Modul Perfect Score Sbp 2009

BAHAGIAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

ADDITIONAL MATHEMATICS

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PAGE 1

ABOUT THIS MODULE

I

2

WE LEARN

II

3

EXAMINATION FORMAT

4

ANALYSIS TABLE OF SPM ADD MATHS QUESTIONS

5

LIST OF FORMULAE AND NORMAL TABLE

6

ADDITIONAL MATHEMATICS NOTES

7

PROBLEM SOLVING STRATEGY

XVII

8

PARTITION

XVIII

II-III IV V-VII VIII-XVI

This module is… 1. … specially planned for students who will be sitting for SPM. 2. … to provide exposure and to familiarize students with the needs of the actual SPM exam questions. 3. … to prepare students with adequate knowledge prior to the examination. 4. … comprises challenging questions which incorporate a variety of questioning techniques and levels of difficulty and conforms to the current SPM farmat.

“That which we persist in doing becomes easier – not that the nature of the task has changed, but our ability to do has increased.”

I


Key towards achieving 1A … Read question carefully Follow instructions Start with your favourite question Show your working clearly Choose the correct formula to be used +(Gunakannya dengan betul !!!) Final answer must be in the simplest form The end answer should be correct to 4 S.F. (or follow the instruction given in the question) π ≅ 3.142

Kunci Mencapai kecemerlangan

Proper / Correct ways of writing mathematical notations Check answers! Proper allocation of time (for each question)

Paper 1 : 3 - 7 minutes for each question Paper 2 : Sec. A : 8 - 10 minutes for each question Sec. B : 15 minutes for each question Sec. C : 15 minutes for each question

III www.cikgurohaiza.com

ANALYSIS TABLE OF SPM ADDITIONAL MATHEMATICS QUESTIONS 2004-2008 AMaths (3472) SPM Chapter 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 8 9

10

Functions Quadratic Equations Quadratic Functions Simultaneous Equations Indices and Logarithms Coordinate Geometry Statistics Circular Measure Differentiation Solution of Triangle Index Number Progressions Linear Law Integration Vectors Trigonometric Functions Permutations / Combinations Probability Probability Distributions Motion Along A Straight Line Linear Programming Total

Paper 2

Paper 1 04 3

05 3

06 2

07 3

08 3

1

2

1

1

1

2

1

1

2

2

04

3

3

2

2

2

1

1

2

2

1

1

1

1

1

1

1

1

1

2

2

3

2

2

4 1 1 2

3 1 1 2

2 1 2 2

3 1 1 2

3 1 1 2

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

25

25

25

25

25

Section A 06 07 1

08

04

05

Section B 06 07

08

Section C 06 07

04

05

08

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

4

4

4

4

4

1 1

2

05

1

1

1

1

1 1

1/2

1

1

1/2

1

½

½ 1

1

1

1

6

6

1

1

1

6

6

IV

1

1

1

1/2

1/3

1

1

1

1/3

1/3

1

1/3 1

1

1

2/3

1

1

1

1

1 1/2 1

1 2/3

1 1

1 2/3 1

1 2/3

1

1

1

1

1

1

6

5

5

5

5

5

SULIT

3472/2

The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. ALGEBRA

− b ± b − 4ac 2a

log c b log c a

2

1

x=

2 am × a n = a m + n am ÷ an = a m - n

3

4 (am) n = a nm 5

loga mn = log am + loga n

6

loga

7

log a mn = n log a m

8

logab =

9

Tn = a + (n-1)d

10

Sn =

11

Tn = ar n-1

n [2a + (n − 1)d ] 2

a(r n − 1) a(1 − r n ) , (r ≠ 1) = r −1 1− r a , r <1 13 S ∞ = 1− r Sn =

12

m = log am - loga n n

CALCULUS

dy dv du =u +v dx dx dx du dv v −u u dy = dx 2 dx , y= , v v dx

1 y = uv ,

2

4 Area under a curve b

∫y

=

b

=

dy dy du = × dx du dx

3

dx or

a

∫ x dy a

5 Volume generated b

=

∫ πy

2

dx or

2

dy

a

b

=

∫ πx a

GEOM ETRY

1 Distance =

( x1 − x 2 ) 2 + ( y1 − y 2 ) 2

2 Midpoint

y + y2 ⎞ ⎛ x1 + x 2 , 1 ⎟ 2 ⎠ ⎝ 2

(x , y) = ⎜

3

r = x2 + y2

4

r=

∧

xi + yj

5 A point dividing a segment of a line ⎛ nx + mx 2 ny1 + my 2 ⎞ , ( x,y) = ⎜ 1 ⎟ m+n ⎠ ⎝ m+n 6. Area of triangle = 1 ( x1 y 2 + x 2 y 3 + x3 y11 ) − ( x 2 y1 + x3 y 2 + x1 y 3 ) 2

x2 + y2

V 3472/2


[ Lihat sebelah SULIT

STATISTICS

1

x =

2

x =

∑x N

7

∑ fx ∑f

8

∑ (x − x )

3 σ =

=

N

σ=

∑ f ( x − x) ∑f

5 M =

⎤ ⎡1 ⎢2N −F⎥ L+⎢ ⎥C ⎢ fm ⎥ ⎦⎥ ⎣⎢

4

∑x

2

N

2

=

9 2

_2

−x

∑ fx ∑f

2

−x

2

10

P(A ∪ B)=P(A)+P(B)-P(A ∩ B)

11

p (X=r) = nCr p r q n − r , p + q = 1

12

Mean , μ = np

13

σ = npq x−μ z= σ

14 6

∑ w1 I1 ∑ w1 n! n Pr = (n − r )! n! n Cr = (n − r )!r!

I=

P I = 1 × 100 P0

TRIGONOMETRY

1 Arc length, s = r θ 2 Area of sector , A =

9 sin (A ± B) = sinAcosB ± cosAsinB 10 cos (A ± B) = cos AcosB m sinAsinB

1 2 rθ 2

3 sin 2A + cos 2A = 1

11 tan (A ± B) =

4 sec2A = 1 + tan2A 12

5 cosec2 A = 1 + cot2 A

tan A ± tan B 1 m tan A tan B

a b c = = sin A sin B sin C

6 sin2A = 2 sinAcosA 2

13 a2 = b2 +c2 - 2bc cosA

2

7 cos 2A = cos A – sin A = 2 cos2A-1 = 1- 2 sin2A 8 tan2A =

14 Area of triangle =

1 ab sin C 2

2 tan A 1 − tan 2 A

VI

3472/2



VII [ Lihat sebelah 3472/2

SULIT

TO EXCEL in

You need to… • set a TARGET • familiar with FORMAT of • • master the

EXAM PAPERS analyse the EXAM QUESTIONS

TECHNIQUES OF ANSWERING QUESTIONS • do

EXERCISES

VIII


ADDITIONAL MATHEMATICS NOTES 1

(c) Absolute Value Function

FUNCTIONS (a)

i. ii iii. iv. v.

y

a b

1

c

3 4

5 4

2

3 2 1

Domain = {a,b,c} Codomain = {1,2,3,4} Range = {1,2,3} Objects of 1 are a and b Images of b are 1,2 and 3.

- 4 - 3 -2 - 1

x

The corresponding range of values of f(x) is 0 ≤ f(x) ≤ 5 The corresponding range of values of f(x) means the range from the smallest value of y to the largest value of y, based on the given domain.

(b) Types of Relations i. One-to-one a b c

0 1 2

1 2 3

(d) Composite Functions g

f x

ii. Many-to-one a b c

1 2

g[f(x)] = gf(x)

gf fg(x) = f[g(x)] In general, gf (x) ≠ fg(x)

iii. One-to-many a

f 2 = ff, f 3 = fff or ff 2

1 2 3

b

(e) Determining one of the functions in a given composite function i. Given f and fg , find g. - Substitute g into f(x) ii. Given f and gf , find g. - Let y= f(x)

iv. Many-to-many a b c

f(x)

1 2 3

(f) To find the Inverse Function : - Let y = f(x), then x = f -1(y).

IX


y

2. QUADRATIC EQUATIONS (a) ax2 + bx + c = 0 − b ± b 2 − 4ac 2a Sum of roots: b α+β= − a Product of roots c αβ = a

x =

(b) Form quadratic equation from 2 given roots: x2 - (sum of two roots)x + product of two roots = 0 3. QUADRATIC FUNCTIONS (a) Types of roots b2 - 4ac > 0 → 2 different (distinct) roots. b2- 4ac = 0 → 2 equal roots b2 - 4ac < 0 →no real roots. b2 - 4ac ≥ 0 → with real roots

0

y

x

b2 - 4ac > 0

b

_

_

+

_

+

+

+

y

a

0

a

_

b

+

4. INDICES & LOGARITHM (a)

x = an Index Form

loga x = n Logarithmic Form

(b) Laws of Indices

y

1. a n × a m = a n+m 2. a n ÷ a m = a n−m

0

x

b2 - 4ac = 0

0

x

3. (a n ) m = a nm

x

b2 - 4ac < 0

Laws of Logarithm 1. logaxy = logax + logay x 2. loga = logax – logay y 3. loga xn = n logax 4. loga a = 1 5. loga 1 = 0 log c b 6. loga b = log c a 1 7. loga b = log b a

(b) Completing the Squares y = a(x - p)2 + q a +ve → minimum point (p, q) a –ve → maximum point (p, q) (c) Quadratic Inequalities (x – a)(x – b) ≥ 0 Range: x ≤ a, x ≥ b (x – a)(x – b) ≤ 0 Range: a ≤ x ≤ b

X


5. COORDINATE GEOMETRY (a) Distance between A(x1, y1) and B(x2, y2)

6. STATISTICS Measure of Central Tendency (a) Mean ∑x x= n for ungrouped data

AB = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 (b) Midpoint of AB ⎛ x + x 2 y1 + y 2 ⎞ M= ⎜ 1 , ⎟ 2 ⎠ ⎝ 2 (c) P divides AB internally in the ratio m : n m : A(x 1 , y1 )

x=

for ungrouped data with frequency.

n

P

B(x2 , y2 )

x=

⎛ nx + mx 2 ny1 + my 2 ⎞ P= ⎜ 1 , ⎟ n+m ⎠ ⎝ n+m (d) Gradient of AB y − y1 m= 2 x 2 − x1 m= −

∑ fx ∑f ∑ fx ∑f

i

for grouped data , xi = midpoint of each class interval (b) Median The centre value of a set of data after the data is arranged in the ascending or descending order.

y-intercept x-intercept

(e) Equation of a straight line

Formula

(i) given m and A(x1, y1) y – y1 = m(x – x1)

n−F ×C fm L = Lower boundary of the Median class n = Total frequency F = Cumulative frequency before the median class fm = Frequency of the median class C = Size of the class interval

M=L+

(ii) given A(x1, y1) and B(x2, y2) y − y1 y 2 − y1 = x − x1 x 2 − x1 (f) Area of polygon x 1 x1 x 2 x3 L= ......... 1 y1 2 y1 y 2 y 3

1 2

From the Ogive (g) Parallel lines m 1 = m2

Cumulative Kekerapan Frequency longgokan n

(h) Perpendicular lines m1 × m2 = -1

n __ 2

0

XI


Median Sempadan atas

Median

Upper Boundary

(c) Mode Data with the highest frequency

For ungrouped data

σ = =

From the Histogram :

σ =

=

n

∑x

2

−x

2

∑ f ( x − x) ∑f

∑ fx ∑f

2

−x

2

2

Mod Sempadan kelas Mode Class Boundary

7. INDEX NUMBERS

Measure of Dispersion (a) Interquartile Range Formulae : 1 n − F1 Q1 = L1 + 4 ×C f Q1 Q3 = L3 +

3 4

(a) Price Index P I = 1 × 100 P0 where Po = price at the base time P1 = price at a specific time

n − F3 ×C f Q3

Ogive:

(b) Composite Index ∑ Iw I= ∑w where I = price index or index number w = weightage

Cumulative Frequency Kekerapan longgokan

3 __ n 4

8. CIRCULAR MEASURE (a) Radian → Degree 180 0 θr=θ×

1 __ n 4

0

2

n For grouped data

KFrequency ekerapan

0

∑ ( x − x)

Q

1

Q 3 Sempadan atas Upper Boundary

π

Interquartile Range = Q3 – Q1

(b) Degree → Radian

θo = θ ×

π

rad 180 (c) Arc length s = jθ (d) Area of sector 1 1 L = j2θ = js 2 2

(b) Variance, Standard Deviation Variance = (Standard Deviation)2

XII


(e) Area of segment 1 L = j2(θ r – sin θo) 2

10. INTEGRATION ax n +1 (a) ∫ ax n dx = +c n +1

(ax + b) n+1 (b) ∫ (ax + b) dx = +c a (n + 1)

9. DIFFERENTIATION (a) Differentiation using the First Principal

n

b

(c)

dy δy = lim dx ∂x→0 δ x

∫ f ( x) + g ( x) dx a

b

=

d (a) = 0, a = constant dx d n (x ) = nxn-1 (c) dx d (d) (axn) = anxn-1 dx

(b)

∫ a

(d)

b

f ( x) dx + ∫ g ( x) dx a

b

c

a

b

c

∫ f ( x) dx + ∫ f ( x) dx = ∫ f ( x) dx a

b

(e)

b

∫ af ( x) dx = a ∫ f ( x) dx a

(e) Product Rule d dv du (uv) = u +v dx dx dx

(f)

a

b

a

a

b

∫ f ( x) dx = − ∫ f ( x) dx

(h) Area under the curve y

(f) Quotient Rule dv d ⎛ u ⎞ v du dx − u dx ⎜ ⎟= dx ⎝ v ⎠ v2

b

A=

∫ y dx a

(g) Composite Function dy du d (ax+b)n = × dx du dx = an(ax+b)n-1 dy =0 (h) Turning point → dx Maximum point: d2y dy = 0 and <0 dx dx 2

a

0 y

b

x

b b

A= a

Minimum point: d2y dy = 0 and >0 dx dx 2

∫ x dy a

0

x

(i) Volume of revolution y

(i) Rate of change dy dy dx = × dt dx dt

b

0

a

(j) Small change : dy δ y ≈ .δ x dx XIII


b

x

V = ∫ π y 2 dx a

y

(a) s = 0 → at the fixed point O (b) v = 0 → stops momentarily → maximum / minimum displacement (c) a = 0 → v constant → v maximum/ minimum

b b

V = ∫ π x 2 dy a

a 0

x

11. PROGRESSIONS Arithmetic Progressions (a) Tn = a + (n - 1)d n (b) Sn = {2a + (n - 1)d} 2 n = (a + l) 2 (c) d = T2 - T1

13. TRIGONOMETRIC FUNCTIONS (a) y y P(x, y) sin θ = r x cos θ = r r y y tan θ = θ x 0

Geometric Progressions (a) Tn = arn-1 a(1 − r n ) (b) Sn = for r < 1 1− r a(r n − 1) Sn = for r > 1 r −1 a (c) S ∞ = for -1 < r < 1 1− r and n ∞ T (d) r = 2 T1

x

x

sin θ cos θ 1 sec θ = cos θ

(b) tan θ =

cosec θ = cot θ =

1 sin θ

1 cos θ = tan θ sin θ

(c)

General (a) S1 = T1 = a (b) Tn = Sn – Sn-1 (c) Sum of terms from Ta to Tb = Sb – Sa-1

12. MOTION ALONG A STRAIGHT LINE ds dv dt dt s → v → a ← ← ∫ v dt ∫ a dt XIV


Sin +ve

All Semua +ve

Tan +ve

Cos Kos +ve

(d) Special Angles 0o 30o θ 0 1 Sin θ 2 1 Cos θ 3 Tan θ

o

45 1

0

60

3 2 1 2

2 1 2

2 1

(f) sin2θ + cos2θ = 1 1 + tan2θ = sec2θ 1 + cot2θ = cosec2θ

o

1

(g) sin(A ± B) = sinA cos B ± cos Asin B cos(A ± B) = cosA cosB m sinA sinB tan (A ± B) tan A ± tan B = 1 m tan A tan B

3

3

θ Sin θ Cos θ Tan θ

90o 1 0 ∞

180o 270o 0 -1 -1 0 0 ∞

360o 0 1 0

(h) sin2A = 2 sinA cosA cos2A = cos2A – sin2A = 2 cos2A – 1 = 1 – 2 sin2A 2 tan A tan 2A = 1 − tan 2 A

(e) Trigonometric Graphs y = a sin bx y

a

0

__ 90 b

180 __ b

270 __ b

__ 360 b

14. VECTORS (a) Addition of Vectors 1. Triangle Law

x

-a b a +

y = a cos bx y

b

a a

2. Parallelogram Law 0

__ 90 b

180 __ b

270 __ b

__ 360 b

x b a +

b

-a

y = a tan bx y

a

3. Polygon Law 0

__ 90 b

180 __ b

270 __ b

__ 360 b

B

x

C

A

E

uuur uuuv uuuv uuuv D uuuv AE = AB + BC + CD + DE XV


(b) Subtraction of vectors a

b

(b) Combination n Cn = 1 n! n Cr = (n − r ) !r !

