∞) . Since
= Rg ⊄ D f = (0, ∞) , so fg does not exist.
RVHS 2010 H2 Math Prelim P2 Solutions
4
(iii)
f -1f ( x) = x, x ∈ D f = (0, ∞)
ff -1 ( x) = x, x ∈ D
f −1
= (1, ∞)
y
y
y = f -1f ( x)
y = ff -1 ( x) (1, 1)
0
x
0
x
The set of values of x such that f -1f ( x) = ff -1 ( x) is (1, ∞) .
Question 6 [3 Marks] Let A be the r.v. denoting number of students out of 120 students who make use of the school’s e-learning portal. A ~ B(120, 85 )
Since n = 120 > 50 is large, np = 120(5 / 8) = 75 > 5 and nq = 120(3 / 8) = 45 > 5 . A ~ N (75, 225 ) approximately. 8
P( A > 80) = P( A ≥ 81) = P( A ≥ 80.5) (by cc) = 0.150 (3.s.f)
Question 7 [4 Marks] The 350 days of ticket sales volume are listed in order (numbered from 1 to 350) with th an interval of selection, 350 50 = 7 . So every 7 day is chosen. The first sample is selected randomly from the first 7 days, say the 2nd night is selected. Then the day 9th, 16th, 23rd,… up to a total of 50 days are chosen. To choose every 7th day means that a particular day sales will be chosen. E.g. if Monday is chosen, then the next 7th day will also be a Monday. But since the movie ticket sales are periodic in nature, the sample will be biased and not be representative of the actual ticket sales. Stratified sampling seems more appropriate here as we should choose samples from Weekends, Beginning of the week and Midweek as our strata so as to get a more representative sample.
RVHS 2010 H2 Math Prelim P2 Solutions
5
Question 8 [8 Marks] (7 − 1)!× 4!×10 = 172800 (i) Case 1: First and last seats occupied by females 4 × 3 × 8! ways
Case 2: First and last seats occupied by males 6 × 5 × 8! ways
Total number of ways = 4 × 3 × 8!+ 6 × 5 × 8! = 1693440 (ii)
10! = 151200 4!
Question 9 [8 Marks] (i) Let X be the random variable for the sales per day. n = 12 ∑ ( x − 450) + 450 = 622.8 + 450 = 501.9 x= 12 12 2 ( x − 450 ) ) ⎤⎥ ( 1 ⎡⎢ 2 ∑ 2 s = Σ ( x − 450 ) − 12 − 1 ⎢ 12 ⎥ ⎣ ⎦ 2 ( 622.8) ⎤ = 4224.970909 1⎡ = ⎢ 78798 − ⎥ 11 ⎣⎢ 12 ⎦⎥ s = 64.99977622 Test against
H0 : H1 :
μ = 452.80 μ > 452.80
at 1% level of significance. Test statistic: Under H0, T =
X − 452.80 ~ t (11) since n = 12 is small. s 12
p-value = P(T > tcalculated ) = 0.011983849 > 0.01 We do not reject H0. We conclude that there is insufficient evidence at 1% level of significance that the food store has underestimated its average daily sales.
Assume that the daily sales follow a normal distribution.
RVHS 2010 H2 Math Prelim P2 Solutions
6
Since H0 is rejected at 5% level of significance, p-value < 0.05 for the one-tailed test. For the two-tailed test to reject H0 at 5% level of significance, 2 x p-value* < 0.05, which may or may not be true. So there may or may not be significant evidence. * the original p-value for the one-tailed test
Question 10 [9 Marks] (a) P(at least 2 of the 4 numbers are the same) (i) = 1 − P(all the 4 numbers are different) = 1 − ( 66 )( 56 )( 64 )( 63 ) 13 = 18
Alternatively, P(at least 2 of the 4 numbers are the same) =P(exactly 2 same, and another 2 same) + P(exactly 2 same and 2 different) + P(exactly 3 same) + P(all 4 same) ⎛6⎞ 4 ⎛ 6⎞⎛ 6⎞ 4 ⎛ 6⎞⎛ 2⎞ 4 ⎛ 6⎞ 4! 1 4 = 2!2! ( 6 ) ⎜ 2 ⎟ + 2!4! ( 16 ) ⎜ 3 ⎟ ⎜1 ⎟ + 3!4! ( 16 ) ⎜ 2 ⎟ ⎜1 ⎟ + ( 16 ) ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ 13 = 18
(a) (ii)
at least one 6)
P(exactly 2 of the 4 numbers are same
P(2 out of the 4 numbers are same ∩ at least 1 six) P(at least one six) P(six appear twice) + P(six appear once) = P(at least one six) P(6, 6, B, C ) + P( A, A, 6, B) = 1 − P(no six) ⎛ 5 ⎞ 4! 1 4 ⎛ 5 ⎞⎛ 2 ⎞ 4! 1 4 2! ( 6 ) ⎜ ⎟ + 2! ( 6 ) ⎜ ⎟⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠⎝1 ⎠ = 4 1 − ( 56 ) =
360 or 0.537 671 4 P( A I B) 4 4 = ⇒ P( A I B) = P ( A) P ( B A) = ⇒ P( A) 7 7 7 P ( A U B ) = P( A) + P( B) − P( A I B ) 4 1 − P( A′ I B′) = P ( A) + [1 − P( B′)] − P( A) 7 4 0.65 = P( A) + 0.5 − P( A) 7 3 0.15 = P ( A) 7 P( A) = 0.35 =
(b)
RVHS 2010 H2 Math Prelim P2 Solutions
7
Question 11 [9 Marks] (i)
(ii)
r = −0.885 This illustrates a moderate linear correlation between the radiation intensity and time which agrees with the scatter diagram, where the points do not seems to lie close to a straight line.
(iii)
Equation of the regression line of x on t : x = 11.610575 − 0.806146t
x = 11.6 − 0.806t (correct to 3 s.f.)
There is a decrease of 0.806 unit in the radiation intensity with every unit increase in time. (iv)
r = −0.995 (3 s.f)
Since the value of the product moment correlation coefficient for ln x and t is closer to −1, the logarithmic model is more suitable than the linear model. (v)
Equation of the regression line of y on t : y = 2.624253 − 0.174726t
When x = 10, y = ln(10) = 2.624253 − 0.174726t , ⇒ t = 1.84098 ≈ 1.84 (correct to 3 s.f.)
Question 12 [10 Marks] (i) Let X and Y be the r.v. denoting the number of disciplinary cases involving long hair and faiure to wear school badge respectively in a given week. X ~ Po(1.12 × 5) and Y ~ Po(4.2) X + Y ~ Po(9.8) P( X + Y ≤ 9) = 0.483188233 ≈ 0.483 (3.s.f) (ii)
P( X + Y ≤ 9 X ≥ 8 ) P( X + Y ≤ 9 ∩ X ≥ 8) P( X ≥ 8) P( X = 8)P(Y ≤ 1) + P( X = 9)P(Y = 0) = P( X ≥ 8) = 0.038145 ≈ 0.0381 (3.s.f)
=
RVHS 2010 H2 Math Prelim P2 Solutions
8
(iii)
Let W be the r.v. denoting the number of disciplinary cases involving long hair recorded in a given day. W ~ Po(1.12) P (W ≥ 4) = 1 − P(W ≤ 3) = 0.0272442116 ≈ 0.0272 Require Prob ⎛ 3⎞ 2 = ⎜ ⎟ [ P(W ≤ 3)][ P(W ≥ 4)] P(W ≥ 4) ⎝ 2⎠
)
(
= 5.87 × 10−5 (3 s.f ) (iv)
The mean number of disciplinary cases involving long hair fluctuates throughout the year, usually having more such cases after the school vacation.
Question 13 [10 Marks] (a) Let C denotes the mass of a randomly chosen chicken. Since n is large, C ~ N (2.5,
0.42 ) n
approximately by Central limit Theorem.
P(C > 2.45) > 0.95 Method 1 1 − P(C ≤ 2.45) > 0.95 P(C ≤ 2.45) < 0.05 Let Z =
P( Z ≤
C − 2.5 0.42 n
~ N (0, 1)
2.45 − 2.5 0.4 n
) < 0.05
− 18 n < −1.64485 n > 13.15882901 n > 173.15 The least value of n is 174.
Method 2
The least value of n is 174.
RVHS 2010 H2 Math Prelim P2 Solutions
9
(b) (i)
X ~ N (2.7, σ 2 ) 50 P( X < 2.9) P( X < 2.5) = 107
50 1 − P( X ≥ 2.9) P( X < 2.5) = 107 [ ] 50 1 − P( X < 2.5) P( X < 2.5) = 107 [ ] 157 P( X < 2.5) = 50 ⇒ P( X 107 107 2.5 − 2.7 50 P( Z < σ ) = 157 50 P( Z < − 51σ ) = 157 − 51σ = −0.4719777448
50 < 2.5) = 157
σ = 0.424 (3.s.f) (Shown)
Alternative method: Plot using GC: Y1=normalcdf(-E99,2.5,2.7,X) Y2=(50/107)normalcdf(-E99,2.9,2.7,X) Sketch the graphs and find point of intersection. (b) (ii)
(b) (iii)
X1 + X 2 ~ N (2(2.7), 2(0.424) 2 )
4( X1 + X 2 ) ~ N (4 × 2(2.7), 42 × 2(0.424)2 ) P(50 − 4( X1 + X 2 ) < 27.5) = P(4( X1 + X 2 ) > 22.5) = 0.354 P(most expensive duck cost less than $11) = P(all ducks cost less than $11 each) 4 = [ P(4 X < 11)] = ⎡⎣ P( X < 11 )⎤ 4 ⎦ = 0.0895
RVHS 2010 H2 Math Prelim P2 Solutions
4
10
SAINT ANDREW’S JUNIOR COLLEGE Preliminary Examination MATHEMATICS Higher 2
9740/01
Paper 1 Wednesday
15 September 2010
3 hours
Additional materials : Answer paper List of Formulae(MF15) Cover Sheet
READ THESE INSTRUCTIONS FIRST Write your name, civics group and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Answer all the questions. Total marks : 100 Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically state otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematic steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages including this page. [Turn over
2 1.
2.
The first 4 terms of a sequence are given by u1 63 , u2 116 , u3 171 , u4 234 . Given that un is a cubic polynomial in n , find un in terms of n . Hence write down the value of u50 .
[3] [1]
Without using a graphic calculator, solve 1 x x 1 . 2 3x
Hence, solve 1 x 2 x 2 1 exactly. 2
[5]
2 3x
3.
Iron Will, the magician, is constructing a prism with an equilateral triangle base for his latest escape act. The edges of the prism are made from iron rods, the rectangular faces of the prism are made of glass panels, and the triangular faces of the prism are made of wooden boards. The volume of the prism is fixed at exactly 2 3 cubic metres. 8 Show that h 2 , where x is the length of the sides of the x triangular base and h is the height of the prism.
[2]
The cost of the iron rods is $1 per metre, the cost of the wooden boards is $ 2 3 per sq. metre and the cost of the glass panels is $2 per sq. metre.
Show that the expression of the total cost C of constructing the prism in terms of x is as follows:
C 3 x 2 6 x 48 x 1 24 x 2
[2]
Using an analytical method, find the minimum cost of constructing this prism.
4.
[4]
(a)
Solve the equation w2 3 4i , expressing your answer(s) in the form x iy .
[3]
(b)
Find the fourth roots of –16, expressing your answers in the form rei , where r 0 and . On an Argand diagram, mark the points that represent the roots clearly.
[5]
[Turn over
3
5.
(a)
Functions f and g are defined by f : x x 2 3 for x 0 g : x 2 x 4 for all x R
[2]
(i) Find f 1 in a similar form. (ii) Sketch the graphs of f, f 1 and f 1f on the same diagram. (iii) Given that gf 1 exists, find gf 1 ( x) . (b)
[1] [1]
The function h is defined by h(x) =
12 x 2 for 0 x 3 3x 6 for 3 x 6
and h(x) = h( x 6 ) for all x R (i) Find h(16) + h(25). (ii) Sketch y = h(x) for 6 x 12 .
6.
A curve C is defined parametrically by the equations x 1 cos t , y t sin t
(i)
(ii)
(iii)
7.
[2] [3]
for
t 2 2
State the exact range of values that x and y can take. Hence sketch the curve C . Label your graph, indicating the axes intercept(s) clearly if any. The tangent to the curve, C at the point P where t
2
[3]
is denoted by l .
Find the Cartesian equation of l .
[3]
Find the range of values of m such that the line y mx intersects C at two points.
[2]
The equations of two planes 1 , 2 are given by 1 : 2 x 4 y z 8 2 : x 2 z 6
(i)
Find the vector equation of the line of intersection l between the planes 1 and 2 .
[2]
(ii)
Find the foot of perpendicular, F1 from the point 6,9, 2 to the plane 1 .
[3]
[Turn over
4 Another plane 3 contains the points F1 and F2 and is parallel to l .
(iii)
(iv)
8.
26 5 Given that OF2 9 , show that the Cartesian equation of the plane 3 is given 2 5 by 15 x 8 y 40 z 22 .
[3]
By moving plane 3 by m units in the direction of vector v, all three planes will meet at l . Find m and v.
[3]
In a “man vs machine” contest, an athlete competes against a robot to see who can pull himself up a 16 m rope faster. The athlete climbs up 0.8 m in his first pull; the height attained by each subsequent pull decreases by a factor of 1/20 as he grows more and more tired. The robot is able to pull itself up by 0.4m each time as it never grows tired.
16m
The athlete and robot both start climbing at the same time from the base of the rope. Assuming they pull at the same rate, after how many pulls will the robot overtake the athlete?
[4]
(ii)
Will the athlete be able to reach the top of his rope? Justify your answer.
[2]
(iii)
The contestants again start climbing from the base but the athlete changes his strategy. He continues pulling at the same rate, but now he climbs up x m in each of his first two pulls; (x – 0.02) m in each of his next two pulls; and every subsequent pair of pulls shows a similar decrease of 2cm from the previous pair.
(i)
Show that the distance he travels after 2n pulls is 2nx 0.02n(n 1) . Given that he won the contest, find, correct to 2 decimal places, the minimum value of x.
[4]
[Turn over
5 1
dy e tan x It is given that , where tan 1 x denotes the principal value. 2 dx 1 x
9.
(i)
Find an expression for y in terms of x given that y = 1 when x = 0.
(ii)
Show that 1 x 2
(iii)
By further differentiation of the result in (ii), find the Maclaurin’s series for y up to and including the term in x3.
(iv)
ddx y (2 x 1) ddyx 0 . 2
[2]
2
[4]
Hence, deduce the series expansion for 1
e tan x (a) (1 x) 2
10. (a)
[2]
(b) e2 x tan
1 x
up to and including the term in x2.
Prove by induction for all n
[4]
that
n
r ! r n 1! 1
[5]
r 1
(b)
The diagram shows part of the graph of y x3 2 x 2 5 x 6 . The 3 real roots of the equation x3 2 x2 5x 6 0 are denoted by , and where < .
0
(i) Find the values of , and .
[1]
A sequence of real numbers x1 , x2 , x3 , x4 ,... satisfies the recurrence relation
xn1 1 3 5xn 2 8 xn 5 for n
. [Turn over
6 (ii) Prove algebraically that if the sequence converges to L, then L is equal to either [3] , or . (iii) Show that if xn 1 xn , then xn or xn .
11. (a)
Given that A(sin + cos ) + B(cos – sin ) 4 sin , Find the values of the constants A and B. Hence find the exact value of
(b)
[3]
(i) Express
1 4 0
4sin d . sin cos
[3]
2t in partial fractions. (t 1) 2
(ii) Hence, find the exact value of
t=
[2]
5 1
[2]
1 dx , using the substitution x 2x 1
2x 1 .
[6]
End of Paper
[Turn over
SAINT ANDREW’S JUNIOR COLLEGE Preliminary Examination MATHEMATICS Higher 2
9740/02
Paper 2 Monday
20 September 2010
3 hours
Additional materials : Answer paper List of Formulae(MF15) Cover Sheet
READ THESE INSTRUCTIONS FIRST Write your name, civics group and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Answer all the questions. Total marks : 100 Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically state otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematic steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages including this page. [Turn over
2 Section A: Pure Mathematics [40 marks]
1
1 The diagram below shows a rectangle tank where AG 2 and AC is parallel to the 4 2 vector 1 . 2
H
G F
E
I D
A
C
B
Find the length of projection of AG onto the line passing through A and C. (ii) Given that GI : IC 3: 2 , find AI . (iii) Find the acute angle between AG and GC .
[2]
(i)
2
[2] [2]
In a chemical reaction a compound X is formed from a compound Y. The mass in grams of X and Y present at time t seconds after the start of the reaction are x and y respectively. The sum of the two masses is equal to 100 grams throughout the reaction. At any time t, the rate of formation of X is proportional to the mass of Y at that time. dx When t = 0, x = 5 and 1.9 . dt
dx 0.02(100 x) . dt
(i)
Show that x satisfies the differential equation
(ii)
Solve this differential equation, obtaining an expression for x in terms of t.
[4]
(iii) Calculate the time taken for the mass of compound Y to decrease to half its initial value.
[2]
(iv)
[2]
Sketch the solution curve obtained in part (ii) and state what happens to [3] compounds X and Y as t becomes very large. [Turn over
3
(a)
Indicate clearly on a single Argand diagram, the locus of z that satisfy 0 arg( z 1 i)
2
and z i 2 .
Hence, find the complex number z , in the form x + iy, for which arg( z 1)
4
,
leaving your answer in exact form.
[6]
3
(b)
4.