-b

C r = nC n − r (c) Binomial Distribution P(X = r) = n C r pr qn-r (d) Mean = μ = np n

a

(c) Vectors in the Cartesan Coordinates

Standard deviation σ = npq

y

(e) Converting Normal Distribution to Standard Normal X −μ Distribution Z =

P(x, y)

σ

r

(f) Probability Distribution Graph

yj

0

xi

x

r = xi + yj

r = x2 + y2 xi + y j r rˆ = = r x2 + y 2

0

a

Z

P(Z > a)

1. P(Z < a) = 1 – P(Z > a) Æuse P 2. P(Z < -a) = P(Z > a) Æ use P 3. P(Z > -a) = 1 – P(Z > a)Æ use R

15. SOLUTIONS OF TRIANGLE (a) Sine Rule

4. 5. 6.

a b c = = sin A sin B sin C (b) Cosine Rule a2 = b2 + c2 – 2bc cos A

P(a < Z < b) = P(Z>a) – P(Z > b) P(-a < Z< b) = 1– P(Z >a) – P(Z > b) P(-a < Z < -b) = P(b < Z < a)

Examples: a) P(Z> 0.1)

b2 + c2 − a2 2bc (c) Area of Triangle 1 L = ab sin C 2

b) P ( Z< 0.1)

cos A =

c) P ( -1.2 < Z < 0.4) Examples: d) P( Z > a ) = 0.3, find a e) P (Z > a) = 0.6, find a

16. PROBABILITY DISTRIBUTIONS (a) Permutation n Pn = n! n! n Pr = (n − r )!

f) P (Z < a) = 0.1, find a g) P ( Z < a ) = 0.73, find a h) P(X > a) = 0.3, given μ = 45, σ =3

XVI


How to Solve a Problem

Understand

Plan your

Do - Carry out

Check your

the Problem

Strategy

Your Strategy

Answers

•Which Topic / •Subtopic ?

•Choose suitable

•Carry out the

•Is the answer

strategy

calculations

reasonable?

1.

PN ZABIDAH BINTI AWANG SM AGAMA PERSEKUTUAN, LABU.

2.

EN AMIRULFAIZAN BIN AHMAD SBP INTEGRASI SELANDAR, MELAKA.

3.

PN ROHANI MD NOR SEKOLAH SULTAN ALAM SHAH, PUTRAJAYA

4.

EN ZUZI BIN SHAFIE SM AGAMA PERSEKUTUAN, KAJANG.

5.

PN SARIPAH BINTI AHMAD SM SAINS MUZAFFAR SYAH, MELAKA.

XVII


Master these questions ……

XVIII


For examiner’s use only

Answer all questions. f

g

1. 3

Set A

-1

6

Set B

Set C

Diagram 1

In Diagram 1, the function f maps set A to set B and the function g maps set B to set C. Determine (a) f (3 ) (b) g(-1) (c) gf (3) [ 3 marks ] Answer : (a) …………………….. (b) ……………………...

1 3

(c).................................... 2. Given function f : x → 3 − 4x and function g : x → x2 − 1, find (a) f −1, (b) the value of f −1g(3). [ 3 marks ]

2 Answer : (a) …………………….. (b) ……………………...


3


3

Given the function f (x) = 4x, x ≠ 0 and the composite function f g(x) =

− 16 . Find x

(a) g(x), (b) the value of x when g(x) = 8.

[3 marks]

3

Answer : .........………………… 3

4

Solve the quadratic equation 2 x ( x − 5) = (2 − x )( x + 3) . Give your answer correct to four significant figures. [ 3 marks ]

Answer : .........…………………

4 3


5


4− y , find the range of x if y > 10. 2 (b) Find the range of x if x2 − 2x ≤ 3.

(a) Given x =

[4 marks]

Answer : ................................. ___________________________________________________________________________ 6

Diagram below shows the graph of a quadratic function y = f (x) . The straight line y = −9 is a tangent to the curve y = f (x) .

y y = f (x)

0

1

x 7 y = -9 Diagram 1

a) Write the equation of the axis of symmetry of the curve. b) Express f (x) in form of ( x + p ) 2 + q , where p and q are constants. [ 3 marks ]

6

Answer : (a) ……........................ (b) ……........................

3472/1


3

For examiner’s use only 7

Solve the equation 324x = 48x + 6 [3 marks]

7 3

Answer : .................................. 8.

Given log5 3 = 0.683 and log5 7 = 1.209. Calculate (i) log5 1.4, (ii) log7 75.

[ 4 marks]

8 4

Answer : ...................................

9.

Solve the equation log

x

16 − log x 2 = 3.

[3 marks]

9 3

Answer : ......................................


10.

The first terms of the series are 2, x , 8. Find the value of x such that the series is a (a) an arithmetic progression, (b) a geometric progression. [2 marks ]


10 2

Answer : ....……………...………..

11. The sum of the first n terms of an arithmetic progression is given by S n = 3n 2 + 13n. Find

(a) the ninth term, (b) the sum of the next 20 terms after the 9th terms. [3 marks]

11

Answer: a)…...…………..…....... b) ....................................

3472/1


4


12.

Given that

1 = 0.166666666..... p = 0.1 + a + b + ............

[ 3 marks ]

Find the values of a and b. Hence, find the value of p.

12

Answer: a =...….… 4

b =….......

p = ........................ ___________________________________________________________________________ 13.

Diagram 2 shows a linear graph of

y against x2 x

y x

●

(4,1)

x2 ●

Given that

(1,-5)

DIAGRAM 2

y = hx2 + k, where k and h are contants. x

Calculate the value of h and k.

[3 marks]

Answer :

h = ………………..…….

13

k = ……………….....…...

3



x y + = 1. Find the equation of a straight line 3 2 that is parallel to PQ and passes through the point (−6 , 3). [3 marks]

The equation of a straight line PQ is

14.

14 3 Answer : .…………………

15

⎛7⎞ Given u = ⎜⎜ ⎟⎟ dan v = ⎝9⎠ case:

⎛ p − 1⎞ ⎜⎜ ⎟⎟ , find the possible values of p for each of the following ⎝ 3 ⎠

(a) u and v are parallel, (b) u = v .

[2 marks] [2 marks]

15 4 Answer : a)………………….. b) ………………………

3472/1



P

16

R

S

Q r O →

→

s

→

→

The diagram above shows OR = r, OS = s, OP and PQ are drawn in the square grid. Express in terms of r and s. (i) (ii)

→

OP uuur PQ .

[ 3 marks ]

uuur Answer: a) OP = …….…………...

16 3

uuur b) PQ =...……………….. ___________________________________________________________________________ 17.

Solve the equation 3 cos2 θ + sin 2θ = 0 for 0 0 ≤ θ ≤ 360 0 .

[ 4 marks ]

17 4

Answer: …...…………..…....... www.cikgurohaiza.com


P

18.

θ Q

O

Diagram above shows a length of wire in the form of sector OPQ, centre O. The length of the wire is 100 cm. Given the arc length PQ is 20 cm, find (a) the angle θ in radian, (b) area of the sector OPQ.

[2 marks] [2 marks]

18

Answer: a)…………………… b) …………………

4

___________________________________________________________________________ 19.

Find the equation of the tangent to the curve y =

5 at the point (3, 4). ( x − 5) 3 [2 marks] 19 2

Answer:………………………

3472/1


For examiner’s use only 20.

A roll of wire of length 60 cm is bent into the shape of a circle. When above the wire is heated, its length increases at a rate of 0.1 cms−1. (Use π = 3.142) (i) Calculate the rate of change of radius of the circle. (ii) Hence, calculate the radius of the circle after 4 seconds.

[2 marks] [2 marks]

20 5

Answer: …...…………..…....... ___________________________________________________________________________ 21.

Given

∫

4 0

f ( x) dx = 5 and

∫

3 1

g ( x) dx = 6.

Find the value (a)

∫

4 0

1

2 f ( x) dx + ∫ g ( x)dx ,

(b) k if

[1 marks]

3

∫

3 1

[ g ( x) − k x] dx =14.

[2 marks]

Answer: a) ……………………..

21

k =.……………..………

3

22. A chess club has 10 members of whom 6 are men and 4 are women. A team of 4 members is selected to play in a match. Find the number of different ways of selecting the team if (a) all the players are to be of the same gender,

22

(b) there must be an equal number of men and women. [3 marks]

3

Answer: p = ……………………. . 3472/1 www.cikgurohaiza.com

11

3472/1


23. (a) Given that the mean for four positive integer is 9. When a number y is added to the four positive integer, the mean becomes 10. Find the value of y. [2 marks] (b) Find the standard deviation for the set of numbers 5, 6, 6, 4, 7. [3 marks]

23

Answer: …a)...…………..….......

5

b) ............................... ___________________________________________________________________________ 24. Hanif , Zaki and Fauzi will be taking a driving test. The probabilities that Hanif , 1 1 1 respectively. Calculate the , and Zaki and Fauzi will pass the test are 2 3 4 probability that (a) only Hanif will pass the test (b) at least one of them will pass the test. [ 3 marks ]

24 3

Answer: ……………………………

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25.

Diagram below shows a standard normal distribution graph.

-k

25 2

k

z

Given that the area of shaded region in the diagram is 0.7828 , calculate the value of k. [ 2 marks ]

Answer: …...…………..….......

END OF QUESTION PAPER

3472/1 www.cikgurohaiza.com

13

3472/1

JAWAPAN

1 2

(a) −1 (a) f

−1

=

(b) 6

(c) 6

3− x 4

(b) −

(a) g(x) =

3

5 4

−4 ,x≠0 x

13

h = 2 , k = −7

14

3y = − 2x − 3

15

1 (b) x = − 2

4

3.562 , -0.5616

5.

(a) x < − 3

6

a) x = 4

(b) −1 ≤ x ≤ 3

(a)

10 3

(b) −10, 12

(ii) − r − 3s

16

(b)(i) 3r + 2s

17

90°, 123° 41’, 270°, 303° 41’

(a)

1 2

(b) 400

b) f ( x) = ( x − 4) − 9

18

7

x=3

19

15x + 16y −109 = 0

8

( i) 0.209 (ii) 2.219

20

( i) 0.01591 cms−1

9

x = 4

21

(a) 4

10

a) 5

22

14 553

11

(a) 64

23

(a ) 14 (b) 1.020

12

a = 0.06 , b = 0.006 , p = 6

24

(a) 9/35

25

k = 1.234

2

3472/1

b) 4 ( b) 2540

(ii) 9.612

(b) k = −2


( b) 5/6


4 THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1) 1

2

3

4

0.4641

4

8

12

16

20

0.4286

0.4247

4

8

12

16

0.3897

0.3859

4

8

12

15

0.3557

0.3520

0.3483

4

7

11

0.3192

0.3156

0.3121

4

7

11

0.2877

0.2843

0.2810

0.2776

3

7

0.2546

0.2514

0.2483

0.2451

3

7

z

0

1

2

3

4

5

6

7

8

9

0.0

0.5000

0.4960

0.4920

0.4880

0.4840

0.4801

0.4761

0.4721

0.4681

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.2

0.4207

0.4168

0.4129

0.4090

0.4052

0.4013

0.3974

0.3936

0.3

0.3821

0.3783

0.3745

0.3707

0.3669

0.3632

0.3594

0.4

0.3446

0.3409

0.3372

0.3336

0.3300

0.3264

0.3228

0.5

0.3085

0.3050

0.3015

0.2981

0.2946

0.2912

0.6

0.2743

0.2709

0.2676

0.2643

0.2611

0.2578

5

6

7

8

9

24

28

32

36

20

24

28

32

36

19

23

27

31

35

15

19

22

26

30

34

15

18

22

25

29

32

10

14

17

20

24

27

31

10

13

16

19

23

26

29

24

27

Minus / Tolak

0.7

0.2420

0.2389

0.2358

0.2327

0.2296

0.2266

0.2236

0.2206

0.2177

0.2148

3

6

9

12

15

18

21

0.8

0.2119

0.2090

0.2061

0.2033

0.2005

0.1977

0.1949

0.1922

0.1894

0.1867

3

5

8

11

14

16

19

22

25

0.9

0.1841

0.1814

0.1788

0.1762

0.1736

0.1711

0.1685

0.1660

0.1635

0.1611

3

5

8

10

13

15

18

20

23

1.0

0.1587

0.1562

0.1539

0.1515

0.1492

0.1469

0.1446

0.1423

0.1401

0.1379

2

5

7

9

12

14

16

19

21

1.1

0.1357

0.1335

0.1314

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

0.1170

2

4

6

8

10

12

14

16

18

1.2

0.1151

0.1131

0.1112

0.1093

0.1075

0.1056

0.1038

0.1020

0.1003

0.0985

2

4

6

7

9

11

13

15

17

1.3

0.0968

0.0951

0.0934

0.0918

0.0901

0.0885

0.0869

0.0853

0.0838

0.0823

2

3

5

6

8

10

11

13

14

1.4

0.0808

0.0793

0.0778

0.0764

0.0749

0.0735

0.0721

0.0708

0.0694

0.0681

1

3

4

6

7

8

10

11

13

1.5

0.0668

0.0655

0.0643

0.0630

0.0618

0.0606

0.0594

0.0582

0.0571

0.0559

1

2

4

5

6

7

8

10

11

1.6

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0485

0..0475

0.0465

0.0455

1

2

3

4

5

6

7

8

9

1.7

0.0446

0.0436

0.0427

0.0418

0.0409

0.0401

0.0392

0.0384

0.0375

0.0367

1

2

3

4

4

5

6

7

8

1.8

0.0359

0.0351

0.0344

0.0336

0.0329

0.0322

0.0314

0.0307

0.0301

0.0294

1

1

2

3

4

4

5

6

6

1.9

0.0287

0.0281

0.0274

0.0268

0.0262

0.0256

0.0250

0.0244

0.0239

0.0233

1

1

2

2

3

4

4

5

5

2.0

0.0228

0.0222

0.0217

0.0212

0.0207

0.0202

0.0197

0.0192

0.0188

0.0183

0

1

1

2

2

3

3

4

4

2.1

0.0179

0.0174

0.0170

0.0166

0.0162

0.0158

0.0154

0.0150

0.0146

0.0143

0

1

1

2

2

2

3

3

4

2.2

0.0139

0.0136

0.0132

0.0129

0.0125

0.0122

0.0119

0.0116

0.0113

0.0110

0

1

1

1

2

2

2

3

3

2.3

0.0107

0.0104

0.0102

0

1

1

1

1

2

2

2

2

0.00990

0.00964

0.00939

0.00914

3

5

8

10

13

15

18

20

23

2

5

7

9

12

14

16

16

21

2

4

6

8

11

13

15

17

19

0.00889

0.00866

0.00842

2.4

0.00820

0.00798

0.00776

0.00755

0.00734 0.00714

0.00695

0.00676

0.00657

0.00639

2

4

6

7

9

11

13

15

17

2.5

0.00621

0.00604

0.00587

0.00570

0.00554

0.00539

0.00523

0.00508

0.00494

0.00480

2

3

5

6

8

9

11

12

14

2.6

0.00466

0.00453

0.00440

0.00427

0.00415

0.00402

0.00391

0.00379

0.00368

0.00357

1

2

3

5

6

7

9

9

10

2.7

0.00347

0.00336

0.00326

0.00317

0.00307

0.00298

0.00289

0.00280

0.00272

0.00264

1

2

3

4

5

6

7

8

9

2.8

0.00256

0.00248

0.00240

0.00233

0.00226

0.00219

0.00212

0.00205

0.00199

0.00193

1

1

2

3

4

4

5

6

6

2.9

0.00187

0.00181

0.00175

0.00169

0.00164

0.00159

0.00154

0.00149

0.00144

0.00139

0

1

1

2

2

3

3

4

4

3.0

0.00135

0.00131

0.00126

0.00122

0.00118

0.00114

0.00111

0.00107

0.00104

0.00100

0

1

1

2

2

2

3

3

4

f

Example / Contoh:

⎛ 1 ⎞ exp⎜ − z 2 ⎟ 2π ⎝ 2 ⎠ 1

f ( z) =

If X ~ N(0, 1), then Jika X ~ N(0, 1), maka

∞

Q(z)

Q ( z ) = ∫ f ( z ) dz

P(X > k) = Q(k)

k

P(X > 2.1) = Q(2.1) = 0.0179

O

k


z

SECTION A

[40 marks] [40 markah] Answer all questions in this section .

1. Solve the equations x2 − y + y2 = 2x + 2y = 10. [5 marks] [ Answer x = 2, y = 3; x =

5 5 ,y= ] 2 2

2 Given kx2 − x is the gradient function for a curve such that k is a constant. y − 5x + 7 = 0 is the equation of tangent at the point (1, −2) to the curve. Find, (a) the value of k, (b) the equation of the curve.

[2 marks] [3 marks] [ Answer k = 6 ] x2 7 [ y = 2x3 − − ] 2 2

3

Diagram 3 Diagram 3 shows a string of length 125π cm is cut and made into ten circle as shown above . The diameter of each subsequent circles are difrent by 1 cm from its previous. Calculate, (i) the diameter of the smallest circle , (ii) the number or a circle if the length of a circle is 400 [6 marks] Answer : (b)(i) 8 (ii) 21 4 Table 2 shows the frequency distribution of the marks of a group of form 4 students in a test. Mark

Number of students

20 – 29

2

30 – 39

10

40 – 49

36

50 – 59

55

60 – 69

k

70 - 79

5


(a) (b)

(c)

It is given that the first quartile score it 44.5. Find the value of k. [ 3 marks ] [ Use the graph paper to answer this question] Using the scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 10 students on the vertical axis, draw a histogram based on the given data. Hence, estimate the mode mark [ 3marks ] Calculate the mean marks. [2 marks ] [ Answer k = 12, mode = 52.25

5

a)

Prove that

mean = 55.81

1 - sec θ = - sin θ tan θ secθ [3 marks]

(b)

Sketch y = 1 − sin 2 x for 0 ≤ x ≤ π . Hence using the same axes , draw a suitable straight line to find the number of solutions of the equation π sin 2 x − x = 0 . State the number of solutions [ line y =

−x

π

+ 1] , 4 number of solution ] [5 marks]

6

5x A

B

4y

P

D

C x → → → In the diagram above, AB = 5x, AD = 4y and DC = x. (i) Express, → (a) AC → (b) BD in terms of x and y. → → → → (ii) Given AP = h AC and BP = k BD . → State AP (a) in terms of h, x and y, (b) in terms of k, x and y. Hence, or otherwise, prove that h = k.

Answer (i)(a) 4y + x, (b) −5x + 4y 3472/2

[2 marks]

[5 marks]

(ii)(a) h(4y + x) (b) ( (5 − 5k)x + 4ky; k =


SULIT

5 6

SECTION B [ 40 Marks ] Answer four equations from this section.

7 Table 7 shows the values of two variables x and y ,obtained from an experiment. Variable x and y are related by the equations y = ab –x , where a and b are constants. One of the value of y is wrongly recorded. 1 2 3 4 5 x y 41.7 38.7 28.9 27.5 20.1 (a) Plot log 10 y against x. (b) By using your graph find, (i) the value of y which is wrongly recorded and determine the correct value (ii) the value of a and the value of b (iii) the value of y when x = 2.5 . 8

y

y=a

y = x2 + 1

Q

1 − 3

1 3

O

x

(a) Refer to the diagram above, answer the following question: (i) Calculate the area of the shaded region. (ii) Q is a solid obtained when the region bounded by the curve y = x2 + 1 and the line y = a is 1 revolved through 180° at the y - axis. If the volume of Q is π unit2 Find the value a. 2 [6 marks] (b) Find the equation of tangent to the curve y = 2x2 + r at point x = k. If the tangent passes through the point (2, 0), find r in terms of k. [4 marks ] [Answer 16. (a)(i)

56 (ii) a = 2 81

(b) y − (2k2 + r) = 4k(x − k); r = 2k2 − 8k ]


Solutions to this question by scale drawing will not be accepted. 9. y U(5, 6) T(0, 4)

x O

V(p, q)

W Diagram above shows the vertices of a rectangle TUVW in a Cartesian plane. (a) Find the equation which relates p and q by using the gradient of UV.

[3maks] 5 (b) Shows that the area of the Δ TUV can be expressed as p − q + 10. [2marks] 2 (c) Hence, calculate the coordinates of V given the area of the rectangle TUVW is 5 unit2. [3marks] (d) Find the equation of the straight line TW in the intercept form. [2marks] 10

Diagram above shows a sector MJKL of a circle centre M and two sectors, PJM and QML, with centre P and Q respectively. Given the angle of the major sector JML is 3.6 radian. Find, (a) the radius of the sector MJKL, [2 marks] (b) perimeter of the shaded region, [2 marks] (c) the area of sector PJM, [2 marks] (d) the area of the shaded region. [4 marks] [ Answer 11. (a) 4.795

3472/2


SULIT

(b) 27.24

(c)

25 ] 2

11 (a) In a centre of chicken eggs incubation, 30% of the eggs hatched are male chickens. If 10 newly born chickens are chosen at random, find the probability (correct to four decimal places) that (i) 4 eggs hatched are male chicken, (ii) at least 9 eggs hatched are female chickens. [4 marks] (b) The mass of guava fruits produced in a farm shows a normal distribution with mean 420 g and standard deviation 12 g. Guava fruits with mass between 406 g and 441 g are sold in market, whereas those with mass 406 g or less are sent to the factory to be processed as drinks. Calculate, (i) the probability (correct to four decimal places) that a guava fruit chosen randomly from the farm will be sold in the market, (ii) the number of guava fruits that has been sent to the factory and also not sold in the market, if the farm produced 2 500 guava fruits. [ Answer (a)(i) 0.2001 (ii) 0.1493

[6 marks]

(b)(i) 0.8383 (ii) 100 ]

Sections C Answer two questions from this section.

12 . A• P

8m

•B Q

In the diagram above, P and Q are two fixed points on a straight line such that PQ = 8 m. At a certain instant, particle A passes the point P with a velocity VA = 2t − 6, whereas particle B passes the point Q with a velocity VB = 5 − t where t is time in second after the particle A and the particle B have passed the point P and the point Q. [Assume direction P to Q is the positive.] (a) Find the distance between the particle A and particle B at the instant when particle A stopped momentarily. (d)

[3marks ] Find the time, t1, when the distance between the particle A and particle B is maximum before the two particles meet.. [ 2 marks ]

(c) For how long the two particles A and B are moving in the same direction? (d)(i) Find the time, t2, when the particles A and B meets. (ii) Hence, find the distance from the point P when the two particles meet..

1 [Answer (a) 27 m 2

(b)

11 s 3

[3 marks ]

(c) 2 s (d)(i) 8 s (ii) 16 m ]

13 A small factory produces a certain goods of A model and B model. In a day, the factory produces x units of A model and y units of B model where x ≥ 0 and y ≥ 0. Time taken to produce one unit A model and one unit B model is 5 minutes and minutes respectively. The production of these goods in a certain day is www.cikgurohaiza.com

restricted by the following conditions: I. The number of units of A model is not more than 60, II. The number of units of B model is more than two times the number of units of A model by 10 units or less. III. The total time for the production of A model and B model is not more than 400 minutes. Write an inequality for each of the above condition.. Hence draw the graphs for the three inequalities. Marks and shades the region R which satisfy the above conditions. Use your graph to answer the following questions: (a) Find the range of the number of units of A model which can be produced if 40 units of B model are produced. (b) Find the total maximum profits which can be obtained if the profit gained from one unit of A model and one unit of B model is RM 6 and RM 3 respectively. (c) If the factory intends to produce the same number of units of A model and B model, find the maximum number of units which can be produced for each o A model and B model. Answer x ≤ 60, y − 2x ≤ 10, 5x + 4y ≤ 400 (a) 15 ≤ x ≤ 48 (b) RM435 14 . Diagram 6 shows a quadrilateral ABCD such that ∠ABC is acute. D 5.2 cm C

9.8 cm

A

40.5°

12.3 cm

9.5 cm

DIAGRAM 6 B (a) Calculate, (i) ∠ABC, (ii) ∠ADC, (iii) area, in cm2, of quadrilateral ABCD. [8 marks] (b) A triagle A’B’C’ has the same measurements as those given for triangle ABC, that is, A’C’ = 12.3 cm, C’B’ = 9.5 cm and ∠B’A’C’ = 40.5°, but which is different in shape to triangle ABC. (i) Sketch the triangle A’B’C’. (ii) State the size of ∠A’B’C’. [2 marks] Answer . (a)(i) 57.21° - 57.25° (ii) 106.07° - 106.08° (iii) 82.37° - 82.39° (b)(i) C (ii) 122.75° - 122.79° A

B

10. Table 2 shows the price indices and percentage usage of four items, P, Q, R, and S, which are the main ingredients of a type biscuits. 3472/2


SULIT

(c) 44

Item

Price index for the year 1995 based on the year 1993

Percentage of usage (%)

P Q R S

135 x 105 130

40 30 10 20

(a) Calculate, (i) the price of S in the year 1993 if its price in the year 1995 is RM37.70 (ii) the price index of P in the year 1995 based on the year 1991 if its price index in the year 1993 based in the year 1991 is 120. [5 marks] (b) The composite index number of the cost of biscuits production for the year 1995 based on the year 1993 is 128. Calculate, (i) the value of x, (ii) the price of a box of biscuit in the year1993 if the corresponding price in the year 1995 is RM 32. [5 marks] [ Answer (a)(i) RM 29

(ii) 162

Section C Alternative

Answer two questions from this section. 12. Diagram 6 shows ΔSTQ such that ST = 12.1 cm and TQ = 9.5 cm.

T

S

Q Diagram 6

The area of the triangle is 45 cm2 and ∠STQ is obtuse. (a)

Find (i) ∠STQ [ ∠STQ = 128.47° or 128°28' ] (ii) the length, in cm, of SQ [19.49 cm] (iii) the shortest distance, in cm, from T to SQ. [4.613] [ 5 marks]


(b)(i) 125

(ii) RM 25 ]

T

13 cm

5 cm

Q 5 cm

R P

Diagram 7

(b) Diagram 7 shows a pyramid TPQR with a horizontal triangular base PQR. T is vertically above Q. Given that PQ = QT = 5 cm, TR = 13 cm and ∠PRQ = 15° .Calculate two possible values of ∠PQR [∠PQR = 126.60o and 23.40o] (c) Using the acute ∠PQR in (i), calculate ( i) the length of PR [7.673] (ii) the value of ∠PTR [29.420] (iii) the surface area of the plane TPR [22.58] [ 5marks]

13. shows the bar chart for the monthly sales of five essential items sold at a sundry shop. Table 3 shows their price in the year 2000 and 2006, and the corresponding price index for the year 2006 taking 2000 as the base year. Cooking Oil Rice Salt Sugar Flour 10 20 30 40 50 60 70 80 90 100

units

Diagram 2

Items Cooking Oil Rice Salt Sugar Flour

3472/2

Price in the Price in the year 2000 year 2006 x RM1.60 RM0.40 RM0.80 RM2.00

RM2.50 RM2.00 RM0.55 RM1.20 z TABLE 4

Price Index for the year 2006 based on the year 2000 125 125 y 150 120


SULIT

(a)

Find the values of (i) x, (ii) y (iii) z.

[x=2.00,y=137.5,z=2.40]

[3 marks]

(b)

Find the composite price index for cooking oil, rice, salt, sugar and flour in the year 2006 based on the year 2000. [131.17] [2 marks]

(c)

Calculate the total monthly sales for those essential items in the year 2006, given that the total monthly sales in the year 2000 was RM 150.[3 marks] [120]

(d) the composite index for the year 2008 based on the year 2000 if the essential items increased by 20% from the year 2006 to the year 2008. [157.40] [3 marks] 14. Use the graph paper provided to answer this question.

monthly sales of those

The Mathematics Society of a school is selling x souvenirs of type A and y souvenirs of type B in a charity project based on the following constraints : I

:

The total number of souvenirs sold must not exceed 75.

II

:

The number of souvenirs of type A sold must not exceed twice the number of souvenirs of type B sold.

III

:

The profit gained from the selling of a souvenir of type A is RM9 while the profit gained from the selling of a souvenir of type B is RM2. The total profit must not be less than RM200.

(a)

Write down three inequilities other than x ≥ 0 dan y ≥ 0 which satisfy the above constraints. Answer x + y ≤ 75 , x ≤ 2y and 9x + 2y ≥ 200] [3 marks]

(b)

Hence, by using a scale of 2 cm to 10 souvenirs on both axes, construct and shade the region R which satisfies all the above constraints. [ 3 marks]

(c)

By using your graph from (b), find

[

the range of number of souvenirs of type A sold if 30 souvenirs of type B are sold. [ 16 ≤ number of A type souvenirs sold ≤ 45] (ii) the maksimum which may be gained. [Answer RM 500]

(i)

[4 marks]

15.

An object, P, moves along a straight line which passes through a fixed point O. www.cikgurohaiza.com

Figure 8 shows the object passes the point O in its motion. t seconds after leaving the point O , the velocity of P, v m s─1 is given by v = 3t2 – 18t + 24. The object P stops momentarily for the first time at the point B. P O

B Figure 8

(Assume right-is-positive) Find: (a) the velocity of P when its acceleration is 12 ms – 2 , [9 ms – 1 ] [3 marks] (b) the distance OB in meters, [20 m] [4 marks] (c) the total distance travelled during the first 5 seconds. [28 m] [ 3 marks]

12. (a)

(i)

Use area formula 1 (12.1)(9.5) sin STQ = 45 2

(ii)

(iii)

∠STQ = 128.47° or 128°28' Using cosine Rule SQ 2 = 12.12 + 9.5 2 − 2(12.1)(9.5) cos STQ SQ = 19.49cm ∠TQS = 29.05° h sin 29.05 = or equivalent 9.5 = 4.613 cm

(b) 5 12 = sin 15° sin p 12 sin p = × sin 15° = 0.6212 5 ∠QPR = 38°24' ,141°36' @ 38.40°, 141.60° ∠PQR = 180o – 15o – 38.40o @ ∠PQR = 180o – 15o – 141.60o ∠PQR = 126.60o and 23.40o (c)

3472/2

(i)

PR 5 = sin 23.4° sin 15° 5 PR = × sin 23.4° sin 15° = 7.672 cm PR = 7.673 cm www.cikgurohaiza.com

SULIT

(ii)

Use Cosine Rule

cos ∠PTR =

13 2 + ( 50 ) 2 − (7.672) 2 2(13)( 50 )

= 0.8710

∠PTR = 29.420 (iii) Area ∆ PVR =

1 (13)( 50 ) sin 29.42° 2

= 22.58 cm2

13.

(a)

(i) x = 2.00 (ii) y = 137.5 (iii) z = 2.40 (b) Use composite index formula − 125(80) + 125(100) + 137.5(50) + 150(60) + 120(30) I= 80 + 100 + 50 + 60 + 30 = 131.17 P2006 × 100 = 131.17 (c) 150 P2006 = RM 196.76 2008 I 2006 = 120 2008 (d) I 2000 =

14. (a)

120 × 131.17 100 = 157.40

The three inequalities are x + y ≤ 75 , x ≤ 2y and 9x + 2y ≥ 200

(b) (c)

refer by graph (i) 16 ≤ number of A type souvenirs sold ≤ 45 (ii)Maximum profit = RM [ 9(50) + 2(25) ] = RM500.


y

100

90 9x + 2y = 200 80

70

60

50 x = 2y 40

30

R y = 30 ( 50 , 25 )

20 x + y = 75 10

O

3472/2

x 10

16 20

30

40

45

50


70

60

SULIT

80


Answer all questions. 1.

Function f is defined by ⎧ 2− x,x ≤ 3 ⎫ ⎪ ⎪ f ( x) = ⎨11 3 ⎬ ⎪⎩ 2 − 2 x, x ≥ 3⎪⎭

Find the range corresponding to the domain 0 ≤ x ≤ 4

[3 marks]

1 Answer : ……………………..

2.

3

x+2 mx + n and fg: x → , 5 5 where m and n are constants , find the value of m and of n, Given the function f: x → 2x + 5 , g : x →

[2 marks ]

2

Answer : m =……………………. n =..............................

Perfect Score 2009 www.cikgurohaiza.com


2


3.

Diagram 1 shows part of the mapping of x to z by the function 12 f : x → ax + b followed by the function g : y → , y ≠ c . Calculate the values of a, b, c y−c and d.