3 3 i sin Consider the complex number w 2 cos cos i sin . 4 4 6 6
(i)
Find the exact values of the modulus and argument of w. Hence, express w in modulus-argument form.
[3]
(ii)
Find the possible value(s) of wn , when n is a multiple of 4.
[2]
The curve C has equation y f x , where f x
8x . x 1 2
Using a non-calculator method, find, in simplest form, the exact coordinates of the turning points of C .
[3]
(ii)
Sketch the curve C , indicating clearly any axis intercept(s) and asymptote(s).
[2]
(iii)
Given that n 0 , show that
(i)
value of (iv)
2 2
f x dx n
0
2 ln n 2 1 . Hence deduce the exact
f x dx .
[3]
Sketch on a separate diagram, the graph of g x f x , labeling clearly any stationary point(s) and asymptote(s). The region R is bounded by the curve y = g(x), the lines x 0 , x 1 and the x-axis. Find, in exact form, the volume generated when R is rotated completely about the x-axis.
[4]
Section B: Statistics [60 marks] 5
ABC College has a student population of 1800, of whom 60% are female students and 40% are male students. The College intends to get a sample of 100 student volunteers to take part in a survey on the College’s National Education Programme. (i)
(ii)
Comment whether such a sample consisting of all the volunteers is likely to give a true picture of the opinions of all the students about the College’s National Education Programme. Suggest a method of obtaining a more representative sample and describe how it may be carried out.
[1] [3]
4
6
James has 9 marbles, 4 of them are red, 3 of them are blue, and 2 of them are yellow. He arranged these marbles on a straight line. All marbles are identical except for their colour. Find the number of ways that the marbles can be arranged if (a)
there are no restrictions,
[1]
(b)
the arrangement is symmetrical about the centre marble, (e.g. R-R-Y-B-B-B-Y-R-R)
[2]
each blue marble is between two marbles of the same colour. (e.g. Y-B-Y or R-B-R)
[4]
(c)
7
(a)
(b)
A & B are two events with non-zero probability. Explain if each of the following statements is necessarily true, necessarily false, or neither necessarily true nor necessarily false. (i)
If A & B are mutually exclusive, then they are independent.
[1]
(ii)
If A & B are independent, then they are mutually exclusive.
[1]
A teacher brings 4 black, 3 blue, 2 red and 1 green markers to the classroom for each of his lessons. Unknown to him, the probabilities that a black, blue, red or 1 1 1 p, p and p respectively, where green marker is out of ink are p , 2 4 8 0 p 1. (i)
Find, in terms of p, the probability that a randomly chosen marker from his set of ten markers is out of ink.
[2]
(ii)
In the classroom, the first marker he tries out is out of ink. Find, in terms of p, the probability that the next marker he tries out is a red marker that is also out of ink.
[3]
(iii) After numerous lessons, the teacher realised that, in general, the first marker he tries out works at least 7 out of 10 times. Find the range of possible values of p.
[2]
[Turn over
5 8
(a)
A secret source claims that 0.3% of the residents in the suburbs of Zozoland are spies from Buzzland. If the claim were true, what is the probability, using a suitable approximation, that there are at least five spies living in a Zozoland suburb [3] of 1200 residents?
(b)
Another independent source (assumed to be reliable) claims that there is an average of 1.3 Buzzland spies and 0.4 Dodoland spies in a typical Zozoland city. (i)
(ii)
9
The investigation bureau will only do extensive combing of a city where there are more than five spies present. Given that there are 10,000 Zozoland cities, estimate the number of cities that will be under investigation.
[3]
The bureau investigated a certain number of cities and caught 23 Buzzland spies and 11 Dodoland spies. Find the most likely number of cities that had been investigated, stating an assumption needed for your calculation in the context of the question.
[4]
Marine biologists discovered that in a remote island off Philippines, a type of algae is growing at an exponential rate and is threatening the marine life in the region. The growth of the algae was collected over a period of 10 years and was recorded as follows (measured as cell density in millions). Year (x)
1
Cell density(y)
(i)
2
3
4
5
6
1.21 1.66 2.83 4.35 4.91 6.55
7
8
9
10
8.01
9.66 12.72 18.01
The relationship between the year, x, and the cell density recorded as y, are related by y ae , where a and b are unknown constants. By plotting a scatter diagram, bx
[2]
comment on the relationship between x and ln y. Find the estimated regression line of ln y on x. Hence calculate estimates of a and b.
[3]
(iii)
Estimate the cell density at Year 12. Comment on the reliability of your answer.
[2]
(iv)
Estimate the year at which the cell density is at 7 million. Comment on the choice of your regression line.
[2]
In a manufacturing company, the mean salary of its employees is S$30,000. The salary structure of the company is such that only 20% of the employees earn higher than the mean salary. Explain whether the use of Normal distribution to model the salary distribution of the employees in this company is appropriate.
[2]
(ii)
10
(a)
[Turn over
6 (b)
Anne travels to work each day by bus. The total time, T A in minutes for Anne’s journey to her office, is a normal random variable with a mean of 55 and a variance of 25. Ben drives to work every day. The total time, T B in minutes, that Ben spends driving to work is a normal random variable with a mean of 53 and a variance of 16. (i) (ii)
Find the probability that the average of the individual times taken by Anne and Ben to travel to work in a day is between 50 to 60 minutes. Find the greatest value of a, if the probability that Ben’s travel time differs from 53 minutes by at most a minutes is not more than 0.6.
(iii) Using a suitable approximation, calculate the probability, that for a randomly chosen period of 60 days, there are at least 43 days on which Anne will take longer than Ben to travel to work.
11
(a)
[2] [2]
[4]
A water treatment plant monitors their drinking water from their storage tanks on an hourly basis. The water must maintain a pH level of 8.5 (they try to maintain an alkaline level) for it to be ‘drinkable’. At 0800 hours, they recorded the following pH level from 11 of their tanks: Tank
1
2
3
4
5
6
7
8
pH level 8.31 8.41 8.51 8.46 8.52 8.48 8.33 8.1
(i)
(ii) (b)
9
10
11
8.39 8.42 8.52
Using an appropriate test, determine at 1% level of significance, if this sample provides sufficient evidence that the mean pH level of the water differs from 8.5? State an assumption necessary for the test in a(i) to be valid.
[3] [1]
At 1500 hours, a random sample was recorded from 80 tanks. Denoting the pH level readings by x, the results are summarized as follows:
( x 8.5) 15.2 , ( x 8.5)
2
232.2
(i)
Find the unbiased estimate of the population mean and variance.
[2]
(ii)
Another sample of n (assume n is large) readings was recorded. Using the unbiased estimate of the population mean and variance found in (b)(i), find the least value of n so that the probability that this sample mean has a pH reading of less than 8.2 is less than 0.3.
[5]
End of Paper [Turn over
1 SERANGOON JUNIOR COLLEGE 2010 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2
Wednesday
9740/1
18 August 2010
Additional materials: Writing paper List of Formulae (MF15) TIME : 3 hours READ THESE INSTRUCTIONS FIRST Write your name and class on the cover page and on all the work you hand in. Write in dark or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
Total marks for this paper is 100 marks. This question paper consists of 6 printed pages and no blank pages.
[Turn Over
2 Answer all questions (100 marks).
1
Without using the calculator solve the inequality
2 x2 + x < 3 .
[3]
Hence solve the inequality 1x
2e x + e 2 < 3 .
2
(i)
Show that
(ii)
Given that
[2]
d 1 for x < 1 . sin −1 x = dx 2 1− x 1
∫0
1
2
1 − x2
π
dx = ∫ 4 a
[2]
( 2 cos2 x ) dx , show that
sin 2a + 2a − 1 = 0 .
3
[3]
A sequence of numbers { xn } is defined by the relation
( )
xn = 9 2n −1 − xn −1 for n = 2,3, 4,... and x1 = 7 .
( )
n
Write down the values of x2 , x3 , x4 and x5 in the form of a bn − ( −1) , where a and b are real numbers.
[2]
Hence make a conjecture for xn and prove the conjecture using Mathematical Induction.[4]
4
Given that e y = 3 e + x + sin x . Show that
3e3 y
d2 y
2
dy + 9e3 y + sin x = 0. 2 dx dx
[3]
Hence, find in terms of e, the Maclaurin’s series for y, up to and including the term in x2. [4]
3 5
The function f is defined by f : x 3 + x − 2 x 2 for x ∈ , x ≥ k . (i) (ii)
6
Find the least value of k such that f has an inverse.
[2]
–1
Using the value of k in part (i), find f (x) and state the domain of f –1
–1
.
[4]
(iii)
Hence, find the exact solution of the equation f (x) = f
(x).
(a)
The diagram, not to scale, shows the graph of y = f(x). The vertical asymptotes are
[2]
x = 0 and x = 2. Sketch the graph of y = f '( x) , showing all corresponding coordinates and asymptotes where possible.
[3]
y
A(-1, 2)
-1/2
B(1, 3)
2
x y= f(x)
(b)
A cylindrical silo with a hemispherical rooftop is constructed to store rubbish.
r
The cost of each unit area of the rooftop is fixed at $21 and the cost of each unit area of the curved surface of the cylinder is fixed at $7. The total cost of the rooftop and the curved surface is fixed at $2100. Given that the radius of the hemisphere is r, show that the volume V of the structure can be expressed as 7π 3 V = 150r − r . 3 Find the exact cost of the hemispherical top when V is at its maximum. 4 [Curved surface area of sphere = 4π r 2 ; volume of sphere = π r 3 ] 3
[3] [4]
[Turn Over
4
7
(a)
A complex number w is such that ww∗ + 64 3 i + 16iw = 0 and Im( w) < 5 , where w∗ is the conjugate of w .
(b)
(i)
Find w in the form x + yi, where x, y ∈ .
[3]
(ii)
Find the integer values of n such that w n is real .
[3]
Find the Cartesian equation of the locus of the point P representing the complex number z where z − 1 − i = 3 1 + iz . Hence sketch the locus of point P.
8
[5]
Lovewell the Frog is stuck at the bottom of a well. It tries to make a series of vertical upward jumps on the well wall to get out. Its first jump is 1 m, but because of fatigue from over9 of the distance of the previous jump. exertion, each subsequent jump is 10 However, as the wall is wet and slippery, Lovewell slips down some distance after every jump is made. It slips down 0.3 m after its first jump, but because it learns to grip better, each subsequent slip is 0.02 m less than that of the previous slip. For example, its first jump gets it 1m up the well, and then it slides down 0.3 m. From its new position it makes a second jump up 0.9 m, and then it slides down 0.28 m. Its third jump sees it go up a further 0.81 m, and it slides down 0.26 m; so on and so forth.
(i)
How many jumps are needed for Lovewell to clear the 4m mark in the well?
[5]
(ii)
How many jumps are needed for Lovewell to clear the 6.5m mark in the well?
[4]
(iii) Determine the highest mark in the well that Lovewell is able to reach?
9
[2]
The position vectors of the points A, B, C and D are given as i + 3j , 2 j + 4k , i + j + k and 4i + 5k respectively.
(i)
Find the vector equation of plane π in the form r ⋅ n = p , that contains the points A, B and C. [3]
(ii)
Find the foot of the perpendicular of the point D to the plane π . Hence find the shortest distance from the point D to the plane π .
[4]
A line l parallel to the vector j + k passes through point D and it meets the plane π at the point N. (iii) Find the position vector of the point N and hence find the vector equation of the reflection of line l about the plane π . [5]
5
10
(a)
x 2 + ax + b where a, b are real constants. If the x+9 two asymptotes of the curve intersect at the point ( −9, −12 ) and the x-axis is a tangent
The curve C has equation y = f ( x ) =
to the curve, find the values of a and b.
[4]
Sketch the curve y = f ( x ) , stating clearly the equations of the asymptotes and the coordinates of any points of intersection with the axes. Hence deduce the values of p and q such that the equation f ( x + p ) − q = kx has no real roots for k ≤ 1 . [3]
(b)
The diagram below shows the sketch of the graph y = f ( x ) for a > 0 .
y
( 5a,3a ) y = ax
( −3a, 2a ) y=0
−a
0
x
2a
x=0
x = 4a
Sketch on separate diagrams the graphs of 1 (i) , y= f ( x)
(ii)
[3]
y2 = f ( x ) ,
[2]
showing clearly the equations of any asymptote(s), the coordinates of any turning point(s) and axial intercept(s).
[Turn Over
6
11
(i)
Solve the equation z5 = −32 , giving your answers in the form reiθ ,where r > 0 and [4] −π < θ ≤ π .
(ii)
Hence, show the roots of the equation w5 = 32i in an Argand diagram. The roots represented by W1 and W2 are such that −
π
< arg (w1) < arg (w2) <
2 Find the area of triangle OW1W2 where O is the origin.
(iii)
π 2
.
[5]
Express z 5 + 32 in the form
( z 2 − p cosα z + 4)( z 2 − p cos β z + 4) ( z + k ) where p, α , β and k are constants to be found.
End of Paper
[4]
1
SERANGOON JUNIOR COLLEGE 2010 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2
Wednesday
9740/2
25 August 2010
Additional materials: Writing paper List of Formulae (MF15) TIME : 3 hours READ THESE INSTRUCTIONS FIRST Write your name and class on the cover page and on all the work you hand in. Write in dark or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. Total marks for this paper is 100 marks. This question paper consists of 9 printed pages and 1 blank page.
[Turn Over
2
Section A: Pure Mathematics [40 marks]
1
The computer company call “Orange” manufactures the latest electronic gadget in town called the iBoard with 3 different storage capacities namely 16GB, 32GB and 64GB. The profit earned from each unit sold is as shown in the table below. Storage Capacity Profit
16GB $x
32GB $y
64GB $z
Within the first week after it was officially launched, the sales from 3 of its outlets for the three different storage capacities is as shown below. Storage Capacity Outlet A Outlet B Outlet C
16GB 75 180 45
32GB 120 230 50
64GB 20 70 10
The total profit collected from outlets B and C are $38 750 and $8750 respectively. If the total profit earned due to both the sales of 16GB and 32GB iBoard is equal to 12 times the total profit earned from the sales of the 64GB iBoard, find the value of x, y and z. [4] Find the total profit collected from outlet A. [1]
2
A curve C has parametric equations x = a sin2 t, y = a cos t, 0 ≤ t ≤
π 2
where a > 0.
(i)
Sketch the curve.
(ii)
Find the equation of the normal at the point P where t =
(iii)
Using a non-calculator method, determine whether the normal at P will meet C again. [3]
[2]
π 3
.
[3]
[Turn Over
3
3 y
• A(4, 3)
1 0
2
C(3,0)
x
B(1, – 1)
The diagram above shows the graph of y = f ( x ) . On separate diagrams, sketch the graphs of (i)
y = f (1 − 2 x ) ,
[3]
(ii)
y = f ( x ),
[2]
(iii)
y = 2f ( x ) + 1 .
[3]
showing in each case, the coordinates of the points corresponding to A, B, C and the equations of the asymptotes.
[Turn Over
4
2
4
(a)
(i)
( ln t ) + C. t +1 Show that ∫ ln t dt = t ln t − t + 2 t
(ii)
A curve C is defined by the parametric equations
[2]
x = t + ln t y = t − ln t ,
t > 0.
The region R, which is bounded by the curve C, the line x = 1 , x = 2 + ln 2 and the x – axis is as shown below.
y
C
R 0
x 1
2 + ln 2
Find the exact area of R.
(b)
[4]
2
The region S is enclosed by the curve D with equation ( y + 2 ) = 4 − x and the line y = x . Find the volume generated when S is rotated through 2π radian about the x – axis, giving your answer correct to 2 decimal places. [3]
[Turn Over
5
5
A disease is found to be present in a protected reserve containing 35 orangutans. The rate at which the number of infected orangutans, x, is increasing at any time t is proportional to the product of the number of infected orangutans and the number that have yet to be infected at that instant. Initially there were 5 animals infected. Form a differential equation that describes this model and show that
x=
where A>0 is to be found.
35 Ae35kt 1 + Ae35kt
, [6]
Deduce the total number of infected orangutans after a long period of time and represent the solution to this model on an appropriate graph. [4]
[Turn Over
6
Section B: Statistics [60 marks] 6
A survey on dining experience was undertaken in a small town with 10 three-star restaurants, 60 two-star restaurants and 30 one-star restaurants. The 100 restaurants in the small town are numbered from 1 to 100. A sample of 10 restaurants is selected by randomly choosing 10 numbers that are assigned to the restaurants.
(i)
Suggest one disadvantage of this sampling method.
(ii)
Suggest a better method of sampling and explain briefly how this could be done.
[1]
[3]
7
1 . For each of the 5 next 2 days, Wednesday and Thursday, the conditional probability that it rains, given that it rained the previous day is α and the conditional probability that it rains, given that it did not rain the previous day is β . The situation is illustrated in the uncompleted tree
In a certain country, the probability that it rains on a given Tuesday is
diagram below.
α 1 5
R
R
β
R’ R
R’ Tuesday
R’ Wednesday
Complete the tree diagram to represent all the possible outcomes up to Thursday. [2] 1 2 For α = and β = , find 3 3 (ii) the probability that it rains on a Thursday, [2]
(i)
(iii)
the probability that it rains on at least two days given that it rains on a Thursday. [3]
[Turn Over
7
8
Paul ordered 9 plates of sushi, namely 3 plates of unagi sushi, 2 plates of uni sushi and 4 plates of ebi sushi.
(i)
Find the number of ways Paul can eat the 9 plates of sushi.
(ii)
Find the number of ways Paul can eat the 9 plates of sushi such that the first plate
[2]
and last plate are different types of sushi.