12 4 6

3

1 d Diagram 1 [ 4 marks]

Answer: a=………b=………c=………d=…………..

4.

If the x-axis is a tangent to the curve x 2 + 3 px = p − 3 , find the values of p. [3 marks ]

4

3 Answer : p =.........………



5.

Given α and β are the roots of 2 x 2 − 4 x + 1 = 0 . Form the quadratic equation with roots α 2 and β 2 . [ 4 marks ]


5

4 Answer : ................................. ___________________________________________________________________________ 6.

Given the quadratic function of f(x) = 6x − 1 − 3x2. a) Express the quadratic function f(x) in the form k + m(x + n)2, where k, m and n are constants. b) write the equation of the axis of symmetry

[ 3 marks ]

6

3 Answer : (a) .……........................ (b) ……........................




7.

Find the range of values of x if

f ( x ) = 3 x 2 + 2 x − 5 always positive. [3 marks]

7

3

Answer : ..................................

8.

Simplify and state your answer in the simplest form 5 3n +1 + 5 3n − 2 − 125 n −1 . [2 marks]

8

Answer : ................................... 3 9.

Solve the equation 9

y +1

= 24 + 9 y . [3 marks]

9

Answer : ....................................

3



10.


Given 2 + log 3 k = log 9 (m + 3) , express k in terms of m. [4 marks]

10

4

Answer : ....……………...……….. 11. Solve the equation log 3 x = log 9 (2 x + 3)

[3 marks]

11

Answer: …...…………..….......



3


12.

Given that the n th term , Tn = 20 − 4n for an arithmetic progression. Find the sum of the first 12 terms of the progression. [3 marks]

12

4 Answer: …...….………..…....... ___________________________________________________________________________ 13. Given the sum of the first 3 terms of a geometric progression is 567 and the sum of the next three terms of the progression is −168. Find the sum to infinity of the progression. [4 marks]

13

Answer :

…………………….

4



14.

Given that the sum of the first three terms of a geometric progression is 13 times the third term of the progression. If the common ratio, r > 0, find the common ratio. [ 2 marks ]


14

Answer : .…………………

15.

Diagram 2 shows the graph of log2 y against log2 x. Values of x and values x2n , where n and k are constants. of y are related by the equation y = k Find the value of n and the value of k. log2 y

*(5, 6)

0

log2 x

(2, 0)

Diagram 2 [4 marks]

Answer : n= ..……k=……….



2


16.

Diagram 3 shows a semicircle KLMN, of diameter KLM , with centre L. y

N (x,y)

•

K

L 0

x

M Diagram 3

x y + = 1 and point N( x , y ) lies 4 3 on the circumference of a circle KLMN , find the equation of the locus of the moving point N. [ 3 marks ] Given that the equation of the straight line KLM is

16

3 Answer: 17.

……..…….…………...

If a = 2i + ( p + 1) j and b = −3i + 6 j , find the value of p if a + b is parallel to the x-axis. [3 marks]

17

4

Answer:

……..…….…………...

___________________________________________________________________________



18. Given that sin 20 0 = a and cos 30 0 = b , express sin 50 0 in terms of a and


[3 marks]

18

Answer: ...………………………

3

___________________________________________________________________________ 19.

Diagram below shows two sectors , OAB and OCD with centre O.

E

D

C

A

B

O Given that ∠ COD = 0.92 rad, BC = 5 cm and perimeter of sector OAB is 20.44 cm. Using π = 3.142 , find the area of the shaded region of ABCED. [ 3 marks ]

19

3 Answer:……………………… Perfect Score 2009 www.cikgurohaiza.com


SULIT For examiner’s use only

3472/1

20. Given that y=

2x − 1 dy and = 2 g(x) where g(x) is a function in x . 2 x dx

1

Find ∫ g ( x)dx .

[3 marks]

−1

20

3

Answer: …...…………..…....... ___________________________________________________________________________ 21.

The gradient of the curve y = hx + and the value of k.

k at the point x2

7⎞ ⎛ ⎜ −1, ⎟ is 2. Find the value of h ⎝ 2⎠ [3 marks]

21

Answer: …………………….. 3

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3472/1


22.

A coach wish to choose a player from two bowlers to represent the nation in a tournament. The following data show the number of pins scored by the two players in six sucessive bowls: Player A: 8, 9, 8, 9, 8, 6 Player B: 7, 8, 8, 9, 7, 9 By using the values of mean and standard deviation, determine the player which qualified to be choosen because the score is consistent. [3 marks]

22

3

Answer: …...…………..…....... ___________________________________________________________________________ 23. In a debate competition, the probability of team A, team B and team C will qualify for 1 1 1 the final are , , respectively. Find the probability that at least 2 teams will qualify 3 4 5 for the final. [3 marks]

23

Answer: …………………………… 3

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3472/1


24. The letters of the word G R O U P S are arranged in a row. Find the probability that an arrangement chosen at random (a) begins with the letter P, (b) begins with the letter P and ends with a vowels. [3 marks]

24

Answer: ( a)…...…………..…....... 3

( b )...................................

25. The life span of certain computer chip has a normal distribution with a mean of 1500 days and a standard deviation of 40 days. a) Calculate the probability that a computer chip chosen at random has a life span of more than 1540 days b) Given that 6% of the computer chips have a life span of more than n days, find the value of n. [4 marks]

25

Answer : (a)…...…………..……..…... 4

(b)..........................................

Perfect Score 2009


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3472/1 END OF QUESTION PAPER

Perfect Score 2009


SULIT

1

Paper 2 Time: Two hours and thirty minutes Instruction : This question paper consists of three sections: Section A, Section B and Section C. Answer all questions in Section A, four questions from Section B and two questions from Section C. Give only one answer/ solution for each question. All the working steps must be written clearly. Scientific calculator that are non-programmable are allowed. Section A [40 marks] 1. Given that (-1, 2k) is a solution for the simultaneous equation x2 + py − 29 = 4 = px − xy where k and p are constants. Find the value of k and of p. [5 marks]

Jawapan: k = 4, p = 4; k = −2, p = −8 2. Given function f : x → 4 − 3x. (a) Find, (i) f 2(x), (ii) (f 2)−1(x).

[3 marks]

(b) Hence, or otherwise, find (f −1)2(x) and show (f 2)−1(x) = (f −1)2(x).

[3 marks]

(c) Sketch the graph of ⏐f 2(x)⏐ for the domain 0 ≤ x ≤ 2 and find it’s corresponding range. [2 marks]

Jawapan: (a)(i) 9x − 8 (ii)

x+8 9

(c) y 10 --------------------------

0 ≤ y ≤ 10

8

0

8/9

x


2

3. Diagram 3 shows five semicircles.

DIAGRAM 3 The area of the semicircles form a geometric progression. Given that area of the 1 smallest semicircle is of the area of the largest semicircle. If the total area of the 16 semicircles is 30 cm 2 , find (a) area of the smallest semicircle (b) area of the largest semicircle [5 marks] Jawapan: (a) 10 (b) 160

3 1 , show that tan x = − . 4 7 0 0 Sketch the graph of y = tan x for 0 ≤ x ≤ 360 .

4. Given that tan( x − y ) = −1 and tan y =

Hence, using the same axes , draw a suitable straight line and find the number of solutions for the equation 3 tan x + x = 6 [6 marks]

Jawapan: Number of solutions = 3


3

5. Diagram 5 shows a parallelogram OABC. O

A

P D

B C DIAGRAM 5 →

→

Given that APD, OPC and DCB are straight lines. Given that OA = 6a, OC = 12c and OP : PC = 3 : 1. (i) (ii)

→

Express AP in terms of a and/or c. Given the area of the ΔADB = 32 unit 2 and the perpendicular distance from A to DB is 4 units, find ⏐a⏐. [5 marks]

Jawapan: (a) − 6a + 9c % % (b)2

6. Cumulative frequency

x (25.5, 80)

(20.5, 74) x x(15.5, 58) (10.5, 26) x x (5.5, 6)

O

0.5 DIAGRAM 6


Length of fish in cm

4

Diagram 5 shows an ogive for the distribution of 80 fishes in a tank when the cumulative frequency is plotted against upper boundaries for a certain classes. O is the origin. (a) Construct a frequency table with a uniform class interval from the information given in the ogive. [2 marks] (b) Draw a histogram and determine the mode. [3 marks] (c) From the frequency table, find (i) the variance, (ii) the median for the length of fish in the tank. [4 marks]

7. Use the graph paper provided to answer this question. An experiment which involves samples of red blood cell used to trace the percentage, P, of the red blood cell which experience creanation when it is added by drops of sodium chloride solution with different concentration, K mol dm3. Table below shows the results of the above experiment. Sodium chloride concentration (K) Percentage of red blood cells which experience creanation (P)

0.50

0.75

1.00

1.25

1.50

1.75

0.4

5.0

14.5

27.6

46.2

68.9

TABLE 7 Variables P and K are related by the equation P = constants. (a) Draw the graph of

4

μ2

(K + A)2 where μ and A are

P against K.

[5 marks] (b) From your graph, find the value of μ and the value of A. [4 marks] (c) Find the value of P when K = 1.4? [1 mark]

Jawapan: (b) μ = 0.33, A = −0.40 (c)37.21 − 38.44


5

8.

y y = x(x − 1)(x + 3)

x

O

(a) Diagram above shows the curve y = x(x − 1)(x + 3). Calculate the area bounded by the curve, x-axis, line x = −2 and line x = 1.

[6 marks]

(b) Diagram below shows the shaded region bounded by the curve y = 2 x + 1 , line x = 1 and line x = k. When the region is revolved 360° at the x-axis, the volume generated is 18π unit3. Find the value of k. [4 marks] [ answer (a)

47 12

(b) k = 4 ]

9. y Jawapan: (a)α 2 + β 2 = 81

P(0, β)

⎛2 ⎞ (b)(ii ) ⎜ ,5.85 ⎟ ⎝3 ⎠ (c)0.35

• R(x, y)

Wall

Q(α, 0)

O

x

Floor Diagram 9 shows the x-axis and the y-axis which represent the floor and the wall. The end of a piece of wood PQ with length 9 m touches the wall and the floor at the point P(0, β) and point Q(α, 0). (a) Write the equation which relates α and β. [1 mark] (b) Given R is a point on the piece of wood such that PR : RQ = 1 : 2. (i) Show that the locus of the point R when the ends of the wood is slipping along the x-axis and the y-axis is 4x2 + y2 = 38. (ii) Find the coordinates of R when α = 2. (iii) Find the value of tan ∠ ORQ when α = 2. [9 marks]


6

10. L J P

Q α rad

K R

O

T

DIAGRAM 10 Diagram 10 shows a circle PQRT, centre O and radius 5 cm. JQK is a tangent to the circle at Q. The straight lines, JO and KO, intersect the circle at P and R respectively. OPQR is a rhombus. JLK is an arc of a circle, centre O. Calculate [ 2 marks] (a) the angle α , in terms of π (b) the length, in cm, of the arc JKL [ 4 marks] (c) the area, in cm2, of the shaded region. [4 marks]

Jawapan: 2 (a) π 3 20 (b) π 3 (c)61.50

11. (a) A study on post graduate students, revealed that 70% out of them obtained jobs six months after graduating. (i) If 15 post graduates were chosen at random, find the probability of not more than 2 students not getting jobs after six months. (ii) It is expected that 280 students will succeed in obtaining jobs after six months. Find the total number of students involved in the study. [5 marks] (b)

The mass of 5000 students in a college is normally distributed with a mean of 58kg and variance of 25 kg2. Find (i) the number of students with the mass of more than 90 kg. (ii) the value of w if 10% of the students in the colleges are less than w kg. [5 marks] Jawapan: (a) (i) 0.1268 (ii)400 (b) (i) 82or 83 (ii) 38.77 or 38.79


7

12. Diagram 12 shows the position and direction of motion for two objects, P and Q, which move along a straight line and passes through two fixed points, A and B respectively. At the instant when P passes through the fixed point A, Q passes through the fixed point B. Distance AB is 28 m. P

Q

A

C

B

28 m DIAGRAM 12 The velocity of P, vp ms , is given by vp = 6 + 4t − 2t 2, where t is the time in seconds, after passing through A, whereas Q moves with a constant velocity of −2 ms−1. Object P stops instantaneously at the point C. (Assume towards the right is positive.) Find, [3 marks] (a) the maximum velocity, in ms−1, for P, (b) the distance, in m, C from A, [4 marks] (c) the distance, in m, between P and Q at the instant when P is at the point C. [3 marks] −1

Jawapan: (a) 8 m/s (b) 18 (c) 4


8

13. . Diagram 13 shows a quadrilateral ABCD such that ∠ABC is acute. 9.8 cm

D

5.2 cm C

A

40.50

12.3 cm

9.5 cm

B

DIAGRAM 13 (a) Calculate, (i) ∠ABC, (ii) ∠ADC, [8 marks] (iii) area, in cm2, of quadrilateral ABCD. (b) A triangle A’B’C’ has the same measurements as those given for triangle ABC, that is, A’C’ = 12.3 cm, C’B’ = 9.5 cm and ∠B’A’C’ = 40.5°, but which is different in shape to triangle ABC. (i) Sketch the triangle A’B’C’. [2 marks] (ii) State the size of ∠A’B’C’.

Jawapan: (a) (i) 57.21-57.25 (ii) 106.07-106.08 (iii)82.37-82.39 (b) (ii) 122.75-122.79


9

14. Use the graph paper provided to answer this question. Cloth T-shirt Slack

Preparation time (minutes)

Sewing time (minutes)

45 30

50 70

A tailor shop received payment only for sewing T-shirt and slack. Preparation time and sewing time for each T-shirt and slack are shown in the table above. The maximum preparation time used is10 hours and the sewing time must be at least 5 hours 50 minutes. The ratio of the number of T-shirt to slack is not more than 4 : 5. In a certain time, the shop is able to complete x pieces of T-shirt and y pieces of slack. (a) Write three inequalities, other than x ≥ 0 and y ≥ 0, which satisfy the above conditions. [3 marks] (b) By using a scale of 2 cm to I unit on the x-axis and 2 cm to 2 units on the y-axis, draw the graphs for the three inequalities. Hence, shades the region R which satisfies the above conditions. [3 marks] (c) Based on your graph, find (i) the minimum number of slacks which can be sewn in that time if 3 pieces of of T-shirt has been sewn.. (ii) maximum total profit received in that time if the profit gained from each piece of T-shirt and slack are RM16 and RM 10 respectively. [4 marks]

Jawapan: (a)45 x + 3 y ≤ 600 50 x + 70 y ≥ 350 5x ≤ 4 y (c)4 (6,11), RM 206


10

15. (a) 105 5−x

Index number, Ii Weightage, Wi

94 x

120 4

The composite index number for the data in the table above is 108. Find the value of x.

[4 marks]

(b) (i) In the year 1995, price and price index for one kilogram of certain grade of rice is RM2.40 and 160 respectively. Based on the year 1990, calculate the price per kilogram of rice in the year 1990. [2 marks] Item Timber Cement Iron Steel

Price index in the year 1994 180 116 140 124

Change of price index from the year 1994 to the year 1996 Increased 10 % Decreased 5 % No change No change

Weightage 5 4 2 1

(ii) Table above shows the price index in the year 1994 based on the year 1992, the change in price index from the year 1994 to the year 1996 and the weightage respectively. Calculate the composite price index in the year 1996. . [4 marks] Jawapan: (a) x=3 (b) (i) 1.50 (ii) RM152.90

End of question paper


-1-

FUNCTIONS 1. Given that f : x → 4 x + m and f

−1

: x → nx +

3 , find the values of m and n. 4

1 4 2. Given that f : x → 2 x − 1 , g : x → 4 x and fg : x → ax + b , find the values of a and b . Answer:- a = 8 ; b = –1 Answer:- m = – 3 ; n =

2 2 3. Given that f : x → x + 3 , g : x → a + bx and gf : x → 6 x + 36 x + 56 , find the values of a and b . Answer:- a = 2 ; b = 6

−1 4. Given that g : x → m + 3 x and g : x → 2kx −

5. Given the inverse function f

−1

( x) =

4 , find the values of m and k. 3 Answer:- k =

1 ;m=4 6

Answer:-(a)

11 1 (b) − 2 2

2x − 3 , find 2

(a) the value of f(4), (b) the value of k if f –1 (2k) = – k – 3 .

6. Given the function f : x → 2 x − 1 and g : x → (a) f –1 (x) , (b) f – 1 g(x) , (c) h(x) such that hg(x) = 6x – 3 . Answer:-(a)

7. Diagram 1 shows the function g : x →

g

x

x − 2 , find 3

x +1 2

(b)

1 1 x− 6 2

(c) 18x + 33

p + 3x , x ≠ 2 , where p is a constant. x−2

p + 3x x−2

7 5 Diagram 1 Find the value of p.


-2Answer:- p = 4 8. x

y 4

4

z 4

2

2

2

0

0

0 −1

−2

−2 −2 Diagram 2 Diagram 2 shows the mapping of y to x by the function g : y → ay + b and mapping 6 b to z by the function h : y → , y ≠ . Find the, 2 2y − b (a) value of a and value of b, (b) the function which maps x to y, (c) the function which maps x to z. 10 − y 18 Answer:- (a)a= –6, b=10 (b) (c) 6 − y − 20 9. In the Diagram 3, function h mapped x to y and function g mapped y to z.

x

h

y

g

z 8

5 2 Diagram 3 Determine the values of, (a) h−1(5), (b) gh(2)

Answer:- (a)2 (b)8 10. Given function f : x → 2 − x and function g : x → kx + n. If composite function gf is given as gf : → 3x2 − 12x + 8, find (a) the value of k and value of n, (b) the value of g2(0). Answer:-(a) k = 3 ,n = –4 (b)44 2


-311.

The following information refers to the functions f and g. g (x) = 4 – 3 x fg (x) = 2 x + 5

Find f (x).

Answer:-

23 − 2 x 3

12. (a) Function f, g and h are given as f : x → 2x 3 g:x→ ,x≠2 x−2 h : x → 6x2 − 2. (i) Determine the function fh(x). At the same axis, sketch the graphs of y = g(x) and y = fh(x). Hence, determine the number of solutions for g(x) = fh(x). (ii) Find the value of g−1(−2). (b) Function m is defined as m : x → 5 − 3x. If p is another function and mp is defined as mp : x → −1 − 3x2, determine function p. Answer:-(a)(i)12x2 – 4 (b) p ( x ) = 2 + x 2

13. Given function f : x → 4 − 3x. (a) Find (i) f 2(x), (ii) (f 2)−1(x). (b) Hence, or otherwise, find (f −1)2(x) and show

(f 2)−1(x) = (f −1)2(x).

(c) Sketch the graph of ⏐f 2(x)⏐ for the domain 0 ≤ x ≤ 2 and find it’s corresponding x +8 Answer:-(a)9x – 8 (b) range. 9 14. A function f is defined as f : x →

p+x , for all values of x except x = h and p 3 + 2x

are constants. (a) Determine the value h. (b) Given value 2 is mapped to itself by the function f. Find the (i) value p, (ii) another value of x which is mapped to itself, (iii) value of f −1(−1). 3 Answer:-(a) h = − (b)(i)p =12(ii)x = –3 (iii)–5 2


-4QUADRATIC EQUATIONS is twice the other root.

1. One of the roots of the quadratic equation Find the possible values of p.

Answer ; p = 5, −7

2. If one of the roots of the quadratic equation . find an expression that relates

is two times the other root, Answer : 2b 2 = 9ac

3. Find the possible value of m , if the quadratic equation roots.

has two equal

Answer ;

4. Straight line y = mx + 1 is tangent to the curve x2 + y2 − 2x + 4y = 0. Find the possible values of m.

Answer : −

1 or 2 2

α β and are roots of the equation k x(x − 1) = 2m − x. 2 2 If α + β = 6 and α β = 3, find the value of k and of m.

5. Given

Answer : k = −

1 3 ,m= 2 16

6. Find the values of λ such that the equation (3 − λ)x2 − 2(λ + 1)x + λ + 1 = 0 has equal roots. Hence, find the roots of the equation base on the values of λ obtained.

Answer : λ = ± 1; roots: λ = 1, x = 1; λ = −1, x = 0


-5QUADRATIC FUNCTIONS 1.

Diagram 1 shows the graph of the function y = −2 ( x − p ) + 5 , where p is constant. y 2

x

0 ( 0, –3 )

( 4 , –3 )

Find, Diagram 1 (a) the value of p , the equation of the axis of symmetry, (b) (c) the coordinate of the maximum point.

Answer:- (a) p = 2 (b) x = 2 (c) ( 2, 5 ) 2.

y

0

( 0, –2 )

x

f ( x ) = ( p − 1) x 2 + 2 x + q

Diagram 2 Diagram 2 shows the graph of the function f ( x ) = ( p − 1) x 2 + 2 x + q . (a) State the value of q . (b) Find the range of values of p .

Answer:-(a) q = – 2 (b) p <


1 2

-63.

y

y = x 2 + bx + c

( 0, 9)

x

K Diagram 3

Diagram 3 shows the graph of the function y = x 2 + bx + c that intersects the y- axis at point ( 0, 9 ) and touches the x- axis at point K. Find, (a) the value of b and c , (b) the coordinates of point K.

Answer:-(a) b = – 6 , c = 9 (b)

4.

( 3, 0 )

y

( 0, 23) ( 2, 3) 0

x

Diagram 4 In Diagram 4 above point ( 2, 3 ) is the turning point on the graph which has equation of the form y = p(x + h)2 + k. Find the, (a) values of p, h and k, (b) equation of the curve formed when the graph as shown is reflected at the x–axis. (c) equation of the curve formed when the graph as shown is reflected at the y–axis. Answer :- (a) p = 5 , h = −2, k = 3

(b) y = −5(x − 2)2 − 3


(c) y = 5(x + 2)2 + 3

-75.

Function f ( x ) = x 2 − 8kx + 20k 2 + 1 has a minimum value of r 2 + 4k , where r and k are constants. (a) (b)

By using the method of completing the square, show that r = 2k − 1 . Hence or otherwise, find the values of k and r if the graph of the function is symmetrical about x = r 2 − 13 . Answer:-(b)

6.

k = 3 , –1 and r = –3 , 5

The function f ( x ) = ( 6 + x )( 2 − x ) + h has a maximum value of 10 and h is a constant. (a) Find the value of h. (b) Sketch the graph of f ( x ) = ( 6 + x )( 2 − x ) + h for the value of h that is determined in (a) above. (c) Write the equation of the axis of symmetry. Answer:- ( a) h = --6 (c) x = –2

7.

Given y = x2 + 2kx + 3k has minimum value 2. (a) Without using the method of differentiation, find the two possible values of k. (b) With these values of k, sketch on the same axis, two graphs for y = x2 + 2kx + 3k. (c) State the coordinates of the minimum point for y = x2 + 2kx + 3k . Answer:- (a) k =1 , 2

(c) (−1, 2), (−2, 2)

****************************************************************************** SIMULTANEOUS EQUATIONS 1. Solve the simultaneous equations 3x + 2y = 1 and 3x2 – y2 = 5x + 3y. Answer: x = -7/3 , y = 4 ; x = 1, y = -1 2. Solve the simultaneous equations

x+ y =

xy − 3 2x − 5 y and = 2 3 5 Answer: x = 41/10, y = 16/3 ; x = 1, y = -5

3. Solve the simultaneous equations 2x - y = 4 and 2x2 + xy - 3x = 7. Give your answers correct to three decimal places. Answer: x = 2.461, y = 0.922; x = -0.711, y = -5.422


-8INDICES AND LOGARITHMS 625 x + 2 =

1

1.

Solve the equation

2.

Solve the equation 2 x .8 x = 4 5 x −3

3.

Show that 3

x+2

+3

x −1

Answer:- x = −

5 x .25 x −1

Answer:- x = 1

( ) is divisible by 13.

−53

2 5

x

( )

Answer : 13 3 x −1

.4.

Solve the equation 64

2 x −3

+ 2 = 34

Answer:5.

23 12

Solve the equation 3 x 2 2 x +1 = 10

Answer: 0.6477 6.

Solve the equation log 4 [log 2 (2 x − 3)] = log 9 3

Answer:3.5 7.

Solve the equation log( x + 2) − log(4 x − 1) = log

8.

Solve the equation log2x - 4 logx16 = 0

1 x Answer :x =1 Answer: 16 ,

9.

( )

1 16

Solve the equation 2 7 x −1 = 5 x

Answer: x = 3.7232 log 2 x

= 81

10.

Solve 3

11.

Solve the equation 32x+1 - 2 (3x+0.5) - 3 = 0

Answer : x = 16 Answer: 0.5

12.

Solve the equation 102x+1 - 7 (10x) = 26

Answer:0.3010 13.

Given that log 3 5 = m and log 9 2 = n , express log 3 50 in terms of m and n Answer:- 2m + 2n

14.

Given that log x 2 = k and log x 7 = h , express log

15.

Answer:- 2h + 2 − 2k Given that 2 log 2 ( x + y ) = 3 + log 2 x + log 2 y , show that x + y = 6 xy

16.

x

3.5 x in terms of k and h

(

2

2

)

If log2a + log2b = 4, show that log4ab = 2 and that log8ab = 4/3. If log2a + log2b = 4, show that log4ab = 2 and that log8ab = 4/3.


-9-

COORDINATE GEOMETRY 1.

The following information refers to the equations of two straight lines, AB and CD which are parallel to each other.

AB : 2y = p x + q CD : 3y = (q + 1) x + 2 Where p and q are constants

Express p in terms of q.

Answer: p =

2 (q + 1) 3

2.

The triangle with vertices A(4,3), B(-1,1) and C(t , -3) has an area 11 unit2. Find the possible values of t. Answer: t = 0 , -22

3.

The points P(3, p), B(-1, 2) and C(9,7) lie on a straight line. If P divides BC internally in the ratio m : n , find (a) m : n , (b) the value of p. Answer:(a) 2 : 3 (b) p = 4

4.

(a) A point P moves such that its distance from point A (1,– 4) is always 5 units. Find the equation of the locus of P. Answer: x 2 + y 2 − 2 x + 8 y − 8 = 0 (b) The point A is (-1, 3) and the point B is (4, 6). The point Q moves such that QA : QB = 2 : 3. Find the equation of the locus of Q. Answer: 5 x 2 + 5 y 2 + 14 x + 102 y − 54 = 0 (c) A point R moves along the arc of a circle with centre A(2, 3). The arc passes through Q(-2, 0). Find the equation of the locus of R. Answer: x 2 + y 2 − 4 x − 6 y + 8 = 0

(d) A point S moves such that its distance from point A(–3,4) is always twice its distance from point B(6,-2). Find the equation of the locus of S. Answer: x 2 + y 2 − 18 x + 8 y + 45 = 0 (e) The point M is (2, –3) and N is (4, 5). The point T moves such that it is always equidistance from M and from N. Find the equation of locus of T. Answer : e) x+4y = 7 (f) Given point A (1,2) and point B (4, –5). Find the locus of point W which moves such that ∠ AWB is always 900.


- 10 -

Answer: x 2 + y 2 − 5 x + 3 y − 6 = 0 Solutions to question no 5, 6 and 7 by scale drawing will not be accepted. 5.

In Diagram 1, the straight line PR cuts y-axis at Q such that PQ : QR = 1 : 3. The equation of PS is 2y = x + 3.

y

R

S

Q( 0, 4 )

P(–3, 0 ) (a) Find

x

O Diagram 1 the coordinates of R, the equation of the straight line RS, the area Δ PRS.

(i) (ii) (iii)

(b) A point T moves such that its locus is a circle which passes through the points P, R and S. Find the equation of the locus of T.

Answer: a)(i) R = (9 , 16) (ii) y = – 2x + 34 (iii) 80 unit 2 b) x2 + y2 – 6x – 16y – 27 = 0 6.

Diagram 2 shows the straight line graphs PQS and QRT in a Cartesian plane. Point P and point S lies on the x-axis and y-axis respectively. Q is the mid point of PS. y

S

y − 3x = 4

Q R(0, 1) P x O T Diagram 2


- 11 (a) Find, (i) coordinates of the point Q, (ii) area of the quadrilateral OPQR. (iii) The equation of the straight line which is parallel to QT and passes through S. (b) Given 3QR = RT, calculate the coordinates of the point T. 1 (c) A point moves in such a way that it’s distance from S is it’s distance from the point T. 2 (i) Find the equation of locus of the point T. (ii) Hence, determine whether the locus cuts the x-axis or not. . 2 5 3 Answer: (a)(i) (− , 2) (ii) (iii) y = − x + 4 (b) (2, −2) (c)(i) 3x2 + 3y2 + 4x − 36y +56 = 0 2 3 3 (ii) No 7.

y K P• J

•Q R• x

O L Diagram 3 1 In Diagram 3, P(2, 9), Q(5, 7) and R(4 , 3) are the mid point of the straight line JK, KL and 2 LJ such that JPQR form a parallelogram. (a) Find, (i) the equation of the straight line JK, (ii) the equation of the perpendicular bisector of the straight line LJ. (b) Straight line KJ is extended until it intersects the perpendicular bisector of the straight line LJ at the point S. Find the coordinates of the point S. (c) Calculate the area of ΔPQR and consequently the area of ΔJKL.

Answer: (a)(i) y = 8x − 7 (ii) 4y = 6x − 15


⎛1 ⎞ (b) ⎜ ,−3 ⎟ ⎝2 ⎠

1 (c) 6 ; 26 2

- 12 STATISTICS

1. Table 1 shows the results obtained by 100 pupils in a test. Marks

< 20

< 30

< 40

< 50

< 60

< 70

< 80

< 90

Number of pupils

3

8

20

41

65

85

96

100

Table 1 (a) Based on Table 1, complete the table below. Marks

10 – 19

Frequency (b) Without drawing an ogive, estimate the interquartile range.

Answer:-(b)Interquartile range = 22.62 2.

The mean and standard deviation of a set of integers 2 , 4 , 8 , p and q are 5 and 2 respectively. (a) Find the values of p and of q . (b) State the mean and variance of the set integers 7, 11, 9 , 2p + 3 and 2q + 3

Answer:-(a) p = 5,q = 6 or p = 6, q = 5 (b) Mean =13 Variance = 16 3. The histogram in Diagram 1 shows the marks obtained by 40 students in Mathematics test. Number of students 10 8 6 4 2 0 15.5 20.5 25.2 30.5 35.5 40.5

Marks

Diagram 1 (a) (b)

Without drawing an ogive , calculate the median mark. Calculate the standard deviation of the marks.


- 13 Answer:(a) 27.17 (b) 6.595

4.

Table 2 shows the frequency distribution of the Chemistry marks of a group of students. Marks 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60

Number of students 2 3 5 10 p 2 Table 2

(a) (b) (c)

If the median mark is 34.5 , calculate the value of p . By using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 2 students on the vertical axis, draw a histogram to represent the frequency distribution of the marks. Find the modal mark. What is the modal mark if the mark of each student is increased by 8 ? Answer:- (a) p = 6 (b) Mode = 36 (c) 44

5.

The scores, x , obtained by 32 students of Class 5 Alfa in a test are summarized as ∑ x = 2496 and ∑ x 2 = 195488. The mean and the standard deviation of the scores, y , obtained by 40 students and Class 5 Beta in the test are 66 and 6 respectively. (i)

∑y

(ii)

∑ y2

(a)

Find

(b)

Calculate the mean and the standard deviation of the scores obtained by all the 72 students. Answer:-(a)(i) 2640 (ii)175680

6.

(b)Mean = 71.33 , S Deviation= 8.194

A set of data consists of 10 numbers. The sum of the numbers is 120 and the sum of the squares of the numbers is 1650. (a)

Find the mean and variance of the set of data,

(b)

A number a is added to the set of data and the mean is increased by 2, find (i) the value of a, (ii) the standard deviation of the new set of data. Answer:-(a)

Mean = 12 , Variance = 21 (b)(i) a = 34 (ii) S Deviation = 7.687


- 14 CIRCULAR MEASURES

1.

A

T B

O

Diagram 1 Diagram 1 shows a circle with centre O and OA = 10 cm. Straight line AT is a tangent to the circle at point A, and AOT is a triangle. Given that the area of triangle OAT = 60 cm2, find the area of sector OAB. Answer:- 43.80 cm 2 2. A

O B

Diagram 2 Given that the area of a sector OAB in Diagram 2 with centre O and radius 20 cm is 240 cm2. Calculate (a) the length of arc AB (b) area of shaded region Answer:- : (a) 24 cm (b) 53.60 3. P

O

θ R

Diagram 3

Q

In the Diagram 3, POQ is a circular sector with centre O and a radius of 17 cm. Point R is on the straight line ORQ such that RQ = 5 cm. Calculate (a) the value of θ in radian (b) the area of the shaded region, in cm2 Answer:- (a) 0.7871


(b) 41.49

- 15 4.