(iii)
[4]
Find the number of ways of arranging the nine plates of sushi on a round table such that no two plates of ebi sushi are placed together.
9
[2]
Miss Curious wants to determine if there is any correlation between the amount of preparation and the results obtained in a recently concluded exam. She asked her friends how much time they spent preparing for the exam (x), with their exam scores (y), and recorded her findings in the table below.
(i)
x (hour)
10
15
22
27
38
46
53
64
y (score)
11
40
51
56
61
62
64
66
Give a sketch of the scatter diagram for the data and find the equation of the least squares regression line of y on x.
[2]
(ii)
State, with a reason, which of the following would be an appropriate model to represent the above data (where a and b are constants and b > 0). b B : y = a + be− x A: y =a+ C : y = a + b ln x [2] x
(iii)
For the appropriate model chosen, find the values of a and b. [1] Explain how this model is a better one than the equation found in part (i). [1]
(iv)
Obtain a good estimate of the score of a student who spent 8 hours studying for the exam and comment on the reliability of your answer.
[2]
[Turn Over
8
10
You are an Intelligence Quotient (IQ) expert. While reading the newspaper, you become interested in a newspaper advertisement that reads as follows:
Increase the IQ of your children by 10 points in just 16 weeks! Subscribe now to Dr. Dune’s Drill (DDD) program and astound your children’s friends, teachers and grandparents! Assure a university education for your children (and security for you in your old age). A scientific study of 15 children from all over Singapore showed an average IQ score of 108* after only six weeks of the fantastic DDD program . *standard deviation is 15
As a concerned IQ expert, you would like to investigate the validity of this advertisement. You know that for the general population of children, the mean IQ is 100. Through close contacts in the industry, you confirmed that the scientific study as stated in the advertisement is valid.
(i)
Test at 5% significance level, whether the mean IQ points of children who participated in the DDD program has increased. State any assumptions that you have to make in carrying out the test. [6]
(ii)
What do you understand by 5% significance level in this context?
[1]
Suppose that the population standard deviation σ is now known and a larger sample size of 50 children is taken.
(iii)
Find the range of values of the sample mean in terms of σ if the conclusion at 5% significance level is now different from that concluded in part (i). [4]
[Turn Over
9
11
12
In a certain country, it is found that on average the number of pairs of twins born weekly are 2.
(i)
Find the most probable number(s) of pairs of twins that are to be born weekly. [1]
(ii)
Find the probability of having at most 7 pairs of twins to be born in a two-week period. [2]
(iii)
Assuming there are 26 two-week periods in a year, estimate the probability that there are less than 73 two-week periods with at most 7 pairs of twins born in 3 [3] years.
(iv)
Using a suitable approximation, find the least number of consecutive weeks such that the probability of having at most 20 pairs of twins born falls below half. You may assume that the number of weeks is more than 5. [3]
(v)
Find the probability that the mean number of pairs of twins born in 50 weeks is less than 1.8. [2]
An ornithologist, who studies the behavior of birds, captures one male and one female hornbill from a forest in Osaka, Japan. The masses of hornbills in that forest are assumed to follow normal distributions with male hornbills having mean 3500g and standard deviation 150g while female hornbills having mean 3000g and standard deviation σ .
(i)
It is found from research that 5% of the female hornbills from the forest have masses exceeding 3.2kg. Show that σ = 122 . [2]
(ii)
Find the probability that the difference in mass between two randomly chosen male hornbills is at least 0.1kg. [3]
(iii)
Find the probability that the mass of 5 randomly chosen female hornbills exceeds twice the mass of 2 randomly chosen male hornbills. [3]
(iv)
Five male hornbills are randomly chosen. Find the probability that the fifth male hornbill is the third hornbill with mass exceeding 3.6kg. [3]
End of Paper
[Turn Over
10 BLANK PAGE
[Turn Over
1
SERANGOON JUNIOR COLLEGE 2010 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2
9740/1
Solutions 1
2x2 + x < 3
−3 < 2 x 2 + x < 3 −3 < 2x 2 + x
and
2 x2 + x < 3
2 x2 + x + 3 > 0
and
2 x2 + x − 3 < 0
1 ⎞ 23 ⎛ ⎜x+ ⎟ + >0 4 ⎠ 16 ⎝
and
( 2 x + 3)( x − 1) < 0
x∈
and
3 − < x <1 2
2
∴−
2e
x
3 < x <1 2 1 x + e2
∴ replace x with 1
2
1 ⎛ 1x ⎞ x < 3 ⇒ 2⎜ e2 ⎟ + e2 < 3 ⎜ ⎟ ⎝ ⎠
1 x e2 1
x x 3 − < e2 < 1 ⇒ 0 < e2 < 1 2 1 x < ln1 2 ∴x < 0
[Turn Over
2
2(i)
Now, − π ≤ sin−1 x ≤ π where x ≤ 1 2
2
Let y = sin−1 x ⇒ sin y = x ⇒ cos y
(since from the graph 1
(ii)
dy =1 dx
⇒
dy 1 = dx cos y
⇒
dy 1 = dx ± 1 − sin 2 y
⇒
dy 1 = dx ± 1 − x 2
⇒
dy 1 = dx + 1 − x 2
dy ≥ 0 for all x ≤ 1 ) dx
π
⎡ sin −1 x ⎤ 2 = 4 (cos 2x + 1)dx ∫a ⎣ ⎦0
π
π
⎡1 ⎤ − 0 = ⎢ sin 2x + x ⎥ 4 4 ⎣2 ⎦a
π
⎛1 π⎞ ⎛1 ⎞ = ⎜ + ⎟ − ⎜ sin 2a + a⎟ ⎠ 4 ⎝ 2 4⎠ ⎝ 2
3
⇒ sin 2a + 2a − 1 = 0
x2 = 9 ( 2 2 −1 ) − x1 = 18 − 7 = 11 = 3 ( 2 2 ) − ( −1)
2
x3 = 9 ( 23−1 ) − x2 = 36 − 11 = 25 = 3 ( 23 ) − ( −1)
3
x4 = 9 ( 2 4 −1 ) − x3 = 72 − 25 = 47 = 3 ( 2 4 ) − ( −1)
4
x5 = 9 ( 25 −1 ) − x4 = 144 − 47 = 97 = 3 ( 25 ) − ( −1)
5
OR Using G.C, 2 x2 = 11 = 3 ( 2 2 ) − ( −1) x3 = 25 = 3 ( 23 ) − ( −1)
3
[Turn Over
3
x4 = 47 = 3 ( 2 4 ) − ( −1)
x5 = 97 = 3 ( 25 ) − ( −1)
4
5
From above, n Conjecture: xn = 3 ( 2 n ) − ( − 1 )
Let Pn be the statement that xn = 3 ( 2 n ) − ( −1) for n ∈ + n
When n = 1, 1 = x1 = 3 ( 21 ) − ( −1) = 7 = R.H.S L.H.S Hence P1 is true. Assume Pk is true for some k ∈ + i.e xk = 3 ( 2 k ) − ( −1) To show that Pk+1 is true, i.e xk +1 = 3 ( 2 k +1 ) − ( − 1)
(
So L.H.S = xk +1 = 9 2(
k +1) −1
) − x(
k
k +1
k +1) −1
k = 9 ( 2k ) − ⎡3 ( 2k ) − ( −1) ⎤ ⎣ ⎦
= 9 ( 2 k ) − 3 ( 2 k ) − ( −1) = 9 ( 2 k ) − 3 ( 2 k ) − ( −1) = 6 ( 2 k ) − ( −1)
k +1
k +1
k +1
= 3 ( 2 ) ( 2 k ) − ( −1) = 3 ( 2 k +1 ) − ( −1)
k +1
k +1
= R.H.S
Hence Pk true => Pk+1 is true. Since P1 is true and Pk=> Pk+1 is also true, by Mathematical Induction, Pn is true for all positive integer n.
[Turn Over
4
4
e y = 3 e + x + sin x
e3 y = e + x + sin x Differentiating w.r.t. x, dy = 1 + cos x 3e3 y dx 2
⎛ dy ⎞ d2 y 3e + 9e3y ⎜ ⎟ + sin x = 0 (shown) 2 ⎝ dx ⎠ dx 3y
When x = 0,
e3 y = e + x + sin x ⇒ y = 3e3 y
1 3
dy dy 2 = 1 + cos x ⇒ = dx dx 3e 2
2
⎛ 2⎞ d2 y d2 y d2 y 4 3 y ⎛ dy ⎞ 3e + 9e + sin x = 0 ⇒ 3e + 9e = 0 ⇒ =− 2 2 2 2 ⎜ ⎟ ⎜ ⎟ ⎝ dx ⎠ ⎝ 3e ⎠ dx dx dx 3e 3y
Hence, the Maclaurin series of y is ⎛ 4 ⎞ ⎜⎝ − 3e2 ⎟⎠ 1 2 y= + x+ x 2 + ... 3 3e 2! 1 2 2 ≈ + x − 2 x2 3 3e 3e
5 (i)
Least value of k = 1.5
(ii)
let y = 2 x 2 − x − 3
⇒
1 25 y = 2( x − ) 2 − 4 8
∴ f −1 ( x ) =
(iii)
⇒
x=
1 8 y + 25 ± 4 16
1 8 x + 25 , x≥0 + 4 16
2 x2 − x − 3 = x ⇒
x=
1+ 7 2
[Turn Over
5
6(a) y
B(1, 0)
A(-1, 0)
x
y= f’(x) x =0
x =2
2 Area = 2π r + 2π rh (where h is the height of the cylinder)
(b)
2100 = 2π r 2 (21) + 2π rh(7)
2100 = 14π r (3r + h) 150 − 3r πr 2 3 π r + π r 2h V= 3 h=
2 ⎛ 150 ⎞ 2 = π r3 + π r 2 ⎜ − 3r ⎟ = π r 3 + 150r − 3π r 3 3 ⎝ πr ⎠ 3 7π 3 r = 150r − 3
dV = 150 − 7π r 2 = 0 dr
⇒r= d 2V dr 2
150 7π
(reject r <0)
= − 14π r < 0 ⇒ Volume is max.
Thus, cost of hemispherical top = 2π r 2 (21) = 2π
150 (21)= $900 7π
[Turn Over
6
7(a)
(i)
ww∗ + 64 3 i + 16iw = 0
( x + yi )( x − yi ) + 64
3 i + 16i ( x + yi ) = 0
x 2 + y 2 + 64 3 i + 16ix − 16 y = 0
(x
2
) (
)
+ y 2 − 16 y + 16 x + 64 3 i = 0
Comparing coefficients,
(16 x + 64 3 ) = 0
and
(x
x = −4 3
2
)
+ y 2 − 16 y = 0
y 2 − 16 y + 48 = 0
( y − 4 )( y − 12 ) = 0 y = 4 since y < 5
∴ w = −4 3 + 4i (ii)
w = 8 , arg ( w ) =
5nπ 5nπ ⎞ 5π ⎛ + i sin , thus wn = 8n ⎜ cos ⎟ 6 6 ⎠ 6 ⎝ 5nπ = kπ , k ∈ 6
( )
Since wn is real, Im wn = 0 , ⇒n=
6k ,k ∈ 5
For n to be an integer, n = 6k , k ∈ . (b)
z − 1 − i = 3 1 + iz
Let z = x + iy , then
x + iy − 1 − i = 3 1 + i ( x + iy ) x − 1 + i ( y − 1) = 3 1 + ix − y
( x − 1) + i ( y − 1) = 3 (1 − y ) + ix
( x − 1)2 + ( y − 1)2
= 3 x 2 + (1 − y )
(
2
x2 − 2 x + 1 + y 2 − 2 y + 1 = 9 x2 + 1 − 2 y + y 2
) [Turn Over
7
8 x 2 + 2 x + 8 y 2 − 16 y = −7 2
1⎞ 9 ⎛ 2 ⎜ x + ⎟ + ( y − 1) = 8⎠ 64 ⎝ 3 ⎛ 1 ⎞ The locus is a circle of centre ⎜ − ,1⎟ of radius units. 8 ⎝ 8 ⎠ y 3 1 8
×
1 5 8
4 − 8
8
1 − 8
0
x 2 8
Total upward distance covered by n jumps =
1(1 − 0.9 n ) =10(1 − 0.9 n ) 1 − 0.9
Total sliding distance after n jumps n = ⎡⎣ 2(0.3) + (n − 1)(−0.02) ⎤⎦ = 0.3n − 0.01n(n − 1) = −0.01n2 + 0.31n 2 Checking after which jump, the frog will not slide back: 0.3,0.28,0.26,...,0.02,0 U n = 0 ⇒ 0.3 + ( n − 1)( −0.02) = 0 ⇒ n = 16 (i)
To clear the 4 m mark, ⎡10(1 − 0.9 n ) ⎤ − ⎡ −0.01n2 + 0.31n ⎤ ≥ 4 ⎣ ⎦ ⎣ ⎦
From GC, n ≥ 8.8715 Therefore, 9 jumps are needed. (ii)
After 15 jumps/slides, there will be no more sliding.
[Turn Over
8
Hence, after 15 jumps/slides, distance climbed = ⎡10(1 − 0.915 ) ⎤ − ⎡ −0.01(15)2 + 0.31(15) ⎤ ⎣ ⎦ ⎣ ⎦ = 5.5411m 15 Subsequent jumps will follow GP with first term 0.9 and ratio 0.9.
0.915 (1 − 0.9 n ) Hence ≥ 6.5 − 5.5411 1 − 0.9 From GC, n ≥ 5.95 Therefore, 6+15=21 jumps are needed.
(iii)
15 Subsequent jumps will follow GP with first term 0.9 and ratio 0.9. Hence
Highest mark that the frog can reach =
0.915 + 5.5411 1 − 0.9
= 7.60 m
9(i)
−1 1 7 → ⎛⎜ ⎞⎟ → ⎛⎜ ⎞⎟ → → ⎛⎜ ⎞⎟ AB = ⎜ −1⎟ , BC = ⎜ −1 ⎟ ⇒ AB× BC = ⎜ 1 ⎟ ⎜4⎟ ⎜ −3 ⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛7⎞ ⎛1⎞ ⎛7⎞ ⎛7⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Equation of plane π : r ⋅ ⎜ 1 ⎟ = ⎜ 3 ⎟ ⋅ ⎜ 1 ⎟ = 10 ⇒ r ⋅ ⎜⎜ 1 ⎟⎟ = 10 ⎜ 2⎟ ⎜0⎟ ⎜ 2⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(ii)
Let foot of perpendicular from D be F. 4 + 7μ ⎞ → ⎛⎜ Now, OF = ⎜ μ ⎟⎟ ⎜ 5 + 2μ ⎟ ⎝ ⎠ ⎛ 4 + 7μ ⎞ ⎛ 7 ⎞ 14 Since F is on π , ⎜⎜ μ ⎟⎟ ⋅ ⎜⎜ 1 ⎟⎟ = 10 ⇒ μ = − 27 ⎜ 5 + 2μ ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠
[Turn Over
9
⎛ 10 ⎞ → ⎜ 27 ⎟ OF = ⎜ − 14 27 ⎟ ⎜⎜ 107 ⎟⎟ ⎝ 27 ⎠ ⎛ 10 ⎞ ⎛ 4 ⎞ ⎛ − 98 ⎞ → → ⎜ 27 ⎟ ⎜ ⎟ ⎜ 27 ⎟ 14 ⇒ DF = 392 = 14 6 − = − DF = ⎜ − 14 0 ⎟ ⎜ ⎟ 27 27 ⎜ ⎟ 27 9 ⎜⎜ 107 ⎟⎟ ⎜ 5 ⎟ ⎜⎜ 28 ⎟⎟ − ⎝ ⎠ ⎝ 27 ⎠ ⎝ 27 ⎠
(iii)
⎛ 4⎞ ⎛ 0⎞ ⎜ ⎟ Equation of line l : r = ⎜ 0 ⎟ + λ ⎜⎜ 1 ⎟⎟ ⎜5⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ 4 ⎞ → ⎛⎜ Thus since ON = ⎜ λ ⎟⎟ lies on plane π , ⎜5 + λ ⎟ ⎝ ⎠ ⎛ 4 ⎞ ⎛7⎞ 28 ⎜ ⎟ ⎜ ⎟ ⋅ = ⇒ = − λ λ 1 10 ⎜ ⎟ ⎜ ⎟ 3 ⎜5 + λ ⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 4 ⎞ → ⎜ ⎟ Therefore ON = ⎜ − 28 3 ⎟ ⎜⎜ 13 ⎟⎟ ⎝− 3 ⎠
⎛ − 88 ⎞ → ⎜ 27 ⎟ → 1⎛ → → ⎞ 28 Now, OF = ⎜ OD + OD ' ⎟ ⇒ OD ' = ⎜ − 27 ⎟ 2⎝ ⎜⎜ 79 ⎟⎟ ⎠ ⎝ 27 ⎠ ⎛ 4 ⎞
⎛ −7 ⎞
⎛ −7 ⎞ ⎜ ⎟ → 28 + β ⎜ 8 ⎟ ⎜ ⎟ 28 r l ' : = − ⎜ Thus ⇒ ND ' = 27 ⎜ 8 ⎟ ⇒ 3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜7⎟ ⎜ 7 ⎟ 13 ⎜− ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ 3⎠
[Turn Over
10
10(a)
x 2 + ax + b 81 − 9 a + b y= = x+ a−9 + x+9 x+9 Equation of asymptotes are: y = x + ( a − 9 ) and x = −9
(
)
When x = −9 , y = −12 So a = 6 When y = 0 , x2 + 6 x + b = 0 Since curve cuts the x-axis only once, Discriminant = 62 − 4b = 0 ∴b = 9 y
y = f ( x) y=x−3
( −3, 0 )
x
( −9, −12 )
x=−9
For the line y = kx to have no intersection with the transformed curve f ( x + p ) − q , the point of intersection has to be shifted to the origin. Hence p = −9 and q = −12
[Turn Over
11
(b)
y y=
1 f ( x)
1 ⎞ ⎛ ⎜ −3a, ⎟ 2a ⎠ ⎝
1 ⎞ ⎛ ⎜ 5a, ⎟ ⎝ 3a ⎠
y=0
4a
0
x
x = 2a
x = −a
y
(5a, ( −3a,
2a
)
3ay 2 = f ( x )
) x
−a 0
y=0
( −3a, −
2a
)
2a
(5a, − 3a ) x = 4a [Turn Over
12
11(i)
z 5 = 32e(
π + 2 k π )i
⎛ π + 2 kπ ⎞ ⎜ ⎟i ⎝ 5 ⎠
z = 2e
π
i
k = 0, ±1, ±2
,
, k = 0, ±1, ±2
3π
i
z = 2e 5 , 2e 5 , 2 e
(ii)
π
− i 5
πi , 2e , 2e
−3π i 5
5 w5 = 32 i ⇒ iw = −32 ⇒ ( iw) = −32 5
Let z = iw ⇒ w = −iz
I
2
R
Area of triangle OW1W2 = 1 × 2 × 2 × sin 2π = 1.90 units2. 2 5
(iii)
π π 3π 3π i ⎞⎛ − i ⎞⎛ i ⎞⎛ − i⎞ ⎛ πi 5 5 5 5 ⎜ z − 2e ⎟⎜ z − 2e ⎟⎜ z − 2e ⎟⎜ z − 2e ⎟ ( z − 2e ) ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ πi 3π i ⎛ 2 ⎞⎛ 2 − ⎞ − ⎛ π5i ⎛ 3π5 i ⎞ ⎞ 5 z − 2z ⎜ e + e 5 ⎟ + 4 ⎟ ( z + 2) ⎜⎜ z − 2 z ⎜ e + e ⎟ + 4 ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎠⎝ ⎝ ⎠ ⎠ ⎝ π 3π ⎛ 2 ⎞⎛ 2 ⎞ z + 4 ⎟ ( z + 2) ⎜ z − 4 cos z + 4 ⎟ ⎜ z − 4 cos 5 5 ⎝ ⎠⎝ ⎠
p = 4,α =
π 5
,β =
3π ,k = 2 5
End of Paper [Turn Over
1
SERANGOON JUNIOR COLLEGE 2010 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2
9740/2
Solutions Section A: Pure Mathematics [40 marks]
1
180 x + 230 y + 70 z = 38750 45 x + 50 y + 10 z = 8750 300 x + 400 y = 12 (100 z ) ⇒ 3 x + 4 y − 12 z = 0 Using G.C, x = 100, y = 75 and z = 50 Profit collected by outlet A = $100 ( 75) + $75 (120 ) + $50 ( 20 ) = $17500
2 (i)
y
a
x a
(ii)
y = a cos t x = a sin2 t, dx dy = 2a sin t cos t , = −a sin t dt dt
− a sin t dy 1 = = − sec t dx 2a sin t cos t 2
At the point P where t =
π 3
,
dy = −1 dx [Turn Over
2
3 1 Equation of normal at P ( a, a ): 4 2 y – y1 = m (x – x1) 1 −1 3 y − a = [ x − a] 2 −1 4 Equation of normal:
(iii)
1 y = x− a 4
Equation of Curve: x = a sin2 t, Equation of normal:
y = a cos t
1 y = x− a 4
4a cos t = 4a sin 2 t − a 4cos2 t + 4cos t − 3 = 0 (2cos t + 3)(2cos t − 1) = 0 1 3 cos t = , cos t = − (rejected) 2 2 π t = (point P) 3
Solving,
Hence, the normal at P does not meet the curve again.