Q

S1 P

O

S

R

Diagram 4 Diagram 4 shows a semicircle with centre O and radius of 10 cm. Given that QS is the length of arc with centre P and ∠ QPS =

π

6

rad . Find

(a) the length of OS. (b) the area of S1

Answer:-(a) 7.32 cm (b) 35.23 cm 2

5.

2r

C

C

Diagram 5 Two identical circles of radius 2r are drawn with their centres, C on the circumference of each circle as shown in the Diagram 5. Show that the area of shaded region A cm 2 , is given by 2 2 r 4π − 3 3 . 3

(

)

6. Q R

O

8cm

P 3cm A

Diagram 6 Diagram 6 shows two circles with centres O and A. The respective radii are 8 cm and 3 cm. A tangent touches the circles at the points Q and R. Given that ∠QOP =

π

3

radians, find

(a) the length of QR (b) the perimeter of the shaded region (c) the area of the shaded region Answer : a) 9.80 cm


b) 24.46 cm

c) 10.96 cm2

- 16 DIFFERENTIATION

1.(a) (i) Given y = 3x2 + 5, find

dy by using the first principle. dx

4 − 3 with the first principle. x

(ii) Differentiate y =

d ⎛ 1 ⎞ ⎟. ⎜ dx ⎝ 2 x + 1⎠ (ii) Given f(x) = 4x(2x − 1)5, find f’(x). (iii) Differentiate 3x2(2x − 5)4 with respect to x.

(b) (i) Find

(iv) Given f(x) = (2x − 3)5, find f ″(x). 1 − 2x3 (v) Given f(x) = , find f ‘(x). x −1 (c) (i) Given h(x) =

(ii) Given

1 , find the value h’’(1). 2 (3 x − 5)

(x f(x) =

− 2) , find f '(0). 1 − 3x 5

2

limit ⎛ n 2 − 4 ⎞ ⎜ ⎟. n→2⎝ n−2 ⎠

(d) (i) Find the limit of (ii) Given f (r) =

4 + 3r . Find the limit of f (r) when r → 1. 5 − 2r

d2y dy +x + 12 in terms of x, in the simplest form. 2 dx dx d2y dy + 12 = 0. Hence, find the value of x which satisfy the equation y 2 + x dx dx

(e) Given y = x(3 − x), express y

Answer : (a) (i) 6x 4 (ii) − x2

(b)( i) −

2 2

(2 x + 1) (ii)[4(2x − 1)4)(12x − 1)] iii )6 x(6 x − 5)(2 x − 5) 3 iv) 40(2x − 3)3

v)

− 4x3 + 6x 2 − 1

( x − 1) 2 27 c) (i) 8 (ii) −96


d) i) 2 7 ii) 3

e) 12 − 3x; x=4

- 17 2. (a) Given the function of the graph f(x) = hx3 + f ‘ (x) = 3x2 −

k x2

, which has a gradient function of

96

, where h and k are constant. Find, x3 (i) the value of h and the value of k, (ii) the coordinate x of the turning point of the graph. (b) The point P lies on the curve y = ( x − 5) 2 . It is given that the gradient of the 1 normal at P is − . Find 4 (i) the coordinates of P. (ii) the equation of the normal to the curve at point P. (c) A curve with the gradient function 2 x −

2 x2

has a turning point at ( k , 8 ) .

(i) Find the value of k . (ii) Determine whether the turning point is a maximum or a minimum point . Answer: 2(a) (i) h=1, k= 48 (ii) 2

(b) (i) (7, 4) (ii)4y + x = 23

(c) i) k = 1, ii) Minimum

2 . x Given that y increases at a constant rate of 4 units per second, find the rate of change of x when x = 2.

3. (a) Two variables, x and y, are related by the equation y = 3x +

(b) On a certain day, the rate of increase of temperature, θ°, with respect to time, t s, is dθ 1 = (12 − t). given by dt 2 (i) Find the value of t at the instant when θ is maximum. (ii) Given θ = 4 when t = 6, find the maximum value of θ. Answer: (a) 1.6 units-1


(b) i) 12

ii) 13

- 18 4. (a) Given y = t − 2t2 and x = 4t + 1. dy in terms of x. dx (ii) If x increases from 3 to 3.01, find the corresponding small increment in t.

(i) Find

(b) Given y = 2x3 − 5x2 + 7, find the value of

dy at the point (2, 3). dx

Hence, find (i) the small change in x which causes y to decrease from 3 to 2.98. (ii) the rate of change of y when x = 2 if the rate of change of x is 0.6 unit per second. (c) Given y = value of

16 dy , find the value of when x = 2. Hence, find the approximate 4 dx x

16 . (198 . )4

Answer: (a) (i)

2− x (ii) 0.0025 4

(b) (i) −0.005 (ii) 2.4 unit s−1

(c) 1.04

5. Diagram 1 shows a composite solid made up of a cone resting on a cylinder with radius x cm.

x cm Diagram 1 16 ⎞ ⎛ The total surface area of the solid, A cm2, is given by the equation A = 3π ⎜ x 2 + ⎟ . x⎠ ⎝ (a) Calculate the minimum value of the surface area of the solid. (b) Given the surface area of the solid is changing at a rate of 42π cm2 s−1. Find the rate of change of radius at the instant when the radius is 4 cm. (c) Given the radius of the cylinder increases from 4 cm to 4.003 cm. Find approximate increment in the surface area of the solid Answer: b) 36π


(c)

2

(d) 0.063π

- 19 -

PROGRESSION 1.

Show that log h, log hk, log hk 2 , log hk 3 ,…… is an arithmetic progression. Then find the common difference of this progression.

2.

An arithmetic progression has 10 terms. The sum of all these 10 terms is 220. The sum of the odd terms is 100. Find the first term and the common difference. Answer : a = 4, d = 4

3.

The sum of the first six terms of an arithmetic progression is 120. The sum of the first six terms is 90 more than the fourth term. Calculate the first term and the common difference. Answer : a = −30, d = 20

4.

Given that the sum of n term of an arithmetic progression is S n = 2n 2 + 3n. Find (a) the n term in terms of n (b) the first term (c) the common difference Answer : (a) 4n + 1 (b) 5( c) 4

5.

An arithmetic progression has 12 terms. The sum of all these 12 terms is 222. The sum of the odd terms is 102. Find (a) the first term and the common difference (b) the last term Answer :(a) a = 2 , d = 3 (b) 35

6.

The n term of an arithmetic progression is 5n − 8 . Find the sum of all the terms from the 5th term to the 8th term. Answer : 98

7.

Estimate the sum to infinity of the geometric progression 9 + 3 + 1 +

1 + ....... 3 Answer :

8.

Write 0.7 + 0.07 + 0.007 + 0.007 + ........ as a fraction.

9.

The sum of the first n terms of a geometric progression is S n = number of terms in the progression that its sum to exceed 60.

10.

13

Answer :

1 2 7 9

(2 2 n +1 ) − 8 . Find the least 3 Answer : n = 4

Find the least number of terms of the geometric progression 4,12,36,……which must be taken for its sum to exceed 1 800. Answer : n = 7


- 20 11.

The sum of the first two terms of a geometric progression is

3 and the sum of the next 4

3 , where the common ratio is positive. Find the sum to infinity of the 16 progression. Answer : S ∞ = 1

two terms is

12.

Diagram 1 Diagram 1 shows four circles. Each circle has a radius that is 2 units longer than that of the previous circle. Given that the sum of the perimeters of these four circles is 120π cm, (i) find the radius of the smallest circle. (ii) the sum of the perimeters from the fifth term to the tenth term. Answer :(i) r = 12 cm(ii) 300π cm 13.

Encik Rahim plans to donate an amount of money to the ‘Rumah Penyayang’ each year from 2008. The amount in 2008 will be RM50 000, and thereafter, the amount each year will be 90% of the amount for the previous year. Calculate (a) the year in which the donation falls below RM 20 000 for the first time

.

(b) the total donation from 2008 to 2015 inclusive Answer : a) n = 10 , 2017 14.

P

b) RM284 766.40

Q

Diagram shows two balls in a tube of length 10 m, moving towards each other. P moves from one end traveling 60 cm in the first second, 59 cm in the next second and 58 cm in the third second. Q moves from the other end traveling 40 cm in the first second, 39 cm in the next second and 38 cm in the third second. The process continues in this manner until the two balls meet. (a) Find the shortest time for the two balls to meet.(give your answer to the nearest second) (b) Calculate the distance traveled by P. (c) Calculate the difference in distance traveled by the two balls. Answer :

a) 11s


b) 605 cm

c) 220 cm

- 21 -

INTEGRATION

∫

1. (a) Find, (i)

∫

(b) Given (c) Find

3

0

(4 − x )(4 + x ) dx x2

(ii)

f(x) dx = 8, find the value of

∫ (2x + 7)

3

∫

3

0

18

∫ (3x − 5)

3

dx

f ( x) + 2 dx. 2

dx

(d) Given kx2 − x is the gradient function for a curve such that k is a constant. y − 5x + 7 = 0 is the equation of tangent at the point (1, −2) to the curve. Find, (i) the value of k, (ii) the equation of the curve. 1 dy d2y (e) Given and = 3. Find y in terms of x. = 4x3 + 1. When x = −1, y = 2 2 dx dx (2 x + 7 )4 + c 3 16 + c (b) 7 ( c) Answer:( a) (i) − − x + c (ii) − x (3 x − 5) 2 8 (d(i)) k = 6 (ii) y = 2x3 −

2.

3.

x2 7 − 2 2

(e) y =

16 x5 x 2 + + 3x + 5 2 5

dy dy = 9, when x = 2 , is directly proportional to x 2 − 1,and that y = 3 and dx dx find the value of y when x = 3 . Answer: 19 2 2 The rate of change of the area, A cm , of a circle is 6t − 2t + 1. Find in terms of if the Given that

area of the circle is 11 cm2 when t = 2 . Answer: 4.

2 . The straight line x = k divides the x2 2 shaded region enclosed by the curve y = 2 , the straight lines x = 1 and x = 5 and x the x-axis into two regions, and . y

(a) Diagram 1 shows part of the curve of y =

A 0

1

B k 5 Diagram 1

x


- 22 Given that the area of region

is five times the area of region , find the value of

(b) Diagram 2 shows part of the curve y = x ( x − 2 ) . y

y = x ( x − 2) x

0

Diagram 2 Find the value of the solid generated when the shaded region is revolved through 360 about the x-axis . Answer: (a) k = 3 (b) 1 5.

(a) Diagram 3 shows a straight line y = 2 x and a curve y = x 2 − 3x y

P x

y = x2 − 3x x

0

Diagram 3 Find (i) the coordinate of the point P, (ii) the area of the shaded region Answer: ( i )( 5,10 ) ( ii ) 6.

Diagram 4 shows part of the curve y =

4

( 2 x − 1)

2

which passes through

y

y=

4

( 2 x − 1)

2

0

x Diagram 4


125 6

- 23 (a) Find the equation of the tangent to the curve at point Q. (b) A region is bounded by the curve, the x-axis and the straight lines

and

(i) Find the area of the region (ii) The region is revolved through 360 about the x-axis. Find the volume generated, in terms of Answer:(a) y = −6 x + 20 (b) i)

,ii )

784 π 10125

4

7.

(a) Evaluate

∫ x(4 − x)dx 0

Diagram 5 (b) Diagram 5 shows the curve y = x(4 − x) , together with a straight line. This line cuts the curve at the origin O and at the point P with x-coordinate k, where 0 < k < 4 . 1 (i) Show that the area of the shaded region, bounded by the line and the curve, is k 3 6 (ii) Find, correct to 3 decimal places, the value of k for which the area of the shaded region is half of the total area under the curve between x = 0 and x = 4 .

8.

Answer : k = 3.175 1 2 Diagram 6 shows, the straight line PQ is normal to the curve y = x + 1 at A(2,3). 2 The straight line AR is parallel to the y – axis. y

P A(2,3)

0

R Diagram 6

Q(k ,0)


- 24 Find (a) the value of k, (b) the area of the shaded region, (c) the volume generated, in terms of π, when the region bounded by the curve, the y – axis and the straight line y = 3 is revolved through 360o about the y-axis. 1 (c) Volume = 4 π 3 9. Diagram 7 below shows the straight line y = x + 4 intersecting the curve y = (x – 2 )2 at the points A and B. y

Answer : (a) k = 8 (b) area = 12

B

y = x+4

y = (x – 2 )2 P k

A

Q

x Diagram 7

Find, (a) the value of k, (b) the are of the shaded region P (c) the volume generated, in terms of π, when the shaded region Q is revolved 360o about the x – axis. 32 Answer : (a) k = 5 (b) area = 20.83 (c) volume = π 5 ****************************************************************************** LINEAR LAW 1.

The data for x and y given in the table below are related by a law of the form

y = px 2 + x + q ,where p and q are constants. x y

1 41.5

2 38.0

3 31.5

4 22.0

5 9.5

(a) Plot y − x againts x 2 , using a scale of 2 cm to 4 unit on both axes. Hence , draw the line of best fit. (b) Use your graph in 1(a) to find the value of (i) p (ii) q


Answer:

- 25 2.

The variables x and y are known to be connected by the equation y = Ca − x . An experiment gave pairs of values of x and y as shown in the table. One of the values of y is subject to an abnormally large area. x y

1 56.20

2 29.90

3 25.10

4 8.91

5 6.31

6 3.35

(a) Plot log y against x, using a scale of 2 cm to 1 unit for x-axis and 2 cm to 0.2 unit for y-axis . Hence , draw the line of best fit. (b) Identify the abnormal reading and estimate its correct value. (c) Use your graph in 2(b) to find the value of (i) C (ii) a Answer: (b)25.1,17.78 (c) C = 100 , a = 1.78 3.

The table shows experimental values of x and y which are known to be related by equation a y = +b x . x 1 2 3 4 5 6 x 2.20 1.74 1.71 1.77 1.86 1.96 y (a) Explain how a straight line graph may be drawn to represent the given equation. (b) Plot xy againts x x , using a scale of 2 cm to 2 unit on both axes . Hence , draw the line of best fit. (c) Use your graph in 3(b) to find the value of (i) a (ii) b Answer : a=1.5 ; b=0.70

4. Table 1 shows the values of two variables, x and y, obtained from an experiment.

Variables x and y are related by the equation y = p k x, where p and k are constants. x y

2 3.16

4 5.50

6 9.12

8 16.22

10 28.84

12 46.77

(a) Plot log y against x by using a scale of 2 cm to 2 units on the x-axis and 2 cm to 0.2 unit on the log10 y-axis. Hence, draw the line of best fit. (b) Use your graph from (a) to find the value of (i) p (ii) k. Answer : p=1.820 ; k=1.309


- 26 5. The variable x and y are related by the equation y =

h . Diagram 1 shows the graph of 2x + k

1 againts x .Calculate the values of h and k. The point P lies on the line. Find the value of r. 1 y y (

x

Diagram 1 Answer:

6. Variables x and y are related by the equation the graph of the

1 1 againts y x

a b + = 2 , where a and b are constants. When x y

is drawn, a straight line is obtained. Given that the intercept on

1 − axis is − 0.5 and that the gradient of the line is 0.75, calculate the value of a and b. y Answer: a = 3, b = – 4

7. Variables x and y are related by the equation 4y = 2(x – 1)2 + 3k where k is a constant. (a) When y is plotted against (x – 1)2, a straight line is obtained, which intersects the y-axis at (0, -6). Find the value of k. (b) Hence, find the gradient and the y intercept for the straight line obtained by plotting the graph of (y + x) against x2.

Answer: (a) y = ½ (x – 1)2 +

3k 11 11 , k = –8 (b) y + x = ½ x2 – , m= ½ , y-intercept = − . 4 2 2


- 27 8. (a) Explain how a straight line graph can be drawn from the equation

y p = + qx , x x

where p and q are constants. log2 y (b) (5, 6)

log2 x O

(2, 0) Diagram 2

Diagram 2 shows the graph of log2 y against log2 x. Values of x and values of y are related by x2n the equation y = , where n and k are constants. Find the value of n and the value of k. k Answer : n = 1, k = 16

9.

y

•(4, 44) • (2, 14) x

0 Diagram 3

Diagram 3 shows part of the curve y against x. It is known that x and y are related by the y linear equation = kx + h, where h and k are constants. x (a) Sketch the straight line graph for the above equation. (b) Calculate the values of h and k. Answer : : (a) h = 3, k = 2


- 28 10.

log10 y Q(8, 7)

log10 x

0 Diagram 4

Diagram 4 shows graph of log10 y against log10 x. Given that PQ = 10 units and the point P lies on the log10 y-axis. (a) Find the coordinates of P. (b) Express y in terms of x. (c) Find the value of y when x = 16. 3

Answer: (a) P(0, 1) (b) y = 10 x 4 (c) y = 80 ****************************************************************************** VECTORS

1. Diagram 1 shows a parallelogram, OPQR, drawn on a Cartesan plane. y Q

R

P x

O Diagram 1 → → → Given that OP = 6 i + 4 j and PQ = −4 i + 5 j . Find PR . ~

~

~

~

Answer: −10i + j

2. Given O(0, 0), A(−3, 4) and B(2, 16), find in terms of unit vector i and j , ~

uuur (a) AB , uuur (b) unit vector in the direction of AB . Answer:


~

( a ) 5i + 12 j

(b)

1 5i + 12 j 13

(

)

- 29 3. Given −2, 6), B(4, 2) and C(m, p), find the value of m and the value of p such that uuur A(uuu r AB + 2 BC = 10 i − 12 j . % % Answer: m = 6, p = −4 4. Given the points A(3,0) , B(7,8) and C (1, k ) (a) Express vector AB in terms of i and j , (b) Find the value of k if vector OC is parallel to vector AB . Answer: ( a ) 4i + 8 j (b) h = →

1 ,k = 2 4

→

5. Given that OABC is a rectangle where OA = 6 cm and OC = 5cm. If OA = a and OB = b ,find ~

~

(a) AC in terms of a and b ~

(b)

~

a+b Answer:(a) −a + b (b) 61

6. Diagram 2 shows vector s , vector t and vector unit a and b . ~

~

~

~

s ~

t

~

b %

a

Diagram 2

~

Given r = 2 s − 3 t , express r in terms a and b . ~

~

~

~

→

~

Answer: 14a + 13b

~

→

→

→

7. Given AB = (k + 1) a and BC = 2 b . If A, B and C are collinear, AB = BC and b = 3 a . ~

~

~

~

Find the value of k.

Answer:k= 5


- 30 8. Given that a = − 2 i + 2 j , b = 2 i − 3 j and c = a − 2b . Find % % % % % % % % % (a ) c

%

(b) unit vector in the direction of c . %

Answer: ( a )10

C

( b)

1 −3i + 4 j 5

(

)

9. G

H

A

B Diagram 3

→ → uuur Diagram 3 shows GH : AB = 3 : 10 and GH is parallel to AB . If AB = 10 a , find GH in ~

terms of a .

Answer: 3a

~

10.

P

Q

U

R T

S

→ → → Diagram 4 Diagram 4 shows PQRSTU is a regular hexagon. Express PQ + PT - RS as a single vector. uuur Answer: PR

11.

→

→

In Δ OPQ, OP = p and OQ = q . T is a point on PQ where PT : TQ=2 : 1. Given that M ~

~

→

is the midpoint of OT, express PM in terms of p and q . ~

~

Answer: −


5 1 p+ q 6 3

- 31 12. Diagram 5 shows triangles OAB. The straight line AP intersects the straight line OQ at R. → → 1 1 It is given that OP = OB , AQ = AB, OP = 6 x and OA = 2 y ~ 3 4 ~ A Q

Diagram 5

R O

B

P

(a) Express in terms of x and/or y : ~

→

(i)

→

→

AP ,

(ii) OQ

~ →

→

(b) (i) Given that AR = h AP , state AR in terms of h, x and y . ~

→

→

~

→

(ii) Given that RQ = k OQ , state RQ in terms of k, x and y . ~

→

~

→

(c) Using AR and RQ from (b), find the value of h and of k

Answer:(a)(i) −2 y + 6 x (ii)

9 3 3 ⎛9 x + y (b)(i) h −2 y + 6 x (ii)k ⎜ x + 2 2 2 ⎝2

(

)

1 1 ⎞ y ⎟ (c) k = , h = 3 2 ⎠

13. Diagram 6, ABCD is a quadrilateral. AED and EFC are straight lines. D

F •

E

C B

A →

→

→

It is given that AB = 20 x , AE = 8 y, DC = 25 x − 24 y, AE = ~

~

(a) Express in terms of x and/or y : ~

~

(i)

~

→

BD ,

1 3 AD and EF = EC 5 4 →

(ii) EC

~

(b) Show that the points B, F and D are collinear →

(c) If x = 2 and y = 3 , find BD ~

~

Answer:(a)(i) −20 x + 32 y (ii) 25 x (c) 104


- 32 TRIGONOMETRIC FUNCTIONS

1. Prove that cosec2x – 2 sin2x – cot2x = cos 2x 2. Prove that tan 2 A sin 2 A ≡ tan 2 A − sin 2 A

3. Prove that

sin 2 x = cot x . 1 − cos 2 x

4. Find all the angles between 0o and 360o which satisfy (a) 3cos2α – 5 = 8 cos α (b) tan 2α tan α = 1 Answer ( a )131.81° , 228.19° (b)30° ,150° , 210° ,330° : 5. Solve the equation 4 sin θ + 3 cos θ = 0 for 0 0 ≤ θ ≤ 360 0 Answer 36.87° , 216.87° : 6. Find all the angles between 0o and 360o which satisfy (a) 3sin 2A = 4sin A (b) 5sin2A = 5 – cos 2A Answer ( a ) 0° ,180° ,360° , 48.19° ,311.81° ( b ) 35.27° , 215.27° ,144.73° ,324.73° :

8 12 , 90o < α < 270o and sin β = − , 90o < β < 270o. 17 13 Calculate the value of (a) sin (α + β ) (b) cos ( β – α ) 140 21 Answer ( a ) ( b) − : 221 221

7. Given that sin α =

8. Given that cos x =

3 and 0o ≤ x ≤ 180o , find sec x + cosec x . 5 Answer

35 : 12

9. Given that sin Ө = k and Ө is acute angle, express in term of k: (a) tan Ө (b) cosec Ө Answer ( a )


k 1− k

2

(b)

1 : k

- 33 -

10. Solve the equation 5sin 2 A + 2cos 2 A − 3 = 0, 0o ≤ A ≤ 360o Answer 35.27° ,144.73° , 215.27° ,324.73° :

11. Solve the equation

sin 2 x = cot x . 1 − cos 2 x

12. (a) Prove that

(b) Given cos

1 + sek 2 x = 3 for 0° ≤ x ≤ 360°. 2 cot x Answer 45° ,135° , 225° ,315° :

θ 2

=

1 1 + p2

,

2p . 1 − p2 ii. hence, find sin 2θ , when p = 2 . i. prove that tan θ =

Answer ( b )( i ) −

24 : 25

13. (a) Prove that tan θ + cot θ = 2 cosec 2θ . 3 x for 0o ≤ x ≤ 2π . 2 (ii) Find the equation of a suitable straight line for solving the equation 3 3 cos x = x − 1 . Hence , using the same axes , sketch the straight line and 2 4π 3 3 state the number of solutions to the equation cos x = x − 1 for 2 4π 0o ≤ x ≤ 2π.

(b) (i) Sketch the graph y = 2 cos

Answer:(b)(ii)3

14. (a) Sketch the graph of y = cos 2x for 0o ≤ x ≤ 180o. (b) Hence , by drawing a suitable straight line on the same axes , find the number of x for 0o ≤ x ≤ 180o. solutions satisfying the equation 2 sin2 x = 2 – 180 Answer:(b)2


- 34 15. (a) Prove that cosec2 x – 2 sin2 x – cot2 x = cos 2x. (b) (i) Sketch the graph of y = cos 2x for 0 ≤ x ≤ 2π . (ii) Hence , using the same axes , draw a suitable straight line to find the number x – 1 for of solutions to the equation 3(cosec2 x – 2 sin2 x – cot2 x ) =

π

0 ≤ x ≤ 2π . State the number of solutions .

Answer:(b)(ii)4

16. (a) Sketch the graph of y = - 2 cos x for 0 ≤ x ≤ 2π . (b) Hence , using the same axis , sketch a suitable graph to find the number of solutions to the equation

π

x

+ 2 cos x = 0 for 0 ≤ x ≤ 2π . Answer:2

17. (a) Given tan A =

3 ,and A is an acute angle. Find the value of cos 2A. 4

(b)(i) Sketch the curve y = sin 2x for 0 ≤ x ≤ 2 π . (ii) Hence, by drawing a suitable straight line on the same axes, find the number of solutions satisfying the equation sin x kos x =

x 1 for − 4π 2

0 ≤ x ≤ 2π .

Answer: ( a )

7 ( b )( i ) 4 25

**************************************************************************** PERMUTATION AND COMBINATION 1. Four girls and three boys are to be seated in a row. Calculate the number of possible arrangements (a) if all the three boys have to be seated together (b) a boy has to be seated at the centre Answer:(a)720 (b)2880

2. Find the number of the arrangement of all nine letters of word SELECTION in which the two letters E are not next to each other Answer:282240

3. Calculate the number of four digit even number can be formed from the digits 3, 4, 5, 6 and 9 without repetitions.

Answer:48


- 35 4. Three alphabets are chosen from the word WALID. Find the number of possible choice if (a) the alphabet A is chosen (b) the alphabet A and D are chosen Answer:(a) 6 (b) 3

5. A bowling team consists of 8 person. The team will be chosen from a group of 7 boys and 6 girls. Find the number of team that can be formed such that each team consists of (a) 3 boys (b) not more than 1 girl Answer:(a) 210 (b) 6

6. Refrigerators

TV

TV

Refrigerators

TV

Pak Adam’s shop has 5 televisions P, Q, R, S and T and 4 refrigerators W, X, Y and Z (a) If a televisions and a refrigerator is chosen randomly, calculate the probability that television P or Q and refrigerator W are chosen. (b) Pak Adam wish to display his goods as shown in the diagram above. Calculate the number of ways the goods can be displayed. 1 Answer: (a) (b) 720 10 7.. Diagram 1 shows 5 letter and 3 digits. A

B

C

D

E

6

7

8

Diagram 1

A code is to be formed using those letters and digits. The code must consist of 3 letters followed by 2 digits. How many codes can be formed if no letter or digit is repeated in each code ? Answer:144

8. A debating team consists of 5 students. These 5 students are chosen from 4 monitors, 2 assistant monitors and 6 prefects. Calculate the number of different ways the team can be formed if (a) there is no restriction (b) the team contains only one monitor and exactly 3 prefects Answer:(a)792 (b)160


- 36 -

PROBABILITY 1. At the place where Lam stays, rain falls in any two days of a week. Out of 75% of the raining days, Lam goes to school in his father’s car. If there is no rain, Lam cycles to to school. For every 5 days Lam goes to school in his father’s car, for 3 days Lam is able to keep his pocket money. In a certain day, find the probability that (a) Lam does not goes to school in his father’s car, (b) Lam keeps his pocket money because he goes to school in his father’s car. 11 9 Answer:- (a) (b) 14 70 2. Rashid and Rudi compete in a badminton game. The game will end when any of the 3 players has won two sets. The probability that Rashid will win any one set is . 5 Calculate the probability that (a) the game will end in only two set, (b) Rashid will win the competition after playing 3 sets. 13 36 (b) Answer:- (a) 25 125 3. A container consists of 4 soya beans, 3 coffee beans and 2 cocoa beans. (a) If a bean is drawn at random from the container, calculate the probability that the bean is not a cocoa bean. (b) Two beans are drawn at random from the container, one after the other, without replacement. Find the probability that only one bean out of the two beans is a cocoa bean. .Answer:- (a) 7/9 , (b) 7/18 4. Bag P contains five card numbered 5, 6, 7, 8 and 9. Bag Q contains three cards numbered 5, 7 and 9. A card is drawn at random from bag P and at the same time, another card is drawn from bag Q. Find the probability that the two numbers drawn have the same value or their product is an even number. Answer:

Box P Q

Green 6 3

3 5

Number of marbles Red Yellow 7 2 5 8 Table 1

5. Table 1 shows the number of marbles of different colours in boxes P and Q. A marble is picked at random from each box. Find the probability that (a) both are of the same colour, (b) both are of different colours, (b) a yellow marble is picked from box Q. Answer: (a) 23/80 (b) 57/80 (c) ½


- 37 6. Diagram 1 shows a board with a grid of 20 squares, of which a few squares are shaded. A dart is thrown at the board.

Diagram 1 (a) Find the probability that it will hit a shaded square. (b) Find how many additional squares need to be shaded if the probability is increased 3 Answer: (a) 7/20 , (b) 5 to . 5 7. A box contains 4 blue balls, x white balls and y red balls. A ball is drawn at 1 and the random from the box. If the probability of getting a white ball is 6 1 probability of getting a red ball is , find the values of x and y . 2 Answer: x = 2, y = 6 8. The letters of the word G R O U P S are arranged in a row. Find the probability that an arrangement chosen at random (a) begins with the letter P, (b) begins with the letter P and ends with a vowels. Answer: (a) 1/6 , (b) 1/ 15 9. A bag contains 6 red balls and 5 green balls. A ball is chosen from the bag and returned. Another ball is chosen and returned again. Find the probability that (a) both balls are red (b) both balls have same color, (c) both balls have different color. Answer: (a) 36/121 , (b) 61/121 , (c) 60/121

10. There are 7 ribbons in a bag. 1 yellow, 3 black and 3 blue ribbons. (a) If a ribbon is taken out and not returned back, find the probability for the ribbon to be black. (b) If two ribbons are taken, find the probability first one to be blue followed by a black if none of the ribbons are returned. (c) If three ribbons are taken, find the probability for first one to be blue, followed by yellow and a black ribbon if none of the ribbons are returned. Answer: (a) 3/7 , (b) 3/14 , (c) 3/70


- 38 -

PROBABILITY OF DISTRIBUTION 1. (a) A study in a district shows that one out of three teenagers in the district join the `Rakan Muda’ program. (i) If 5 teenagers are chosen randomly from the district, find the probability that 2 or more of them join the `Rakan Muda’ program. (ii) If they are 2 490 teenagers in the district, calculate the mean and the standard deviation of the number of teenagers who join the `Rakan Muda’ program. (b) From a study, it is found that the mass of a deer from a certain jungle shows a normal distribution with mean 55 kg and variance 25 kg2. (i) If a deer is caught randomly from the jungle, find the probability that the deer has a mass more than 60 kg. (ii) Find the percentage number of dears with mass between 45 kg and 60 kg. 131 or 0.5391 (ii) 830, 23.52 (b)(i) 0.1587 (ii) 81.85% 243 2. (a) Usually, when fishing, Wan will get fish as many as 60% from the total number of his throws.. Calculate, (i) the probability Wan will get at least 4 fishes in 5 throws, (ii) the minimum number throws made by Wan so that the probability of getting at least a fish is greater than 0.87. Answer: (a)(i)

(b) The mass of students in a school has a normal distribution with a mean of μ kg and a standard deviation σ kg. It is known that 10.56% of the above students have mass more than 50 kg and 15.87% of them have mass less than 32 kg. Find the value of μ and the value σ.

1053 (ii) 3 (b) σ = 8, μ = 40 3125 3. (a) In a game of guessing, the probability of guessing correctly is p. (i) Find the number of trials required and the value of p, such that the mean and 3 the standard deviation of success are 15 and 5 respectively. 2 (ii) If 10 trials are done, find the probability of guessing exactly 3 correct. Answer: (a)(i) 0.3370 or

(b) The volume of 600 bottles of mineral water produced by a factory follow a normal distribution with a mean of 490 ml per bottle and standard deviation of 20 ml. (i) Find the probability that a bottle of mineral water chosen in random has a volume of less than 515 ml. (ii) If 480 bottles out of 600 bottles of the mineral water have volume greater than k ml, find the value of k. 1 Answer: (a)(i) p = , n = 60 (ii) 0.2503 (b)(i) 0.8944 (ii) 473.16 4


- 39 4. (a) It is known that for every 10 lemon in the box, two are rotten. If a sample of 7 are chosen randomly, calculate the probability that (i) exactly 3 lemons are rotten, (ii).at least 6 lemons are not rotten. (b) The masses of the members of the English Society of School M are normally distributed with a mean of 48 kg and a variance of 25 kg2. 56 of the members have masses between 45 kg and 52 kg. Find the total number of members in the English Society. Answer: (a) (i) 0.1147

(ii) 0.5767

(b) 109

5. Diagram 1 shows a standard normal distribution graph. f(z) 0.3485

k 0 Diagram 1

z

The probability represented by the area of the shaded region is 0.3485. (a) Find the value of k. (b) X is a continuous random variable which is normally distributed with a mean of 79 and a standard deviation of 3. Find the value of X when the z-scores is k. Answer: (a) 1.03

(b) 82.09

6. Diagram 4 shows a probability distribution graph of the continuous random variable x that is normally distributed with a standard deviation of 8. The graph is symmetrical about the vertical line PQ. Q

P m

55

45 Diagram 4


x

- 40 (a) If the standard score found by using the value of x = m is

−3 , find the value of m. 4

(b) Hence, find the area of the shaded region in Diagram 4. (c) If x represents the marks obtained by 180 Form 5 students in an examination, calculate the number of students whose marks are less than 33. Answer: (a) m = 39 , (b) 0.0668, (b) 12 7. (a) A football team organizes a practice session for trainees on scoring goals from penalty kicks. Each trainee has ten goals to score. The probability that a trainee scores a goal is k. After the practice, it is found that the mean number of goals scored for a trainee is 6. (i) Find the value of k. (ii) If a trainee is chosen at random, find the probability that he scores at least two goals. (b) The masses of students of a school are normally distributed with a mean of 56 kg and a standard deviation of 10 kg. (i) If a student is chosen at random, calculate the probability that his mass is less than 50 kg. (ii) Given that 1.5% of the students have masses of more than p kg, find the value of p. (iii) If 75% of the students have masses of more than h kg, find the value of h. Answer(a)(i) k = 0.6 (ii) 0.9983 (b)(i) 0.2743 (ii)77.7 (iii) 49.25 ************************************************************************

SOLUTION OF TRIANGLES 1.