3 (i)
y
A’(‐3/2, 3) •
1 •
C’(-1/2,0) -1/2
x 1/2 • B’(0, -1)
[Turn Over
3
(ii)
y A’’(‐4,3) •
A’(4, 3) •
1 • C’’(-3,0) -2
0 2
• B’’(-1, -1)
x
•
•
C’(3,0)
B’(1, -1)
y
(iii)
A’(4, 7) • 3
• C’(3,1)
1
x
2 B’(1, -1) •
4(a)(i)
⎛ t +1 ⎞ ⎛ 1⎞ ⎟ ln t dt = ∫ ⎜1 + ⎟ ln t dt t ⎠ ⎝ t⎠ 1 = ∫ ln t dt + ∫ ln t dt t
∫ ⎜⎝
= t ln t − ∫
( ln t ) 1 dt +
2
2
( ln t ) = t ln t − t + 2
+c
2
+c
[Turn Over
4
(a)(ii)
Area R = ∫
2 + ln 2
1
y dx
2 ⎛ 1⎞ = ∫ ( t − ln t ) ⎜1 + ⎟ dt 1 ⎝ t⎠ 2⎡ ⎛ t +1 ⎞ ⎤ = ∫ ⎢t + 1 − ⎜ ⎟ ln t ⎥ dt 1 ⎝ t ⎠ ⎦ ⎣ 2
2 2 ln t ) ⎤ ⎡t2 ⎤ ⎡ ( = ⎢ + t ⎥ − ⎢t ln t − t + ⎥ 2 ⎦⎥ ⎣ 2 ⎦1 ⎢⎣ 1 2 ⎤ ln 2 ) ( 1 ⎞ ⎡ ⎛ = ⎜ 2 + 2 − − 1⎟ − ⎢ 2 ln 2 − 2 + − ( −1) ⎥ 2 ⎠ ⎢⎣ 2 ⎝ ⎥⎦ 2 ⎤ ln 2 ) ( 1 ⎡ = 3 − − ⎢ 2 ln 2 − 2 + − ( −1) ⎥ 2 ⎢⎣ 2 ⎥⎦
( ln 2 ) 7 = − 2 ln 2 − units2 2 2 2
(b)
y
−5
x
4
S
(
0 V = π ∫ ⎡ −2 − 4 − x −5 ⎢ ⎣ = 327.25 units3
)
2
(
4 − x 2 ⎤ dx + π ∫ ⎡ −2 − 4 − x ⎥⎦ 0 ⎢ ⎣
) − ( −2 + 2
)
2 4 − x ⎤ dx ⎥⎦
[Turn Over
5
x is the number of infected orangutans at any time t.
5
dx dx ∝ x (35 − x ) ⇒ = kx (35 − x ), k > 0 dt dt 1 ⇒∫ dx = ∫ k dt x (35 − x ) 1 ⎛1 1 ⎞ ⇒ ⎜⎝ + ⎟ dx = k ∫ dt ∫ 35 x 35 − x ⎠ ⇒ ln x − ln (35 − x ) = 35kt + c ⎛ x ⎞ ⇒ ln ⎜ = 35kt + C ⎝ 35 − x ⎟⎠ ⇒
x = e35kt + c = Ae35kt , A > 0 35 − x
⇒x=
35 Ae35kt
x
1 + Ae35kt
When t=0, x=5, ⇒ A =
⎡ ⎤ 6 x = 35 ⎢1 − ⎥ ⎣ 6 + e35kt ⎦ As t → ∞, x → 35
35 1 6
5
0
t
Section B: Statistics [60 marks] 6
(i) As there are only 10 three-star restaurants, there is a high chance that they might not get selected at all during random sampling. OR The numbers of the respective types of restaurants selected might not be representative of the restaurants in the town. (ii) Stratified sampling. Set up the 3 respective sub-groups. - Use population proportions to determine the numbers required for each sub-group: Three-star(1), two-star(6), one-star(3). - Randomly select the restaurants based on the respective numbers determined [Turn Over
6
α
7(i)
α
R
R
1−α
1 5
4 5
β
R’ Tuesday
1− β
R
1−α
β
R’ R
1− β
R’
α
R
R’
R
1−α
R’
β
R
R’
Wednesday 1 − β
(ii)
P(rains on a Thursday)
R’
Thursday
1 1 1 1 2 2 4 2 1 4 1 2 = ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ 5 3 3 5 3 3 5 3 3 5 3 3 21 7 = = 45 15 (iii)
P(rains on at least 2 days given it rains on a Thursday) = P(A\B) =
P(A ∩ B) P(B)
1 1 1 1 2 2 4 2 1 13 ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ 13 = 5 3 3 5 3 3 5 3 3 = 45 = 21 21 21 45 45
8(i)
(ii)
9! = 1260 2!3!4! Case 1:uni and unagi 7! No. of ways = ( 2 ) = 210 2!4! Case 2:uni and ebi 7! No. of ways = ( 2 ) = 280 3!3!
No. of ways =
[Turn Over
7
Case 3:ebi and unagi 7! No. of ways = ( 2 ) = 420 2!2!3! Total no. of ways = 910 (iii)
- Unagi sushi - Uni sushi Since there are only 2 ways of arranging the uni and unagi sushi plates on the round table, No. of ways = 5C4 ( 2 ) = 10
9
(i)
Least squares regression line: y = 0.785x + 24.4 (ii)
Model C. Possible reasons: - Shape of the points follow the shape of an logarithmic graph - y increases as x increases (other 2 choices have p decreasing as x increases)
(iii)
a = - 37.4, b = 26.3 The value of r increases after transformation which indicates there is a better linear correlation between ln x and y instead of just x and y [Turn Over
8
(iv)
y = a + b ln x
∴ when x = 8, y = 17.31 Extrapolation done in calculating y when x = 8. It might not be reliable as there may not be a linear relationship between ln x and y outside the range of data
10(i)
Assuming that the IQ points of children is normally distributed. Ho: μ = 100 H1: μ > 100 (test that the average IQ points of students who have participated in the DDD program has increased) 1-tailed test at 5% significance level.
X−μ ~ t(n − 1) . S n n 3375 15 = 241.07143 Where s 2 = ( sample variance )= (15 )2 = n −1 14 14
Under Ho ,
By using GC, ttest = 1.996 and p-value = 0.0230. Since p-value=0.0230<0.05, we reject Ho and conclude that here, we have sufficient evidence at 5% significance level that the IQ points of children in the DDD program has increased. (ii)
5% significance level means there is a probability of 0.05 that we conclude that the mean IQ points have increased when in fact it did not (reject Ho when Ho is true).
(iii)
Population standard deviation σ is now known and a larger sample size of 50 children is taken. ⎛ σ2 ⎞ Under Ho , X ~ N ⎜100, ⎟ by Central Limit Theorem (since n=50 is large) ⎜ 50 ⎟⎠ ⎝
At 5% significance level, Ho is not rejected when ztest < 1.64485 . Thus
x − 100
σ
< 1.64485 ⇒ x < 100 + 0.233σ
50
[Turn Over
9
11(i)
Let X be the r.v “number of pairs of twins born weekly”, X
Po ( 2 )
Using G.C,when x = 1 and 2, P ( X = 1) = P ( X = 2 ) = 0.27067 give the highest probability. Hence the most probable numbers are 1 pair or 2 pairs of twins.
(ii)
(iii)
In a two-week period, X1 + X 2 ~ Po(4) P ( X1 + X 2 ≤ 7) = 0.948866 ≈ 0.949 Let Y be the r.v “number of two-week period with more than 7 pairs of twins born in 78 two-week period” Y B ( 78, 0.051134 ) Since n = 78 is large, np = 78 ( 0.051134 ) = 3.9884 ( < 5 ) So Y
Po ( 3.9884 ) approximately
P ( less than 73 two-week periods with at most 7 pairs of twins born )
= P (Y ≥ 6 ) = 1 − P (Y ≤ 5 ) = 0.21306 ≈ 0.213
(iv)
Let n be the number of consecutive weeks X 1 + X 2 + ... + X n Po ( 2n ) Since 2n > 10 (Q n > 5 ) , so X 1 + X 2 + ... + X n
N ( 2n, 2n ) approximately
P ( X 1 + X 2 + ... + X n ≤ 20 ) < 0.5 ⎯⎯→ P ( X 1 + X 2 + ... + X n < 20.5 ) < 0.5 C .C
Using G.C,when n =10, P ( X 1 + ... + X 10 < 20.5 ) = 0.50997 when n =11, P ( X 1 + ... + X 11 < 20.5 ) = 0.47282 Thus least n is 11 50
(v)
X=
∑ Xi
⎛ 2 ⎞ ~ N ⎜ 2, ⎟ approximately, by CLT since n is large. 50 ⎝ 50 ⎠
i =1
Thus P ( X < 1.8 ) = 0.15865 = 0.159(3 s.f) [Turn Over
10
12 (i)
Let M denote the r.v of the mass of a male hornbill. Let F denote the r.v of the mass of a female hornbill. M
(
N 3500,150 2
)
F
(
N 3000, σ 2
)
P ( F > 3200 ) = 0.05 P ( F ≤ 3200 ) = 0.95
3200 − 3000 ⎞ ⎛ P⎜ Z ≤ ⎟ = 0.95 σ ⎝ ⎠ 200 = 1.64485 ⇒ σ = 121.59 = 122 (3s.f.) σ
(ii)
M1 − M 2 ~ N (0, 2 × 150 2 )
P ( M1 − M 2 ≥ 100 )= P (M1 − M 2 ≥ 100 ) + P (M1 − M 2 ≤ −100 ) = 2P ( M1 − M 2 ≤ −100 )
=0.36265=0.363 (3 s.f.) (iii)
Let T = F1 + F2 + F3 + F4 + F5 − 2 ( M 1 + M 2 ) T
(
N 5 × 3000 − 4 × 3500,
5 × 121.49 2 + 23 × 150 2
)
P (T > 0 ) ≈ 0.976
(iv)
Probability required =
4! [ P(M > 3600)]3 [ P( M ≤ 3600)]2 2!2!
=0.053967 =0.0540 (3 s.f.)
End of Paper
[Turn Over
TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination Higher 2
MATHEMATICS
9740/01
Paper 1
15 September 2010
Additional Materials:
Answer paper List of Formula (MF15)
3 hours
READ THESE INSTRUCTIONS FIRST Write your Name and Civics Group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 5 printed pages and 1 blank page.
[ Turn over
© TJC 2010
1
1
The points A, B and C have position vectors given respectively by 2i +3k , 2j − k and αi − (α – 2)j + (2α – 1)k where α ∈ . Show that A, B and C are collinear.
[3]
Hence or otherwise, find α such that the C divides the line segment AB in the ratio 2:1.
[3]
2
In a chemical reaction, a compound A is converted to a compound B at a rate proportional to square of the amount of A that has yet to be converted. Initially, 120 g of A is present. After 1 hour, 20 g of A remains. If x denotes the amount of A that is pt converted after t hours, show that x = , where p is an integer to be determined. [6] 5t + 1
3
The complex number z satisfies the equation z 3 + az 2 − az − 1 =0 , where a ∈ .
(i)
Verify that z = 1 is a root of the equation.
[1]
(ii)
Given that one of the complex roots is 2i, find a.
[2] [3]
(iii) Find the third root of the equation. 4
A rectangular cuboid ABCDEFGH has a constant height of 2 cm and a base length of x cm which increases with time t (measured in seconds). The cuboid, which is initially a cube, is expanding at a constant rate of 104 cm3s−1. E
F
G
H 2 D
C x
A (i) Show that = AF
B
x
2x2 + 4 .
[1]
(ii) Find the rate of increase of the distance AF when t = 3.
2
[6]
5
The functions f, g and h are defined by f : x x2 + 4 g : x ln ( − x )
x ≥ −2 , x ∈ , x < 0, x∈ ,
h : x x − 4 +λ
x > 4, x∈ .
(i)
Given that the composite function fh ( x ) exists, find the least value of λ .
[3]
(ii)
Find g −1 ( x ) and state the domain of g −1 .
[2]
(iii) On a clearly labeled diagram, sketch the graphs of y = g ( x ) , y = g −1 ( x ) and
y = g −1g ( x ) . 6
[2]
It is given that for all x ∈ ,
f ( x ) = x2 + 4x + 6 , g ( x ) = x 2 − x − 2 and h ( x ) = ax3 + bx 2 + c . (a)
Solve algebraically the inequality Hence solve the inequality
(b)
f ( x) ≤ 1. g ( x)
f ( e− x )
g ( e− x )
≤ 1.
[3] [2]
If = y h ( x ) + f ( x ) passes through the points ( −1, −18 ) , (1, −14 ) and ( 3,30 ) , evaluate h (101) .
[4]
2r − 1 . r r =1 2 n
7
It is given that S n = ∑ (i)
a Express S1, S2, S3 and S4 in the form 3 − , where a, b ∈ . Hence obtain a b conjecture for Sn in terms of n.
(ii) Prove your conjecture using mathematical induction. N
(iii) Hence find
∑ r= 0
2r + 1 in terms of N. 2 r +1
[3] [4]
[2]
3
8
(i)
Expand (1 − 2x )
−
5 2
as a series of ascending powers of x up to and including the
term in x 2 . State the range of x for which the expansion is valid. (ii)
3 2
dy 3y . = dx By repeated differentiation of this result, show that the Maclaurin’s series expansion of y in ascending powers of x, up to and including the term in x3 , is
Given that y=
(1 − 2 x )
−
, show that (1 − 2 x )
[3]
[1]
1 + 3x + px 2 + qx 3 , where the numerical values of p and q are to be determined. (iii) Explain briefly how the result in (i) can be used as a check on the correctness of your answer in (ii). 9
[4]
[2]
(a) A geometric sequence { xn } has first term a and common ratio r. The sequence of numbers { yn } satisfies the relation yn = log 3 ( xn ) for n ∈ + . 20
(i)
If the product of x5 and x16 is 81, find the value of
∑ log3 xk .
[4]
k =1
(ii) Show that { yn } is an arithmetic sequence. (b) The output of a coal mine in any year is 10% less than in the preceding year. Prove that the output of the coal mined cannot exceed ten times the output in the first year.
[2]
[2]
It is decided to close the mine when the total output exceeds nine times the output in the first year. Determine the maximum number of years the mine will be in [3] operation, and give your answers correct to the nearest year. 10
The plane p1 has equation x + y − 3 z = 6 and the point A has position vector i + 2k. Given that the point B is the foot of perpendicular from A to p1, find the position vector [5] of B. 1 0 Another plane p2 has equation r= λ 0 + µ −1 , λ , µ ∈ . Given that p2 intersects 1 1 p1 at the line l, find the vector equation of l and show that the shortest distance from A to 77 l is . [6] 6 A plane p3 passes through the points A and B. Given that p1, p2 and p3 do not have a common point, find the equation of p3.
4
[3]
11
(a)
Find ∫ sec x cosec x dx .
[2]
1 (b) By using the substitution u = ln x , find ∫ ln ( ln x ) dx . x
[4]
(c) The diagram shows the region R bounded by the circle x 2 + y 2 = 4 , the curve 1 y = and the line x = −1 . x
1 intersects the circle at a point A, and the line x = −1 cuts the x circle at the point B. Find the coordinates of the points A and B.
(i) The curve y =
[2]
(ii) Calculate the area of the region R.
[3]
(iii) Find the volume of the solid formed when R is rotated through 4 right angles about the y –axis.
[4]
END OF PAPER
5
TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination Higher 2
MATHEMATICS
9740/02
Paper 2 Additional Materials:
20 September 2010 Answer paper Graph paper List of Formula (MF15)
3 hours
READ THESE INSTRUCTIONS FIRST Write your Name and Civics Group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages.
[ Turn over
© TJC 2010
1
Section A: Pure Mathematics [40 marks]
1
A curve C is represented by the parametric equations x= t 2 − 2t , y= t 3 − 9t for t ≤ 2 .
Find the equation of the tangent to the curve C which is parallel to the y-axis.
[4]
The figure below shows the region R bounded by the curve C and the x-axis.
2
Show that the curve C intersects the x-axis at the point (15, 0).
[1]
Hence find the exact value of the area of R.
[4]
By sketching the graphs of y = e x and = y 2 x + 2 on the same diagram, find the number x of roots of the equation e − 2 x − 2 = [2] 0. (i)
Given that the root β lies in the interval (a, 0), state the largest integer value of a.
[1]
A sequence of negative numbers, xn , satisfy the relation = xn + 1 (ii)
1 xn (e − 2) , for n ∈ + . 2
If the sequence converges, show that it converges to β .
[2]
It is given that β < xn < 0 . (iii) By considering xn +1 − xn , and with the aid of the sketch in (i), show that xn +1 < xn .
[2]
(iv)
Show that xn +1 > β .
[2]
(v)
Describe the behaviour of the sequence for the case when β < x1 < 0 .
[1]
2
3
3
4
z 3+i 0. = + z 3+i
(a)
Find the roots of the equation
(b)
A complex number z satisfies iz + 3 = 5.
[4]
(i)
Sketch the locus of the points which represents z in an Argand diagram.
[2]
(ii)
Find z when z + 3 is a minimum.
[4]
(a)
= y 6x + 4
y
y = f ( x) 4 −2
1
x
x = −1
The diagram shows the graph of y = f ( x ) with asymptotes x = −1 and = y 6x + 4 . On separate diagrams sketch the graphs of (i)
y = f ′( x)
[2]
(ii)
y2 = f ( x) .
[2]
(b) The curve C has equation y =
− x2 − 2x + 5 . x −1 2 can be transformed to C. x
(i)
Describe how the curve with equation y = x − 4 −
(ii)
State the equations of all the asymptotes of C and hence sketch C. (You are not required to find the x and y intercepts.)
[3] [2]
(iii) By adding a suitable graph to the sketch in (ii), deduce the number of roots of the equation
( x − 1) 22
2
2
− x2 − 2 x + 5 + + 4 = 1. x −1
3
[2]
Section B: Statistics [60 marks]
5
Through-Train Junior College (TTJC) has four levels of students: IP1, IP2, JC1 and JC2. The population breakdown for the year 2010 is as follows: Level
IP1
IP2
JC1
JC2
Number of students
78
70
692
760
Mr Koh wants to find out how students spend their time after lessons. He plans to get 10% of the student population to do a survey. His method of conducting the survey is to spend one hour each day at the canteen and ask if the students entering the canteen are willing to do a survey for him. He continues to do that until he obtains the number of surveys required. (i)
State a flaw in his sampling method.
[1]
Mr Koh’s friend, Mr Toh, suggests he modify his sampling method to collect 40 surveys from each level instead. (ii)
State the name of this sampling method.
[1]
Mr Koh finally decided to obtain a stratified sample across different levels of students. (iii) Describe how a sample of size 160 might be obtained. 6
[2]
The time taken, in minutes, for John, a typist, to type 5000 words is normally distributed with mean 34 and standard deviation 2.1. John attends a training course to improve his typing speed. After the course, John records the times he take to type 5000 words on 8 separate occasions, and obtains the following times (in minutes): 36.5
33.4
27.7
31.5
33.0
34.9
30.7
29.5
Assuming that the initial standard deviation has not changed, test at the 2% level of significance whether John’s typing speed has improved after the training course. [4]
Explain what is meant by 2% level of significance in the context of the question.
4
[1]
7
An Extreme Ironing Club consists of 11 members: 6 male and 5 female. It is also known that 4 of the male members are local while 3 of the female members are foreigners. (a)
Find the number of ways for all 11 members to stand in a row so that no two foreigners are adjacent. [2]
(b)
4 members are to be chosen to enter a competition. In how many ways can this be done if: (i)
there are no restrictions,
[1]
(ii)
there must be at least 1 male member and at least 1 female member,
[2]
(iii) there must be at least 1 female member and at least 1 foreign member (a female member who is a foreigner will satisfy both conditions). 8
[3]
Temasek United Football Club (TUFC) is participating in a football competition. 3 points are awarded for a win, 1 point is awarded for a draw and no points are awarded for a loss. In this competition, TUFC plays three matches in the first round. Being a good team, it may be assumed that for any match in the first round, the probability that TUFC wins is 1 and the probability that TUFC loses is 1 . All the matches are played independently. 2 5
(i)
Find the probability that TUFC draws a match.
[1]
(ii)
Find the probability that after three games, TUFC has exactly three points.
[3]
(iii) To qualify for the second round, TUFC needs to get at least five points. Given that TUFC does not win the first game, find the probability that they will still qualify for the second round. [4] 9
The times taken, in seconds, for two swimmers, A and B, to complete a 100-metre freestyle race are independent and normally distributed with means 48.0 and 47.2 and standard deviations 0.5 and 0.8 respectively. The two swimmers compete in a 100metre race for which the world record is 46.9 seconds. (i)
Show that the probability that at least one of the two swimmers breaks the world record during the race is 0.363 correct to 3 significant figures. [3]
(ii)
Find the probability of Swimmer B beating Swimmer A.
[2]
(iii) If A and B are to meet 20 times for the 100m freestyle race, how many times do you expect A to beat B? Give your answer correct to the nearest integer. [2] (iv) Find the probability that the total sum of four randomly chosen timings of A is more than 4 times a randomly chosen timing of B. [3] 5
10
At a stall in a fun-fair, games of chance are played. The probability that a participant wins a prize in each game is 0.05. (a) If the stall holder wants the probability of more than 5 prizes to be won on a particular day to be less than 0.1, find the largest number of games that can be played on that day. [3] (b) On another day, N games are played. Using a suitable approximation, find the probability that prizes are won in at most 5% of the games if : (i)
N = 60,
[3]
(ii)
N = 200.
[3]
(c) The stall is open for 70 days and on each day, 60 games are played. Find the probability that the average number of prizes won each day is between 2.5 and 3 inclusive. [4] 11
Mary recorded the length of time, y minutes, taken to travel to her office when leaving home x minutes after 7 am on nine selected mornings. The results are as follows. x y
0 20
5 23
10 29
20 33
25 35
30 39
40 40
50 48
60 51
(i)
State, giving a reason, which of the least squares regression lines, x on y or y on x, should be used to express possible relation between x and y. [1]
(ii)
Find the equation of the regression line chosen in part (i) and interpret the slope of your regression line. [3]
(iii) Calculate the product moment correlation coefficient and interpret your value in the context of this question. [2] Mary needs to arrive at her office no later than 8.30 am. The number of minutes by which Mary arrives at her office early, when leaving home x minutes after 7 am, is denoted by z. (iv)
Write z in terms of x and y.
[1]
(v)
Estimate, to the nearest minute, the latest time that Mary can leave home without arriving late at office. [3] Comment on the reliability of this value. [2]
6
TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination Higher 2
MATHEMATICS
9740/01
Paper 1
15 September 2010
Solutions 1
⎛ 0 ⎞ ⎛ 2 ⎞ ⎛ −2 ⎞ ⎛1⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = ⎜ 2 ⎟ − ⎜ 0 ⎟ = ⎜ 2 ⎟ = −2 ⎜ −1⎟ ⎜ −1⎟ ⎜ 3 ⎟ ⎜ −4 ⎟ ⎜2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ α ⎞ ⎛ 2⎞ ⎛ α − 2 ⎞ ⎛1⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AC = ⎜ −α + 2 ⎟ − ⎜ 0 ⎟ = ⎜ −α + 2 ⎟ = (α − 2 ) ⎜ −1⎟ ⎜ 2α − 1 ⎟ ⎜ 3 ⎟ ⎜ 2α − 4 ⎟ ⎜2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Therefore AB is parallel to AC and since A is a common point, we have A, B and C collinear. From above, we have
uuur 2 − α uuur AC = AB 2
uuur 2 uuur Since C divides the line segment AB in the ratio 2:1, AC = AB . 3 2 −α 2 2 ∴ = ⇒α = 2 3 3 Alternative 1for last part: Since C divides the line segment AB in the ratio 2:1, we have AB:BC = 2:1 uuur uuur ⎡⎛ 2 ⎞ ⎛ 0 ⎞ ⎤ ⎛ α ⎞ uuur OA + 2OB 2 ⎜ ⎟ 1 ⎢⎜ ⎟ ⎜ ⎟ ⎥ ⇒ ⎜ 2 − α ⎟ = ⎢⎜ 0 ⎟ + 2 ⎜ 2 ⎟ ⎥ ⇒ α = ∴ OC = 3 3 ⎜ 2α − 1⎟ 3 ⎢⎜ 3 ⎟ ⎜ −1⎟ ⎥ ⎝ ⎠ ⎣⎝ ⎠ ⎝ ⎠ ⎦ 2
dx 2 = k (120 − x ) dt
∫ (120 − x )
−2
dx = ∫ k dt
1
(120 − x )
−1
= kt + c
When t = 0, (120 − 0 ) = k ( 0 ) + c ⇒ c = −1
When t =1, ( 20 ) = k (1) + −1
1 120
1 1 ⇒ k= 120 24
1 1 1 5t + 1 = t+ = 120 − x 24 120 120 x=
600t 5t + 1
So p = 600 . 3
(i)
When z = 1, LHS = (1) + a (1) − a (1) − 1 = 0 = RHS 3
2
Hence z = 1 is a root of the equation. (ii) Since 2i is a root,
∴ ( 2i ) + a ( 2i ) − a ( 2i ) − 1 = 0 3
2
−8i − 4a − 2ai − 1 = 0 a ( −4 − 2i ) = 1 + 8i
a=
1 + 8i 3 = −1 − i −4 − 2i 2
(iii) Let z = b be the other root.
( z − b )( z − 1)( z − 2i ) = z 3 + az 2 − az − 1 Comparing the constant term, we have 1 −2bi = −1 ⇒ b = − i 2 4
(i) Clearly AF = AB 2 + BC 2 + CF 2 = 2 x2 + 4
2
(ii) Let V = 2 x 2 ⇒
dV dx dx 26 ⇒ = - - - - - - (1) = 4x dt dt dt x
Let y = AF . Differentiate (i) wrt x :
dy 2x = - - - - - - (2) dx 2 x2 + 4
When t = 3 , V = 23 + 104(3) = 320 ⇒ x = 4 10 or x = −4 10 (NA since x > 0)
So
dy dy dx 52 26 = ⋅ = = dt dx dt 18 9
Rate of increase of AF at t = 3 is
26 cms−1. 9
Alternative solution: dV = 104 dt ⇒ V = 104t + c When t =0, V = 8. This gives us c = 8. V = 104t + 8. Volume of cuboid = 2x 2 . From (i), let AF = y = 2x 2 + 4 = V + 4
(
)
1 − dy 1 = V +4 2 . ⇒ dV 2 52 dy ⎛ dy ⎞ ⎛ dV ⎞ 104 . =⎜ = = ⎟ ⎜ ⎟ dt ⎝ dV ⎠ ⎝ dt ⎠ 2 V + 4 V +4
When t = 3, V = 104(3) + 8 = 320. dy 52 26 = = cm s-1. Therefore, dt 9 320 + 4
3
5
(i)
fh ( x ) exist ⇒ Rh ⊆ Df ⇒ ( λ , ∞ ) ⊆ [ −2, ∞ ) ⇒ least λ = −2
(ii)
For g −1 ( x ) , Let y = ln ( − x )
⇒ − x = e y ⇒ x = −e y ⇒ g −1 ( x ) = −e x x∈ y
(iii) y = g ( x)
y =0
x y = g −1 ( x )
y = g g ( x) −1
x =0
6
(a)
f ( x) ( 5x + 8) ≤ 0 x2 + 4x + 6 ≤1⇒ 2 −1 ≤ 0 ⇒ g ( x) x − x−2 ( x − 2 )( x + 1)
⇒ ( 5 x + 8 )( x − 2 )( x + 1) ≤ 0 [ x ≠ −1, 2]
⇒x≤− f ( e− x )
g (e
−x
)
8 or − 1 < x < 2 5
≤ 1 ⇒ e− x ≤ −
8 or − 1 < e− x < 2 5
⇒ 0 < e− x < 2 1 ⇒ x > ln 2 (b)
y = h ( x) + f ( x) ⇒ y = ax3 + bx 2 + c + x 2 + 4 x + 6 = ax 3 + ( b + 1) x 2 + 4 x + ( c + 6 )
Since it passes through the points ( −1, −18 ) , (1, −14 ) and ( 3,30 ) a ( −1) + ( b + 1)( −1) + 4 ( −1) + ( c + 6 ) = −18 ⇒ −a + b + c = −21 − − − (1) 3
2
a (1) + ( b + 1)(1) + 4 (1) + ( c + 6 ) = −14 ⇒ a + b + c = −25 − − − (2) 3
2
a ( 3) + ( b + 1)( 3) + 4 ( 3) + ( c + 6 ) = 30 3
2
⇒ 27 a + 9b + c = 3 − − − (3) Solving (1),(2) and (3) gives a = −2, b = 10, c = −33 h ( x ) = −2 x 3 + 10 x 2 − 33 ⇒ h (101) = −1958625
4
7
(i)
(ii)
S1 = 1 = 3 − 5 , S2 = 5 = 3 − 7 , S3 = 15 = 3 − 9 2 2 4 4 8 8 3. Conjecture for S n = 3 − 2n + n 2
and
S4 = 37 = 3 − 11 . 16 16
3 ” , for all n ∈ + . Let Pn be the statement “ S n = 3 − 2n + 2n 2(1) + 3 1 When n = 1, LHS of P1 = S1 = 1 , RHS of P1 = 3 − = . ∴ P1 is true. 2 2 21 3. . i.e. S k = 3 − 2k + k 2 k +1 2r − 1 = ∑ r r =1 2 +
Assume that Pk is true for some k ∈ When n = k + 1, LHS of Sk +1
=
k
∑ r =1
2r − 1 2 ( k + 1) − 1 + 2r 2 k +1
2 k + 3 2k + 1 + k +1 2k 2 4 k + 6 − 2k − 1 = 3− 2 k +1 = 3−
2 ( k + 1) + 3 = RHS of Sk +1 2 k +1 is true and since P1 is true, by mathematical induction, = 3−
Hence, Pk is true ⇒ Pk +1 Pn is true for all n ∈ 2r + 1 N +1 2r − 1 ∑ r +1 = ∑ r r= 0 2 r=1 2 N
(iii) 8
(i)
(ii)
+
.
= 3−
2 ( N + 1) + 3 2N + 5 = 3 − N +1 . N +1 2 2
⎛ 5 ⎞⎛ 7 ⎞ ⎜ − ⎟⎜ − ⎟ 35 2 ⎛ 5⎞ (1 − 2 x ) = 1 + ⎜ − ⎟ ( −2 x ) + ⎝ 2 ⎠ ⎝ 2 ⎠ ( −2 x ) + ... ≈ 1 + 5 x + x 2 2! 2 ⎝ 2⎠ 1 1 Expansion is valid if − < x < 2 2 5 − 2
y = (1 − 2 x ) ⇒ (1 − 2 x )
−
3 2
⇒
5 5 dy 3 − − −1 = − (1 − 2 x ) 2 ( −2 ) = 3 (1 − 2 x ) 2 = 3 y (1 − 2 x ) 2 dx
dy = 3 y − − − (1) dx
Differentiate (1) wrt x,
5
⇒ −2
dy d2y dy d2y dy + (1 − 2 x ) 2 = 3 ⇒ (1 − 2 x ) 2 = 5 − − − (2) dx dx dx dx dx
Differentiate (2) wrt x, d2y d3y d2y d3y d2y ⇒ −2 2 + (1 − 2 x ) 3 = 5 2 ⇒ (1 − 2 x ) 3 = 7 2 − − − (3) dx dx dx dx dx When x = 0, y = 1,
dy d2y d3y = 3, 2 = 15, 3 = 105 dx dx dx
Hence, y ≈ 1 + 3x + p=
15 2 105 3 15 35 x + x = 1 + 3x + x 2 + x3 2! 3! 2 2
15 35 ,q = 2 2
(iii) From (ii), (1 − 2 x )
−
3 2
15 2 35 3 x + x 2 2
≈ 1 + 3x +
5 15 35 − ⎛ 3⎞ Diff wrt x, ⎜ − ⎟ (1 − 2 x ) 2 ( −2 ) ≈ 3 + ( 2 x ) + ( 3 x 2 ) 2 2 ⎝ 2⎠
⇒ (1 − 2 x ) Alternative 1: Now (1 − 2 x )
−
5 2
−
5 2
≈ 1 + 5x +
= 1 + 5x +
⇒ (1 − 2 x )
−
3 2
35 2 x same as (i) 2
35 2 315 3 x + x + ... 2 8
= (1 − 2 x )
−
5 2
(1 − 2 x )
35 315 3 ⎛ ⎞ = ⎜1 + 5 x + x 2 + x + ... ⎟ (1 − 2 x ) 2 8 ⎝ ⎠ 15 35 = 1 + 3x + x 2 + x3 + ... same as (ii) 2 2 Alternative 2: (1 − 2 x )
−
5 2
= (1 − 2 x )
−
3 2
(1 − 2 x )
−1
15 35 ⎛ ⎞ = ⎜ 1 + 3x + x 2 + x3 + ... ⎟ (1 + 2 x + 4 x 2 + 8 x3 + ...) 2 2 ⎝ ⎠ 35 2 = 1 + 5 x + x + ... same as (i) 2
6
9
(a)(i) x5 ( x16 ) = 81 ⇒ ar 4 ( ar 15 ) = 81 ⇒ a 2 r 19 = 81 20
∑ log3 xk = log
k =1
3
x1 + log 3 x2 + log 3 x3 + ... + log 3 x20 = log 3 ( x1 x2 x3 ... x20 )
(
= log 3 a ( ar ) ( ar 2 ) ... ( ar19 )
)
= log 3 ( a 20 r1+ 2+...+19 ) 19( 20 ) ⎛ ⎞ = log 3 ⎜ a 20 r 2 ⎟ ⎜ ⎟ ⎝ ⎠
= log 3 ( a 2 r19 )
10
= log 3 ( 81) = 10(4) = 40 10
(ii)
yn − yn −1 = log 3 ( xn ) − log 3 ( xn −1 ) = log 3
= log 3
xn xn −1
ar n −1 = log 3 r is a constant free from n. ar n − 2
Hence, { yn } is an arithmetic sequence. (b)
Let the amount of coal mined in the first year be a. The maximum total amount of coal mined = a + 0.9a + 0.92 a + 0.93 a + ....
=
a = 10a 1 − 0.9
Let n be the number of year at which the mine will be in operation. a(1 − 0.9n ) > 9a 1 − 0.9 ⇒
0.9n < 0.1
⇒
n>
lg 0.1 ≈ 21.854 lg 0.9
The mine will be in operation for at most 22 years.
7
10
⎛1⎞ The vector equation of p1 is r ⎜⎜ 1 ⎟⎟ = 6 ---(1) ⎜ −3 ⎟ ⎝ ⎠
The vector equation of the line passing thru A and perpendicular to p1 is ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ----(2) r = ⎜0⎟ + λ ⎜ 1 ⎟, λ ∈ ⎜ 2⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ Sub (2) into (1): ⎛ 1 + λ ⎞⎛ 1 ⎞ ⎜ ⎟⎜ ⎟ ⎜ λ ⎟⎜ 1 ⎟ = 6 ⎜ 2 − 3λ ⎟⎜ −3 ⎟ ⎝ ⎠⎝ ⎠ 1 + λ + λ − 6 + 9λ = 6 ⇒ λ = 1
⎛1⎞ ⎛ 1 ⎞ ⎛ 2 ⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Therefore OB = ⎜ 0 ⎟ + ⎜ 1 ⎟ = ⎜ 1 ⎟ ⎜ 2 ⎟ ⎜ −3 ⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛1⎞ ⎛ 0 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Q ⎜ 0 ⎟ × ⎜ −1⎟ = ⎜ −1⎟ and origin is in p2, ⎜ 1 ⎟ ⎜ 1 ⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Therefore p2 has Cartesian equation x − y − z = 0 ---(3) And p1 has Cartesion equation x + y − 3 z = 6 ---(4) Using GC to solve (3) and (4): We have x = 3+2λ, y = 3+λ, z = λ. ⎛ 3⎞ ⎛ 2⎞ ⎜ ⎟ Therefore the equation of l is r = ⎜ 3 ⎟ + λ ⎜⎜ 1 ⎟⎟ , λ ∈ ⎜0⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠
8
⎛⎛ 3⎞ ⎛ 1⎞⎞ ⎛ 2⎞ ⎜⎜ ⎟ ⎜ ⎟⎟ ⎜ ⎟ ⎜⎜ 3⎟ − ⎜ 0⎟⎟×⎜ 1⎟ ⎜⎜0⎟ ⎜ 2⎟⎟ ⎜1⎟ ⎝⎝ ⎠ ⎝ ⎠⎠ ⎝ ⎠
Shortest distance =
22 + 12 + 12
⎛ 2 ⎞ ⎛ 2⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟×⎜1⎟ ⎜ −2 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ = = 6
⎛ 5⎞ ⎜ ⎟ ⎜ −6 ⎟ ⎜ −4 ⎟ ⎝ ⎠ 6
5 2 + ( −6 ) + ( − 4 ) 2
=
2
6
=
77 6
⎛1⎞ ⎜ ⎟ Since p3 passes through the points A and B, therefore p3 // ⎜ 1 ⎟ . ⎜ ⎟ ⎝ −3 ⎠ ⎛ 2⎞ Since p1, p2 and p3 do not have a common point, therefore p3 // ⎜⎜ 1 ⎟⎟ . ⎜1⎟ ⎝ ⎠ ⎛1⎞ ⎛1⎞ ⎛ 2⎞ ⎜ ⎟ ⎜ ⎟ An equation of p3 is r = ⎜ 0 ⎟ + α ⎜ 1 ⎟ + β ⎜⎜ 1 ⎟⎟ , α , β ∈ ⎜ 2⎟ ⎜ −3 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 11
(a)
1
∫ sec x cosec x dx = 2∫ 2sin x cos x dx
.
= 2∫ cosec 2 x dx
= − ln ( cot 2 x + cosec2 x ) + c
Alternative 1:
1 sec 2 x ∫ sec x cosec x dx = ∫ sin x cos x dx = ∫ tan x dx = ln | tan x | +c
Alternative 2:
∫ sec x cosec x dx = ∫
sin 2 x + cos 2 x dx = ∫ tan x dx + ∫ cot x dx sin x cos x = ln | sec x | + ln | sin x | +c
(b)
Let u = ln x ⇒
du 1 = dx x
9
1
∫ x ln ( ln x ) dx = ∫ ln u du = u ln u − ∫ u
1 dx = u ln u − u + c u = ( ln x ) ( ln ( ln x ) − 1) + c
(c) (i) Solving x 2 + y 2 = 4 and y =
)
(
1 ⇒ A is − 2 + 3 , − 2 − 3 x or (−1.931852,−0.5176381) from GC.
[2] [3]
Solving x 2 + y 2 = 4 and x = −1 ⇒ B is (−1,− 3 ) or (−1,−1.73205). [4]
2
− 2+ 3
2 − 2+ 3 −1
− 3
−1
(ii) Area =
∫
−1
4 − x 2 dx −
− 2+ 3
∫
− 2+
1 dx x 3
= 0.546 [From GC]
(iv) Volume = π
− 2− 3
∫
4 − y d y − π (1) ( 3 − 1) − π 2
2
− 2− 3
∫
−1
− 3
1 dy y2
= 4.74
[From GC]
END OF PAPER
10
TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination Higher 2
MATHEMATICS
9740/02
Paper 2
20 September 2010
Solutions Section A: Pure Mathematics [40 marks] dy dy dx 3t 2 − 9 = / = dx dt dt 2t − 2
1
For tangent parallel to the y-axis, 2t − 2 = 0 ⇒ t = 1 Equation of tangent is x = −1 When y = 0, t 3 − 9t = 0 ⇒ (t + 3)t (t − 3) = 0 ∴ t = −3 , 0 or 3 [NA since t ≤ 2 ]
When t = −3 ⇒ x = ( −3) − 2 ( −3) = 15, 2
∴
y=0
C intersects the x-axis at the point (15, 0). 15
Area of R = ∫ y dx 0 −3
=
∫ (t
3
− 9t ) ( 2t − 2 ) dt
0
0
= − ∫ ( 2t 4 − 2t 3 − 18t 2 + 18t ) dt −3
0
⎡ 2t 5 t 4 ⎤ = −⎢ − − 6t 3 + 9t 2 ⎥ = 105.3 2 ⎣ 5 ⎦ −3 2
ex − 2x − 2 = 0 ⇒ ex = 2x + 2 From the diagram, there are two intersection points. ∴ e x − 2 x − 2 = 0 has two roots.
1
(i)
(ii)
Since graph of y = 2 x + 2 cuts x-axis at x = − 1, β ∈ ( −1, 0 ) , ∴ a = −1 . 1 Given that xn + 1 = (e xn − 2) , xn < 0 , for n ∈ + . 2 If the sequence converges to a number L, then when n → ∞ , xn → L , xn+1 → L . 1 ⇒ L = (e L − 2) 2 L ⇒ e − 2L − 2 = 0 ∴ L = β as xn < 0 , for all n ∈ + .
β < xn < 0 ,
(iii) If
xn > β
(iv)
1 xn +1 − xn = (e xn − 2) − xn 2 1 = e xn − 2 xn − 2 2 1 = e xn − ( 2 xn + 2 ) 2 <0
(
⇒ e xn > e β as y = e x is an ↑ function. ⇒ e xn − 2 > e β − 2 1 xn 1 β ⇒ e −2 > e −2 2 2 1 ⇒ xn +1 > ( 2 β ) from (ii) 2 ⇒ xn +1 > β
)
(
(
)
as e xn < 2 xn + 2 from the above diagram
(
)
)
Hence, xn +1 < xn .
From (iii) and (iv), 0 > x1 > x2 > x3 > .... > β . It is a decreasing sequence which converges to β . 3
3
(a) Mtd1
3
⎛ z ⎞ ⎛ z ⎞ 3+i 3+i =0 ⇒ ⎜ ⎜ ⎟ + ⎟ =− z z ⎝ 3+i⎠ ⎝ 3+i⎠
⇒ z4 = − ⇒
(
3+i
z 4 = 24 e
⇒ z = 2e
(
i 23π +π
)
4
( ) iπ
= − 2e 6
4
) = 24 ei 53π = 24 ei( − π3 ) = 24 ei( − π3 + 2 kπ )
⎛ − π + 2 kπ i⎜ 3 ⎜ 4 ⎝
⎞ ⎟ ⎟ ⎠
= 2e
i
( 6 k −1)π 12
, k = 0, ± 1, 2
Mtd2 3
4
⎛ z ⎞ ⎛ z ⎞ 3+i i ( π + 2 kπ ) iπ =0 ⇒ ⎜ ⎜ ⎟ + ⎟ = −1 = e = e z ⎝ 3+i⎠ ⎝ 3+i⎠ ( i z z = π =e i 3+i 2e 6
⇒ ⇒
z i
π
=e
i
( π + 2 kπ ) 4
=e
π + 2 kπ )
k = 0, ± 1, − 2
4
π
± i 4
,e
±
3π i 4
2e 6 ⇒ z = 2e
−
π 12
i
, 2e
5π i 12
, 2e
−
7π i 12
2
, 2e
11π i 12
Mtd3 3
⎛ z ⎞ 3+i =0 ⇒ ⎜ ⎟ + z ⎝ 3+i⎠ z ⇒ = − 2 + 2i, − 3+i ⇒ z=
(
( (
)(
4
⎛ z ⎞ ⎜ ⎟ +1 = 0 ⎝ 3+i⎠ 2 − 2i, 2 + 2i, 2 − 2i
)(
3 + i − 2 + 2i ,
) ( ) (
)(
) ( 3 + i )( 2 + 2i ) , ( 6 ) + (− 6 − 2 ) i⎤ , ⎦ 2 ) − ( 6 − 2 ) i⎤ . ⎦
3 + i − 2 − 2i ,
)
⇒ z = ⎡ − 6 − 2 + 6 − 2 i⎤ , ⎣ ⎦ ⎡ 6 − 2 + 6 + 2 i⎤ , ⎣ ⎦
)
( (
⎡ 2− ⎣ ⎡ 6+ ⎣
3+i
)(
2 − 2i
)
Or −1.93 + 0.518i, − 0.518 − 1.93i, 0.518 + 1.93i, 1.93 − 0.518i . Mtd4 3
⎛ z ⎞ 3+i =0 ⇒ ⎜ ⎟ + z ⎝ 3+i⎠
z3 3+i =− ⇒ z 4 = −8i 8i z
⇒ z = 2e
⎛ − π + 2 kπ i⎜ 3 ⎜ 4 ⎝
⎞ ⎟ ⎟ ⎠
= 2e
i
(
( 6 k −1)π 12
)
3 + i = 8 − 8 3i = 16e
π
− i 3
, k = 0, ± 1, 2
Mtd5 3
⎛ z ⎞ 3+i = 0 ⇒ z4 = − ⎜ ⎟ + z ⎝ 3+i⎠
(
3+i
)
4
⇒ z2 = ± −
(
3+i
)
4
= ±4e −0.5236i
⇒ z = ± ±4e −0.5236i = 1.93 − 0.518i, −1.93 + 0.518i, 0.518 + 1.93i, −0.518 − 1.93i y (b)
iz + 3 = 5 ⇒
z − 3i = 5
3• P
Mtd1
Q θ • •
O
S R Let Q represent the number − 3 in the Argand Diagram.
x
Minimum of z + 3 is equal to the length QR. To find z, we find the complex number represented by R in the above Argand Diagram. 3 π From diagram, tan θ = = 3 ⇒ θ= . 3 3 5 ⎛1⎞ x-coordinates of R is −5cos θ = −5 ⎜ ⎟ = − 2 ⎝2⎠ ⎛ 3 ⎞ 6−5 3 y-coordinates of R is 3 − 5sin θ = 3 − 5 ⎜⎜ ⎟⎟ = 2 2 ⎝ ⎠ 5 6−5 3 Therefore, required z = − + i 2 2
3
Mtd2 By similar triangle:
5 5 3 ; RS = 2 2 ⎛ 5 ⎛5 5 ⎛5 ⎞⎞ ⎞ So P = ⎜ − , − ⎜ 3 − 3⎟ ⎟ ⇒ z = − − ⎜ 3 − 3 ⎟ i or z = −2.5 − 1.33i 2 ⎝2 ⎠⎠ ⎠ ⎝ 2 ⎝2 Mtd3 2 R is the intersection of Line: y = 3x + 3 and Circle: ( y − 3) + x 2 = 52 . PQ = 2 3 ; QR = 5 − 2 3 ; PS =
⎛ 5 ⎛5 5 ⎛5 ⎞⎞ ⎞ Solving gives P = ⎜ − , − ⎜ 3 − 3⎟ ⎟ ⇒ z = − − ⎜ 3 − 3 ⎟ i or z = −2.5 − 1.33i 2 ⎝2 ⎠⎠ ⎠ ⎝ 2 ⎝2 4
(a)(ii)
y y2 = f ( x)
−2
0
1
x
x = −1 2 − x2 − 2x + 5 Note: y = (b) = −x − 3 + x −1 x −1 2 A 2 2 = −x − 4 + Mtd1 y = x − 4 − ⎯⎯ → y = −x − 4 − x −x x 2 2 B = −x − 3 + ⎯⎯ → y = − ( x − 1) − 4 + x −1 ( x − 1) A: a reflection about y-axis B: a translation of 1 unit in the positive direction of x-axis. Mtd2 A: a translation of 1 unit in the negative direction of x-axis. B: a reflection in y-axis. Mtd3 A: a translation of 1 unit in the positive direction of x-axis. B: a reflection in x-axis. C: a translation of 8 units in the negative direction of y-axis.
4
Mtd4 A: a translation of 1 unit in the positive direction of x-axis. B: a translation of 8 units in the positive direction of y-axis. C: a reflection in x-axis. x =1
− x2 − 2x + 5 2 = −x − 3 + , x −1 x −1 Asymptotes are: x = 1 and y = −x−3
y=
(ii)
y = −x − 3
− x2 − 2 x + 5 y= x −1 Adding graph of
(iii)
( x − 1)
2
2
( x − 1) 2
2
2
+ ( y + 4) = 1
+ ( y + 4) = 1 2
2 to part (ii), there are 4 roots for the equation
( x − 1)
2
22
2
⎛ − x2 − 2 x + 5 ⎞ +⎜ + 4 ⎟ = 1. x −1 ⎝ ⎠
Section B: Statistics [60 marks] 5
(i)
−
The sample may not represent all levels of college’s population.
−
Students who do not go to the college canteen during that one hour will be excluded from the sample.
− The sample is not representative because it excludes students who are not willing to take part in the survey. (ii)
Quota sampling.
(iii)
Use Stratified sampling: Draw random samples from each level as follows: Level Number of students
IP1 78 × 160 1600 = 7.8 ≈ 8
IP2
JC1
JC2
70 × 160 1600 =7
692 × 160 1600 = 69.2 ≈ 69
760 × 160 1600 = 76
5
2
6
H o : μ = 34 H1 : μ < 34 Level of significance: 2% X −μ ~ N (0, 1) Test statistic: σ/ n 32.15 − 34 = −2.4917 (5 s.f.) , p –value = 0.00636 < 0.02 Assume H o is true, zcal = 2.1/ 8 Since the p –value is less than the level of significance, we reject H o at 2% level of significance. There is sufficient evidence to conclude that John’s typing speed has improved after the training course. 2% level of significance means there is 2% chance that we wrongly support the claim that John has improved his typing speed after the training course.
7
Local Foreign
Male 4 2
Female 2 3
(a) Number of ways = 6! × 7P5 = 1814400 (b) (i) Number of ways = 11C4 = 330 (ii) Number of ways = No. w/o restriction − No. w/o males − No. w/o females = 11C4 − 5C4 − 6C4 = 310 (iii) No. of ways = No. w/o restriction − No. w/o females − No. w/o foreigners + No. with only local males 11
= C4 − 6C4 − 6C4 + 4C4 = 301. 8(i) P(TUFC draws a match) = 1− (ii)
1 1 3 − = 2 5 10
P(three points after 3 games) = P(WLL) + P(DDD) 2
1⎛1⎞ ⎛ 3⎞ = ⎜ ⎟ ( 3) + ⎜ ⎟ 2⎝5⎠ ⎝ 10 ⎠ = 0.087 (iii)
3
P(qualifies for second round | TUFC did not win the first game) =
P ( qualifies and did not win the first game ) P ( did not win the first game ) 6
P ( DWW ) +P ( DWD ) +P ( DDW ) +P ( LWW ) 1 1− 2 2 2 2 3 ⎛1⎞ ⎛ 3 ⎞ ⎛1⎞ 1⎛1⎞ ⎜ ⎟ + 2⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ 10 ⎝ 2 ⎠ ⎝ 10 ⎠ ⎝ 2 ⎠ 5 ⎝ 2 ⎠ = 1 2 = 0.43 =
9
Let A be the timing of swimmer A for the 100m freestyle. A ~ N (48.0, 0.52 ) Let B be the timing of swimmer B for the 100m freestyle. B ~ N (47.2, 0.82 ) (i) Mtd1 Required Probability = 1− P(non among the two broke the world record) = 1 − P ( A ≥ 46.9 ) P ( B ≥ 46.9 )
= 1 − ( 0.986097 )( 0.646170 ) = 0.36281 ≈ 0.363 Mtd2 Required Probability = P(A breaks record) + P(B breaks record) – P(both break record) = P ( A < 46.9 ) + P ( B < 46.9 ) − P ( A < 46.9 ) P ( B < 46.9 ) ≈ 0.363 Mtd3 Required Probability = P(A breaks record, B don’t break record) + P(A don’t break record, B breaks record) + P(both break record) = P ( A < 46.9, B ≥ 46.9 ) + P ( A ≥ 46.9, B < 46.9 ) + P ( A < 46.9 ) P ( B < 46.9 ) ≈ 0.363
A − B ~ N (0.8, 0.89) P ( A > B ) = P ( A − B > 0 ) = 0.801781 ≈ 0.802
(ii)
20 × (1 − 0.801781) ≈ 4
(iii)
Expected times out of 20 is 4 times. Let W = A1 + A2 + A3 + A4 − 4 B ~ N (3.2,11.24)
(iv)
P ( A1 + A2 + A3 + A4 > 4 B ) = P (W > 0 ) = 0.83008 ≈ 0.830
10
(a)
Let X be the number of prizes won out of N games. P ( X > 5 ) < 0.1 ⇒
P ( X ≤ 5 ) ≥ 0.9
From G.C., When N = 63, P ( X ≤ 5 ) = 0.90551 > 0.9 When N = 64, P ( X ≤ 5 ) = 0.89990 < 0.9 Therefore, largest number of games is 63. 7
X ~ B ( N , 0.05 )
(b)(i)
Let Y be the number of prizes won out of 60 games.
Y ~ B ( 60, 0.05 )
Since 60 × 0.05 = 3 < 5 , Y ~ Po ( 3) approximately
P (Y ≤ 5% ( 60 ) ) = P (Y ≤ 3) = 0.64723 = 0.647 (3 s.f.)
(ii)
Let S be the number of prizes won out of 200 games. S ~ B ( 200, 0.05 ) Since 200 × 0.05 = 10 > 5 , 200 × 0.95 = 190 > 5 S ~ N (10, 9.5 ) approximately P ( S ≤ 5% ( 200 ) ) = P ( S ≤ 10 ) ≈ P ( S < 10.5 )
= 0.564434 = 0.564 (3 s.f.)
(c) Mtd1: Y ~ B ( 60, 0.05 ) , E(Y) = 60 × 0.05 = 3 , Var(Y) = 60(0.05)(0.95) = 2.85 Since sample size 70 is large, by CLT, ⎛ 2.85 ⎞ Y ~ N ⎜ 3, ⎟ , i.e. Y ~ N ( 3, 0.04071) ⎝ 70 ⎠
(
)
P 2.5 ≤ Y ≤ 3 = 0.49340 = 0.493 (3 s.f.) Mtd2: Let T be the total number of prizes won out of 70 × 60 games. T ~ B ( 4200, 0.05 )
T ⎛ ⎞ ≤ 3 ⎟ = P (175 ≤ T ≤ 210 ) = 0.51333 ≈ 0.513 P ⎜ 2.5 ≤ 70 ⎝ ⎠ 11 (i)
The least squares regression line of y on x should be used as x is an independent variable and y depends on x.
(ii)
Equation of the regression line of y on x is y = 21.9 + 0.504x The slope of this regression line means Mary’s travelling time will increase by 0.504 minutes for each minute she leaves home after 7 am.
(iii)
The product moment correlation coefficient, r = 0.988 which indicates the time that Mary leave home after 7 am and the time that she takes to travel are strongly positively linearly correlated.
(iv)
z = Time available – Time taken = 90 – x – y.
(v)
For Mary not to be late, x + y ≤ 90 ⇒ x + 21.881 + 0.504x ≤ 90 ⇒ x ≤ 45.292 The latest time that Mary can leave home is 7.45am. This value is reliable as the value of r is close to 1 and x = 45 is within the given set of data. The estimate is an interpolation. 8
VICTORIA JUNIOR COLLEGE Preliminary Examination Higher 2
MATHEMATICS (Paper 1)
9740/ 01 September 2010 3 hours
Additional materials:
Answer paper Graph Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 5 printed pages © VJC 2010
VICTORIA JUNIOR COLLEGE
[Turn over
1
The equation of a curve C is x 2 3 xy y 2 5 . Show that every line parallel to the x-axis cuts C at two distinct points. [3] Without differentiating, explain, giving a reason, if there is any point on the curve at which the tangent is parallel to the x-axis. [1]
2
In the game of TapFarm, a player is given plots of land to grow tomatoes, pumpkins and cherries. The player can only plant a type of fruits on a plot of land each time and each plot of land produces 1 kg of fruits. The time taken from planting to harvesting the fruits, cost price and selling price for 1kg of fruits are as follows:
Tomatoes Pumpkin Cherries
Time required (Hours) 3 5 16
Cost price ($) 15 a 45
Selling price ($) 35 50 100
Tommy has only 1 plot of land. He used $545 and spent 154 hours playing the game before harvesting a total of 23 kg of fruits and earned a profit of $735. Showing your working clearly, find the value of a. [4] [You are to assume that Tommy harvests the fruits once they are ready and there is no time lapse between harvesting and planting new fruits.]
3
4
Find the exact value of
3 1
x2 dx by using the substitution x tan . ( x 2 1) 2
1 , using the substitution y 2 x 1, show algebraically that 2 22 8 2 x 1 2 x is always positive, [2]
(i)
Given that x
(ii)
Hence, solve the inequality 22 x 8 x 2 x 1 2 x 2
x k x m
2
0,
where k and m are real and 0 < k < m.
5
(a)
[5]
[4]
Solve the equation i( z 2) 4 (1 i) 0 giving the roots in the form z p re i , where p is a real number, r > 0 and . [4]
(b)
Given that 1 i 2 is a root of the equation 3z 3 az 2 bz 3 0 , find the values of the real numbers a and b.
2
[3]
6
7
2 in partial fractions. [2] (r 1)r (r 1) n 1 (ii) Hence, find . (There is no need to express your answer as a single r 2 ( r 1) r ( r 1) algebraic fraction.) [4] n 1 (iii) Given that k , for all n 2 , find the least value of k, showing your r 2 ( r 1) r ( r 1) working clearly. [2] (i) Express
Positive odd integers, starting at 1, are grouped into sets containing 1, 3, 9, … odd integers , as indicated below, so that the number of odd integers in each set after the first is thrice the number of odd integers in the previous set. {1}, {3, 5, 7}, {9, 11, 13, 15, 17, 19, 21, 23, 25}, … Find, in terms of k, (i) the number of terms in the kth set, (ii) the number of terms in the first k sets, (iii) the first integer in the kth set, (iv) the last integer in the kth set, (v) the sum of all integers in the first k sets.
8
[1] [2] [1] [2] [3]
The planes 1 and 2 have equations r. i j k 6 and r. 2i 4 j k 12 respectively. The point A has position vector 9i 7 j 5k . (i) Find the position vector of the foot of perpendicular from A to 2 . (ii) Find a vector equation of the line of intersection of 1 and 2
[3] [2]
The plane 3 has equation
r i j k i 3 j 3k i 9 j bk ,
where , are real parameters and b is a constant. Given that 1 , 2 and 3 have no point in common, find the value of b. [3] 3 meets 1 and 2 in lines l1 and l2 respectively. Without finding the equations of l1 and l2 , describe the relationship between l1 and l2 , giving a reason. [2] 9
A sequence of real positive numbers u1 , u2 , u3 , ... satisfies the recurrence relation un 1 un 4 4 un 1, n and u1 15 . (i) Prove by induction that un 4 n 1 1 for all n . (ii) Determine, giving your reasons, if this is a converging or diverging sequence. 2
The sequence is modified to 1 1 vn 1 4 4 1, vn vn
n and v1
As n , vn . (iii) Find the value of correct to 3 decimal places. (iv) Show, graphically or otherwise, that if vn , vn vn 1 . 3
[4] [2]
1 . 15 [2] [3] [Turn over
10
(a)
(b)
Sketch the curve given by the equation x2 y 2 1, 0 k 3. [2] 9 k2 Given that m is a constant and the equation x 2 m2 x 2 2 1 9 k has 2 real roots, use your sketch above to find, in terms of k, an inequality satisfied by | m | . [3] A curve C has equation y (i) (ii) (iii)
11
3x( x 4) . x2
dy , show that the gradient of C is always positive. dx Find the equations of the asymptotes of C. Sketch C. By considering
[3] [2] [1]
y
(a)
1 y=1 1
y
2
e x 1 e x
R
y=0
(i)
x
O
e x . The region R is 1 e x bounded by C, the positive x-axis and the positive y-axis. Find the exact area of R. [3]
The diagram shows the curve C with equation y
(ii) Denoting the answer you have obtained in part (i) by q, write down, in terms of q, the area of R if C is scaled by factor 2 parallel to the y-axis. [1] x e 1 (iii) Sketch the graph of y 2 , indicating clearly the equations of the x 1 e 2 asymptotes and the shape of the curve for points near y = 0. [3]
4
11
y
(b)
y = ln x
B
A (a, ln a) x
O
A girl intends to design a bowl by rotating a section AB of the curve y ln x completely about the y-axis. She wants the bowl to hold 300 cm3 of fluid. Given that the diameter of the rim is twice that of the base, write down, in terms of a, the coordinates of the point B. Hence find the exact value of a. [5]
12
Verify that the general solution of the differential equation, x
y Axe x 1 .
dy ( y 1)(1 x) is dx [1]
(i) Sketch three members of the family of solution curves, one for positive A, one for negative A and one for A = 0. [3] (ii) It is given that A = 3 and k is a positive constant.
Find
k
3 xe x dx in terms of k. Hence, state the value of
0
3 xe x dx .
[4]
0
Indicate, on a clearly labeled diagram, the region whose area is given by
3 xe x dx .
[1]
0
(iii) A point P is conditioned to move along the curve y 3xe x 1 such that the xcoordinate of P increases at a constant rate of 2 units per second. dy (a) State the range of values of x for which 0. [1] dt (b) Find the x-coordinate of the point on the curve at which P is moving such dy that 2 . [3] dt
5
1
VICTORIA JUNIOR COLLEGE Preliminary Examination Higher 2
MATHEMATICS Paper 2
9740/ 02 September 2010 3 hours
Additional materials:
Answer paper Graph paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages © VJC 2010
VICTORIA JUNIOR COLLEGE
[Turn over
2 Section A: Pure Mathematics [40 marks]
1
Sketch, on an Argand diagram, the locus representing the complex number z for which
z 4 3i 2.
[2]
(i)
Given that a is the least possible value of z , find a.
(ii)
The complex number p is such that p 4 3i 2 and p a.
[2]
State the exact value of arg p .
[1]
z (iii) Deduce the greatest value of arg , giving your answer correct to 2 decimal p places. [2]
2
(i)
A curve C has equation y f ( x ). y 2 f ( x)
y
y=–x–4 –1
O 3
yf x
y
x
y=x–4 –3
O
3
x
(–1, –1) (1, –1) x = –3
The diagram shows the curve C 1 with equation y 2 f ( x ) and the curve C 2 with
equation y f x . The line x 3 is an asymptote to C 1 and C 1 passes through the points ( 1, 0) and (3, 0). The lines y x 4 and y x 4 are asymptotes to C 2. C 2 passes through the 2 points ( 3, 0), (3, 0) and 0, . The minimum points on C 2 have coordinates 3 1, 1 and 1, 1 . 7 Given that f ( x) 0 for 1 x 0 and the point , 9 is the only other 2 turning point on C, sketch C, indicating clearly the intercepts, the equations of the asymptotes and the coordinates of the turning points. [4]
(ii)
Find the x-coordinates of the stationary points on the curve y f ( x) . 2
[3]
3
3
1 1 1 The plane and the line l have equations r. 1 4 and r 2 0 respectively, 2 1 a where is a real parameter and a is a constant. (a)
(b)
4
It is given that a 1 . Find (i) the acute angle between and the plane z 0, (ii) the exact perpendicular distance of the point (1, 3, 2) from . It is given that a 3. Find the acute angle between l and .
1
up to and including the term in x 4 .
[2] 1 x (ii) Given that x is small such that powers of x above x3 could be ignored, use your answer in part (i) to show that sin 1 x x bx3 where b is a constant to be found. [3] 1 (iii) State the equation of the tangent to the curve y sin x at the origin. On a single (i)
Obtain the expansion of
[2] [3] [3]
2
diagram, sketch this tangent and the graph of y x bx3 . You should make clear the relationship between the two graphs for points near the origin. [3]
5
The functions f, g and h are defined by
1 f :x x , x 2 g : x x , 2010 h:x , x (i) (ii) (iii) (iv) (v)
x
, x 0,
x
,
x
, x 0.
Sketch the graph of y = f(x). Define f 1 in a similar form. Use a non-calculator method to solve f 1 ( x) 4. State, giving a reason, whether fg exists. Find h 21 ( x).
[2] [3] [2] [2] [1]
[Turn over
4 Section B: Probability and Statistics [60 marks]
6
A circular disc is divided into n equal sectors which are labeled 1, 2, 3, … n –1 and a “skunk”. A game is played by spinning a pointer pivoted at the centre of the disc at most three times. If the result is a “skunk”, the game ends and the player loses all his previous winnings. If the result from a spin is k where k = 1, 2, 3, … or n – 1, then the player wins $k. He continues spinning and adding to his winnings until a maximum of three spins or a “skunk” is spun.
1 . n3 Find, in terms of n, the probability that the player (i) wins nothing when the game ends, (ii) wins nothing when the game ends, given that he wins $3 in his first spin.
Show that the probability of winning $3 is
[1] [2] [2]
[There is no need for you to simplify your answer in both cases.]
7
In a statistical investigation, a researcher wants to find out how a person’s maximum walking speed varies with age. He selects a random sample of 12 males of certain ages and measures their individual maximum walking speeds. Their ages, t years and maximum walking speeds, w ms–1 are shown in the table below.
t
20
25
30
35
40
45
50
55
60
65
70
75
w
2.59
2.55
2.85
2.62
2.48
2.43
2.32
2.27
2.34
2.28
2.19
2.10
(i) (ii)
Draw a scatter diagram for the data. [1] State, giving a reason, whether a regression line of w on t or t on w could be used to estimate the age of a male who has a maximum walking speed of 2.65 ms–1. (There is no need to do any calculations.) [1]
The researcher decides to study the maximum walking speed of males between the age of 30 and 55 inclusive. It is given that the correlation coefficient for the six data points is 0.965 . (iii) State, giving a reason, whether the regression line stated in (ii) is suitable for this study. [1] 1 (iv) For this study, the variables y is defined by y . For the variables y and w, t (a) calculate the product moment correlation coefficient and comment on its value, [2] (b) calculate the equation of the appropriate regression line, [1] (c) determine the best estimate that you can of the maximum walking speed when the age is 43. [1]
5 8
The distribution of the masses of Perayaan balls has a mean of 420 g and a standard deviation of 6 g. Find the probability that 50 randomly selected Perayaan balls have a combined mass which is more than 20.9 kg. [2] State, with a reason, whether it is necessary to assume that the masses of Perayaan balls follow a normal distribution. [1] The masses of Jubalani balls follow a normal distribution with a mean of 440 g and a standard deviation of 1 g. Find the probability that the combined mass of 50 randomly selected Perayaan balls differs from 50 times the mass of one Jubalani ball by less than 900 g. [3] State the assumption that you have made in arriving at your answer. [1]
9
The number of arrivals per minute at a fast food drive-through outlet has a Poisson distribution with mean . On a weekend evening, 0.8 . Find the probability that in a 10-minute interval, there will be at most 10 arrivals given that there are more than 5 arrivals. [3] In view of space constraints, the management wants to control the number of arrivals during the peak period which spans over a 30-minute interval. By using a normal distribution to approximate the Poisson distribution, find, to 4 decimal places, the largest value of such that the probability of having more than 30 arrivals during the peak period is less than 0.05. [5]
10
Annabel has 7 tiles each lettered A, N, N, A, B, E, L respectively. A code-word is formed when some tiles are picked and arranged to form a “word”. (a)
Find the number of different ways in which a 4-letter code-word can be formed (i) if the first letter is N and last letter is E, [3] (ii) if there are no restrictions. [5]
(b) Annabel picks up 4 tiles and arranges them in a random order. Find the probability that the tiles spell ANNA. [2] 11
The weekly earnings, in dollars, at two casinos are modeled by independent normal distributions with means and standard deviations as shown in the table. Mean Earnings Standard Deviation casino 1 600 000 50 000 casino 2 700 000 75 000 (i)
Find the probability that in 2 randomly chosen weeks, the total earnings at casino 2 exceed 1 500 000 dollars. [2]
(ii) Find the probability that in a 12-week period, the weekly earnings at casino 1 exceeds $650 000 in at least 3 weeks. [3] (iii) The government imposes a weekly tax on the earnings at casino 1 and 2 at a rate of 7% and 10% respectively. Find the probability that the tax exceeds $99 000 in any given week. Hence, by using a suitable approximation, find the probability that in a year consisting of 52 weeks, the weekly tax received by the government from the two casinos exceeds $99 000 in at least 45 weeks. [6] [Turn over
6 12
(a)
Club Gunpla emailed all of its 1600 members to find out which of the four Celestial Being mechs was the most popular. The members were asked to select their favourite mech. 226 members responded and it was concluded that Gundam Exia is the most popular of the four mechs among the club members. Explain if the sampling method used is a random one.
[2]
(b) A large department store wants to find out how much its customers spend on Gundam model kits. From a random selection of 100 transactions, the results are summarized by
x 3500 ,
x
2
220 400 ,
where $x is the amount spent on Gundam model kits in a single transaction. The distributor claimed that the mean amount a customer spent on Gundam model kits is $40. Test whether the distributor has overstated his claim at the 5% significance level. [6] State, giving a reason, whether any assumption is needed for the test to be valid. [1] (c)
In testing the mean breaking strain of a type of fishing line, a researcher measured the breaking strain of 80 fishing lines. He carried out a t-test at the 5% significance level and, based on the sample results, he concluded that the population mean breaking strain is significantly different from 0 kN. If the researcher had carried out a z-test instead, determine which of the following 2 statements is correct, giving clear reasons to support your claim. (I)
The researcher would have concluded that there is significant evidence at the 5% significance level that the population mean breaking strain is different from 0 kN.
(II) It is not possible for the researcher to conclude, by using only the information given, whether there is significant evidence at the 5% significance level that the population mean breaking strain is different from 0 kN . [3]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 1
1
f ( x) , where f(x) is a quadratic function. Given x +1 that the curve passes through the points (1, 4), (−3, −12) and (−2, −14), find the equation of the curve.
A curve has equation given by y =
Hence sketch the curve, showing clearly the coordinates of the turning points and the equations of the asymptotes. [5]
2
(i)
Show that
(ii)
Find
(2n + 1)(2n + 3) (2n − 1)(2n + 1) 2(2n + 1) . − = (n + 1)(n + 2) n(n + 1) n(n + 1)(n + 2) (2n + 1)
N
∑ n(n + 1)(n + 2) , in terms of N.
[2]
[3]
n =1
(iii)
3
4
9 11 5 7 + + + + ... 2 × 3× 4 3× 4 × 5 4 × 5× 6 5× 6 × 7
Hence, find the value of
a
The curve C has equation y =
x − bx 2
[2]
, where a and b are positive constants.
(i)
Find, by differentiation, the coordinates of the turning point(s) of C.
(ii)
Sketch the graph of C, showing clearly the coordinates of any turning point(s) and the equations of the asymptotes. [2]
(iii)
Hence, find the range of values of k, where k is a constant, for which the equation k x 2 − bx − a = 0 has no real root. [2]
(iv)
On separate diagrams, draw sketches of the graphs of
(
[3]
)
(a)
y=
(b)
y2 = −
(x + b )
2
a
− b( x + b )
a x 2 − bx
,
[2]
.
[2]
The functions f and g are defined by f : x ( x − 3) + 8 , x ∈ ℜ , x ≤ 3 , 2
g : x e2x + 1,
x∈ℜ.
(i)
Show that f −1 exists and define f −1 , giving its rule and domain.
(ii)
Determine, with a reason, whether the composite function f −1g exists or not. [2]
(iii)
If α and β are real numbers such that α < β ≤ 3 and gf(α) > gf(β), show that α < 6 − β. [3] 1
[4]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 1 5
6
The nth term of a sequence is given by un = n2n for n ≥ 1 and the sum of the first n terms is denoted by Sn. (i)
Write down the values of S1, S2, S3, S4, S5.
[2]
(ii)
By considering Sn – 2, find a conjecture for the general term Sn in the form of Sn = p2(p + 2) + 2 , where p is in terms of n. [2]
(iii)
Prove the conjecture by mathematical induction for all positive integers n. [4]
(i)
By using the substitution x = a sin θ , find, in terms of a and π , a
⌠ 2 a − x 2 dx . ⌡0
[4] n
(ii)
⌠ Find, in terms of n and e, ( x + n ) e nx dx , where n ≠ 0 . ⌡0
(iii)
Hence, find the exact area of the region R bounded by the line x = 1 and by the curves y = 1 − x 2 and y = ( x + 1)e x .
[2]
6x
7
A curve C has equation y 2 =
8
Adam has many marbles that he wants to put in boxes.
9
[4]
. The region R is enclosed by C and the line 1 + 2x 2 x = 2. Another region S is enclosed by C and the lines x = 2 and x = k (k > 2). The volume of solid formed by region R is equal to the volume of solid formed by region S when both R and S are rotated completely about the x-axis. Find the exact value of k. [4]
(i)
If he puts 13 marbles in the first box and for each subsequent box, he puts double the number of marbles he puts in the previous box, he will need (2k + 1) boxes for all his marbles. Given that he has 104 marbles in the kth box, find the number of marbles he has. [4]
(ii)
If he puts 13 marbles in the first box and for each subsequent box, he puts 13 marbles more than what he puts in the previous box, how many boxes will he need and what is the number of marbles in the last box? [4]
(a)
The complex numbers z and w are such that z = 1 + i p, w = 1 + i q, where p and q are real and p is positive. Given that zw = 3 – 4i, find the exact values of p and q. [4]
(b)
A complex number a is given by a = 1 + i 3 . By using De Moivre’s theorem, express 1 + a + a 2 + a 3 + .... + a 9 in the form x + i y, where x and y are exact values to be determined. [4]
2
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 1 10
It is given that y = ln (1 + sin x ) . d2 y dy dy in terms of x and show that (1 + sin x ) 2 + cos x + sin x = 0 . [2] dx dx dx
(i)
Find
(ii)
By further differentiation of the result in (i), find the Maclaurin’s series for y, up to and including the term in x3. [3] sin x e , up to and including the term Deduce the Maclaurin’s series for ln + x 1 sin in x3. [3]
(iii)
11
The rate at which a substance evaporates is proportional to the volume of the substance which has not yet evaporated. The initial volume of the substance is A m 3 and the volume which has evaporated at time t minutes is x m 3 . Given that it takes 1 − t (2ln2) minutes for half of its initial volume to evaporate, show that x = A 1 − e 2 . Find the additional time needed for three quarters of the substance to evaporate, giving your answer in exact form. [6]
12
The diagram shows a prism with the horizontal rectangular base PQRS. The triangular planes APS and BQR are vertical and AB is horizontal. A
B
θ
S k
θ
j
R
2 P
i
3
Q
Given that PQ = SR = AB = 3 units, PS = QR = 2 units and each of the planes ABQP 3 and ABRS is inclined at an angle θ to the horizontal, where tan θ = . 4 The point P is taken as the origin for position vectors, with unit vectors i and j parallel to PQ and PS respectively and unit vector k perpendicular to the plane PQRS. → (i) Find PA . [1] (ii)
Find the exact value of the cosine of the angle PAR.
[3]
(iii)
Find a vector equation of the line AR.
[1]
(iv)
Show that the foot of perpendicular from P to the line AR has coordinates − 21 162 132 [3] , , . 169 169 169 3
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 1 13
r h
The diagram shows the cross-section of a cone of radius r and height h which is inscribed in a sphere of fixed radius R. Show that 1 V = πh 2 (2 R − h ) , 3 where V is the volume of the cone. Prove that, as r and h varies, the maximum value of V is obtained when h 2 = 2r 2 . [8]
~ End of Paper ~
4
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 2 Section A: Pure Mathematics [40 marks] 1
A curve has parametric equations x = 3(1 − t ) , y = (i) (ii)
2
Find
1 for t ≠ 0 . t3
dy in terms of t and deduce that the curve is an increasing function. dx
[2]
1 Find the equation of L1, the tangent to the curve at the point 3 − 3t , 3 . t Hence, find the coordinates of point P on the curve at which L1 passes through the origin O. [4]
(iii)
The line L2 is another tangent to the curve which is parallel to L1. Find the equation of L2. [3]
(iv)
The line L2 cuts the y-axis at Q. Find the exact area of triangle OPQ.
(i)
The equation 2x2 – x – ln(x + 1) = 0 has 2 real roots α and β , where α <β. Find the values of α and β, giving any non-exact answers correct to 3 decimal places. [1]
(ii)
A sequence of positive real numbers x1 , x2 , x3 , satisfies the recurrence
[2]
ln( x n + 1) + x n for n ≥ 1 . 2 Prove algebraically that, if the sequence converges, then it converges to β . [3] relation xn+1 =
3
(iii)
If x1 = 2, write down the values of x2 and x3 . ln( x + 1) + x Sketch the graphs of y = and y = x on the same axes. 2 Illustrate on your diagram how the sequence x1 , x2 , x3 , will converge to β starting with x1 = 2. [3]
(a)
Solve the equation z 4 = −4 , expressing each of the roots in the form a + ib , where a and b are real. [4] Hence write down the solutions of the equation ( w − 2 i) 4 = −4 .
(b)
[2]
On a single, clearly labelled diagram, sketch the loci of w and z defined by z − 1 = z − 5 and w + 1 − i = 2 . [2] (i) (ii)
State the minimum value of z − w . Find the exact values of the modulus and argument of w for which arg(w + 1 − i) = −
π
[1]
. 4 Hence, express w in the form x + iy , where x and y are exact real values. [3]
1
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 2 4
The plane p1 has equation x + y – 2 z = 4. (i)
A plane p2 with equation 2x + ay + bz = −4 (where a and b are constants) is parallel to p1, find the values of a and b and also the exact distance between p1 and p2 . [3]
(ii)
Another plane p3 contains a point with coordinates (2, 4, 1) and a line with 1 2 equation r = 1 + s 3 , where s ∈ℜ. Show that the equation of p3 can be 1 − 3 expressed as 3x – y + z = 3. Find also a vector equation of the line l where p3 meets p1.
(iii)
[4]
The plane p4 has equation 5x + y + λ z = µ, where λ and µ are constants. (a)
If all three planes p1 , p3 and p4 intersect in the line l , find the values of λ and µ . [2]
(b)
Deduce the geometrical relationship of the three planes p1, p3 and p4 if λ = −3 and µ ≠ 11. [1] Section B: Statistics [60 marks]
5
A Residents’ Committee wants to propose improvements to the recreational facilities in Punggol. In order to find out the needs of the adults and children of both genders, a survey is to be carried out on a sample of 120 residents who are at least 5 years old. Given that of the 6525 male residents who are at least 5 years old, there are 1450 children. On the other hand, there are 7975 females who are at least 5 years old, out of which 2175 are children. Describe how the sample can be obtained using quota sampling. [1] Name and describe another method of sampling in which each group is represented proportionately. [3] State an advantage of quota sampling over the above method.
6
[1]
A recent research study indicates that 16% of the total population are aged 60 years or more and that 18% of the total population have a measurable hearing defect. Furthermore, 65% of those aged 60 years or more have a measurable hearing defect. Find the probability that a randomly chosen person from the population (i)
has a measurable hearing defect, given that he is less than 60 years old,
[3]
(ii)
is either aged 60 years or more, or has a measurable hearing defect, or both. [2]
State, with a reason, whether or not the events ‘a person is 60 years old or more’ and ‘a person has a measurable hearing defect’ are independent. [1] Find, correct to 3 decimal places, the probability that, if two persons are chosen at random from the population, at least one of them will be aged 60 years or more and at least one of them will have a measurable hearing defect. [4] 2
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 2 7
A manufacturer claims that the breaking strength of a climbing rope is normally distributed with mean 1702 N and standard deviation 105 N. A random sample of 10 climbing ropes is tested and the mean breaking strength of the sample is x N. A test is then carried out at the 5% significance level to determine whether the manufacturer’s claim is valid. Given that the null hypothesis is rejected in favour of the alternative hypothesis, find the range of possible values of x . [4] The manufacturer believes that adding a special chemical to the ropes will increase the mean breaking strength and change the standard deviation. A random sample of 10 such ropes is found to have a mean breaking strength of 1724 N and a standard deviation of 35 N. Carry out a significance test at the 5% level to decide whether this result provides sufficient evidence to confirm the manufacturer’s belief that the mean breaking strength has increased. [4]
8
An experiment carried out to see how the intensity of radiation from a particular radioactive source, I, varies with time, t. In the following table, the values of t may be considered to be exact, while the values of I are subject to experimental errors. I t
22.5 44.0
28.0 33.5
30.5 28.0
38.0 18.0
40.5 13.6
42.5 15.0
48.0 10.3
54.5 9.0
55.0 6.3
70.0 4.0
(i)
Sketch the scatter diagram for the given data.
(ii)
State, with a reason, which of the following models is more appropriate to fit the data points: A: B:
9
25.0 42.0
I = a +bt2, where a > 0 and b < 0; I = atb, where a > 0 and b < 0.
[2]
[2]
(iii)
For the appropriate model, find the product-moment correlation coefficient for the transformed data. Find an estimate for a, correct to 3 decimal places. [2]
(iv)
Hence, estimate the time when the value of intensity of radiation is 35.0 and comment on the reliability of the estimate found. [2]
After production, blocks of butter are wrapped by a machine. Each block is supposed to weigh 220 g but the blocks produced have masses which are normally distributed. A worker then packs every six dozen randomly chosen blocks of butter in a box of mass 175 g. It is known that the total mass of each box of butter follows a normal distribution with mean 16.375 kg and standard deviation 0.149 kg. (i)
Three such boxes of butter are chosen at random. Find the probability that each box of butter weighs more than its mean mass. [1]
(ii)
Find the percentage of blocks of butter which weigh at least 220 g.
(iii)
To increase the consumers’ confidence, the company wishes to adjust the mean mass of a block of butter such that more than 75 % of the blocks will have a mass of not less than 220 g while the standard deviation remains unchanged. Find the least value of the mean mass after the adjustment.
3
[3]
[3]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 2 10
(a)
(b)
11
Find the number of 8-letter code-words that can be formed using the letters A, B, C, D, E if (i)
there are no restrictions,
[1]
(ii)
each vowel (A, E) appears once and each consonant (B, C, D) appears twice, [1]
(iii)
each letter occurs at least once and the letters appear in alphabetical order. [3]
At a particular reception, 9 guests are to stand at 3 identical round cocktail tables. How many ways can this be done if there must be at least two people at each table? [3]
A multiple choice test consists of 20 questions, each with four possible answers, of which only one is correct. If a student randomly chooses the answer to each question, find the probability of getting at least four but less than nine correct answers. [2] Suppose that each correct answer is awarded five marks and each incorrect answer carries a penalty of one mark, what is the expected score obtained by a student? [2] 50 students took the test. Using a suitable approximation, find the probability that the mean score is more than 12 marks. [3]
12
At a post office, the number of customers purchasing postal products in a half-hour period during peak hours follows a Poisson distribution with mean 5. During the offpeak period, the number of customers purchasing postal products in a two-hour period is an independent Poisson distribution with mean 10. The peak period for the post office is from 12 pm to 1.30 pm and the post office is open from 8 am to 5 pm. (i)
Find the probability that there are more than 15 customers from 12.30 pm to 1.30 pm. [2]
(ii)
Using suitable approximations, find the probability that the total number of customers during off-peak hours is less than thrice the total number of customers during peak hours on a particular day. [5]
~ End of Paper ~
4