P

8 cm 6.5 cm Diagram 1

50° Q

R

Diagram 1 shows a ΔPQR. (a) Calculate the obtuse angle PRQ. (b) Sketch and label another triangle different from ΔPQR in the diagram above, so that the lengths of PQ and PR and the angle PQR remain unchanged. (c) If the length of PR is reduced whereas the length of PQ and angle PQR remain unchanged, calculate the length of PR so that only one ΔPQR can be formed Answer: (a) 109° 28’ or 109.47°


(c) 6.128 cm

- 41 2.(a) Diagram 2 shows a pyramid VABCD with a square base ABCD. VA is vertical and the base ABCD is horizontal. Calculate, (i) ∠VTU, (ii) the area of the plane VTU. V 8 cm A

B

Diagram 2

6 cm

U 2 cm D

4 cm

T

C

Answer: (a)(i) 84° 58’ or 84.97° (ii) 22.72 cm2

(b) . Q

R

4 cm K L P

Diagram 3

S

6 cm 8 cm

J

M

(i) ∠JQL, (ii) the area of ΔJQL.

Diagram 3 shows a cuboid. Calculate,

Answer: (i) 75° 38’ or 75.640

3.

A

.O

5 cm D

50° 4 cm

B

8 cm C

Diagram 4


(ii) 31.24 cm2

- 42 Diagram 4, ABCD is a cyclic quadrilateral of a circle centered O. Calculate (a) the length of AC, correct to two decimal places, (b) ∠ ACD, (c) the area of quadrilateral ABCD. Answer: (a) 11.85 cm

(b) 115.01°′ (c)36.08cm2

4. Diagram 5 shows two triangles PQT and TRS. Given that PQ = 24 cm, TS = 12 cm, ∠TPQ = 320 , PT = TQ and PTS and TRQ are straight lines. P

32o

24 cm Diagram 5 R

T

Q

12 cm S

(a) (b) (c) (d)

Find the length, in cm, of PT, If the area of triangle PQT is three times the area of triangle TRS, find the length of TR. Find the length of RS. (i) Calculate the angle TSR. (ii) Calculate the area of triangle QRS. Answer: (a) 14.15 cm, (b) 5.563 cm, (c) 10.79 cm, (d)(i) 27.60o (ii) 46.31 cm2

5. Diagram 6 shows a triangle PQR. R

P 10 cm

7 cm 75

Diagram 6

o

Q (a) Calculate the length of PR. (b) A quadrilateral PQRS is now formed so that PR is the diagonal, ∠PRS = 40o and PS = 8 cm. Calculate the two possible values of ∠PSR . (c) Using the obtuse ∠PSR in (b), calculate (i) the length of RS, (ii) the area of the quadrilateral PQRS. Answer: (a) 10.62 cm (b) 58.57o ; 121.43o (c) (i) 3.964 cm ; (ii) 47.34 cm2


- 43 6. Diagram 7 shows a camp of the shape of pyramid VABC. The camp is built on a horizontal triangular base ABC. V is the vertex and the angle between the inclined plane VBC with the base is 60°. V C A

Diagram 7

B Given VB = VC = 25 cm and AB = AC = 32 cm and ∠BAC is an acute angle. Calculate (a) ∠BAC if the area of ΔABC is 400 cm2, (b) the length of BC, (c) the lengths of VT and AT, where T is the midpoint of BC, (d) the length of VA, (e) the area of ΔVAB Answer: (a) 51.38o

(b) 27.74o

(c) 20.80 cm; 28.84 cm (d) 25.78 cm (e) 315.33 cm2

******************************************************************************

INDEX NUMBER 1. Table 1 shows the price indices and percentage usage of four items, P, Q, R, and S, which are the main ingredients of a type biscuits. Item P Q R S

Price index for the year 1995 based on the year 1993 135 x 105 130 Table 1

Percentage of usage (%) 40 30 10 20

Calculate, (a) (i) the price of S in the year 1993 if its price in the year 1995 is RM37.70 (ii) the price index of P in the year 1995 based on the year 1991 if its price index in the year 1993 based in the year 1991 is 120. (b) The composite index number of the cost of biscuits production for the year 1995 based on the year 1993 is 128. Calculate, (i) the value of x, (ii) the price of a box of biscuit in the year1993 if the corresponding price in the year 1995 is RM 32. Answer: (a)(i) RM29 (ii) 162 (b)(i)125 (ii) RM25


- 44 2.

Daily Usage ( RM) 47 34 22 12 3

V

W

X

Y

Z

Component

Diagram 1 A technology product consists of five components, V, W, X, Y and Z. Diagram 1 shows a bar chart showing the daily usage of the components used to produce the technology product. The following table shows the prices and the price indices of the components.

Component V W X Y Z

(a)

Price in the year 2001 Price in the year 2003 Price index in 2003 (RM) (RM) based on 2001 13.00 16.25 y 12.50 17.25 138 2.50 106 x 14.90 22.35 150 24.50 140 z

Find the values of x, y and z.

(b) Calculate the composite index representing the cost of the technology product in the year 2003 using the year 2001 as the base year. (c) If the total monthly cost of the components in the year 2001 is RM1.5 million, find the total monthly cost of the components in the year 2003. (d) If the cost of each component rises by 23% from the year 2003 to 2004, find the composite index representing the cost of the technology product in the year 2004 based on the year 2001. Answer:


- 45 3. (a) In the year 1995, price and price index for one kilogram of certain grade of rice is RM2.40 and 160 respectively. Based on the year 1990, calculate the price per kilogram of rice in the year 1990.

Item Timber Cement Iron Steel

Price index in the year 1994 180 116 140 124

Change of price index from the year 1994 to the year 1996 Increased 10 % Decreased 5 % No change No change Table 2

Weightage 5 4 2 1

(b) Table 2 shows the price index in the year 1994 based on the year 1992, the change in price index from the year 1994 to the year 1996 and the weightage respectively. Calculate the composite price index in the year 1996. . Answer : (a) 1.50 (b) ITimber = 198, ICement = 110.2; 152.9 4. Table 3 shows the price indices and the weightages of Azizan’s monthly expenses in the year 2005 based in the year 2004.

(a) (b) (c) (d)

Expenses

Price index in 2005 based on 2004

Weightage

Rental

108

3

Food

120

4

Car installment

102

2

Miscellaneous

112

1

Table 3 If the expenses for miscellaneous in the year 2005 was RM 1 456 , find the miscellaneous expenses in the year 2004. If the rental increases by 10% from the year 2005 to the year 2006,find the price index for the rental in the year 2006 based on the year 2004. Calculate the composite index for the expenses in the year 2005 based on the year 2004. The price index for food in the year 2006 based on the year 2205 is 105. If the expenses on food in the year 2006 were RM3150, find the expenses on food in the year 2004. Answers : a) 1300


b) 118.8

c) 112

d) 2500

- 46 -

LINEAR PROGRAMMING 1. An institution offers two computer courses, P and Q. The number of participants for course P is x and for course Q is y. The enrolment of the participants is based on the following constraints: I: The total number of participants is not more than 100. II : The number of participants for course Q is not more than 4 times the number of participants for course P. III : The number of participants for course Q must exceed the number of participants for course P by at least 5. (a) Write down three inequalities, other than x ≥ 0 and y ≥ 0, which satisfy the above constraints. (b) By using a scale of 2 cm to 10 participants on both axes, construct and shade the region R that satisfies all the above constraints. (c) Using your graph from (b), find (i) the range of the number participants for course Q if the number participants for course P is 30. (ii) the maximum total fees per month that can be collected if the fees per month for course P and Q are RM50 and RM60 respectively. Answer:-(a) x + y ≤ 100, y ≤ 4x, y – x ≥ 5

2.

(c) (i) 35 ≤ y ≤ 70 (ii) Point (20, 80), RM5 800

A food analysts is supplied with two containers of food, Whiskers and Friskies. The comparison of one scoop of food from each of the two containers is shown in the following table.

Food 1 scoop Whiskers 1 scoop Friskies

Fat Carbohydrate Fibre 8 gm 48 gm 10 gm 16 gm 32 gm 10 gm Table 1 The analysts knows that an animal requires at least 96 gm of protein, 80 gm of fat, 288 gm of carbohydrate and not more than 100 gm of fibre each day. (a) (b)

(c)

Protein 24 gm 8 gm

If the analysts mixed x scoops of Whiskers with y scoops of Friskies, write down the system of inequalities satisfied by x and y. Hence, by using 2 cm to 2 unit on both axes construct and shade the region R that satisfies all the above constraints. If 1 scoop of Whiskers costs RM2 and 1 scoop of Friskies costs RM3, find the mixture that provides (i) the cheapest food. (ii) the most expensive food. Could the animal be fed on a satisfactory diet using food from Whiskers only, (i) food from Friskies only. (ii) Give your reason.

Answer:-(a) 3 x + y ≥ 12 , x + 2y ≥ 10 , 3x + 2y ≥ 18 , x + y ≤ 10 (b) (i)RM 17 (ii) RM 29


- 47 3.

An air craft company is going to purchase planes of type Wing and X – far . They will purchase x units of X – far and y units of Wing planes. The company has set the condition below:I: II : III:

X – far plane consume 100 liters of fuel for a single month. Wing planes consume 70 liters of fuel. Total fuel consumption for one month is at most 3500 liters. X – far planes can take in 200 passengers while Wing planes can take in 100 passengers. Total passengers the planes must take at any time must be at least 3000. Total number of planes purchased must at least 20 units.

(a) (b)

State the inequality that defines the condition above other than x ≥ 0 and y ≥ 0 . Construct the graphs and mark the region R that represents the conditions above. Use a scale of 2 cm for 5 Wing planes and 2 cm for 5 X – far planes.

(c)

The company makes a profit of RM220 for X – far and RM165 for Wing planes from its sales. Identify the minimum amount of profit that the company will obtain. Answer:-(a) 10 x + 7y ≤ 35, 2x + y ≥ 30, x + y ≥ 20 (c) RM3870

4.

A furniture workshop produces tables and chairs. The production of tables and chairs involve two processes , making and shellacking. Table 3 shows the time taken to make and to shellack a table and a chair. Product

Time taken (minutes) Making Shellacking Table 60 20 Chair 40 10 Table 3 The workshop produces x tables and y chairs per day. The production of tables and chairs per day is subject to the following constraints. I: II: III: (a) (b) (c)

The minimum total time for making tables and chairs is 600 minutes. The total time for shellacking tables and chairs is at most 240 minutes. The ratio of the number of tables to the number of chairs is at least 1 : 2.

Write three inequalities that satisfy all of the above constraints other than x ≥ 0 and y ≥ 0 . By using a scale of 2 cm for 2 units of furniture on both axes , construct and shade the region R which satisfies all of the above constraints. By using your graph from (b), find, the maximum number of chairs made if 8 tables are made. (i) the maximum total profit per day if the profit from one table is RM30 and from (ii) one chair is RM20. Answer:-(a)

3 x + 2y ≥ 30, 2x + y ≤ 24, y ≤ 2 x (c)(i) 8


(ii)RM420

- 48 **************************************************************************** MOTION ALONG THE STRAIGHT LINE 1. A particle moves in a straight line and passes through a fixed point O. Its velocity, v ms −1 , is given by v = t 2 − 6t + 5 , where t is the time, in seconds, after leaving O . [Assume motion to the right is positive.] (a) Find (i) the initial velocity of the particle, (ii) the time interval during which the particle moves towards the left, (iii) the time interval during which the acceleration of the particle is positive. (b) Sketch the velocity-time graph of the motion of the particle for 0 ≤ t ≤ 5 . (c) Calculate the total distance traveled during the first 5 seconds after leaving O. Answer: (a) (i) v= 5 (ii) 1 < t < 5 (iii) t > 3 (c) 13 m

2.

Diagram 1 shows the object, P, moving along a straight line and passes through a fixed point O. The velocity of P, v m s─1 , t seconds after leaving the point O is given by v = 3t2 – 18t + 24 . The object P stops momentarily for the first time at the point B. P O

B

Diagram 1 (Assume right-is-positive) Find: (a) the velocity of P when its acceleration is 12 ms – 2 , (b) the distance OB in meters, (c) the total distance travelled during the first 5 seconds. Answer: (a) 9 ms – 1 (b) OB = 20 m (c) 28 m 3. The velocity of an object which moves along a straight line, v ms−1, t s after passing through a fixed point O is v = pt − qt2, where p and q are constants. It is known that 1 the object moves through a distance of 7 m in the 2nd second of it’s motion and 3 experiences a retardation of 4 ms−2 when t = 3. (a) Find the value of p and of q. (b) It is also known that the object moved with a velocity of 6 ms−1 initially at the point A and again at the point B. Find the time taken for the object to move from A to B. (c) Hence if the object stops momentarily at the point C, find the distance between the point B and the point C.


- 49 Answer: (a) p = 8, q = 2

4.

(b) 2 s

1 (c) 3 m 3

Diagram 2 shows the object P and Q moving in the direction as shown by the arrows when the objects P and Q pass through the points A and B respectively. P A

Q 60m

B

Diagram 2 The displacement of the object P from A is represented by sP and the displacement of the object Q from B is represented by sQ. Given sP = t2 + 4t and sQ = t2 - 8t, where t is the time, in seconds , after P and Q pass through point A and B respectively and simultaneously . Given AB = 60 m. (a) Find the time and position where the objects meet. (b) Find the time and position of object Q when it reverses its direction of motion. (c) Find the velocity of object P when object Q reverses its direction of motion. (d) Sketch the graph of displacement - time for object Q for 0 ≤ t ≤ 10. (e) Find the time interval when the object Q moves to the left. Answer: a) 5 s, 45 m on the right of A b) 4 s, 16 m on the left of B. c) 12 m s-1 (e) 0 ≤ t < 4.


- 50 -

Permutation and Combination 1 a)

b)

4 Girls Answer :

3 Boys 5! , 3! = 720

6 • 5 • 4 • 3 • 3 • 2 • 1 = 2160 Centre

2 3

9 ! - ( 8! x 2! ) = 282240 4 x 3 x 2 x 2 = 48 Centre

4

W A L I D a) 1 x 4 C 2 = 6 b) 1 x 1 x 3 C 1 = 3

5

7 boys, 6 Girls a) 7 C 3 x 6 C5 = 210 b) 6 C 1 x 7 C7 = 6

6

a) (

5

b)

7 8

5

1 1 1 1 + )× = 5 5 4 10 P3 x

P3 x

a)

12

c)

6

3

4

P 2 = 720

P 2 = 360

C 5 = 792 C 1 x 7 C7 X 2 C 1 = 160


Number 1

Solution and marking scheme a −

1 ≤ f ( x) ≤ 2 2

Sub Marks

Full Marks

1

3

GRAF

2

1

2

2

a) m = 2 and n = 29 fg(x) =

2 x + 29 5

1

2

3 a = 2, b = 4, c = 8, d = −6

4

a = 2, b = 4 or c = 8, d = −6

3

4a + b = 12 and a + b = 6 or

12 12 = 3 and =d 12 − c 6−8

Either one equation correct

3

2

1

0.9537 , −1.3981

3

Using formula or other method

2

4 3 x + 3 px − p + 3 = 0

1

4 x 2 − 12 x + 1 = 0

4

2

5

α 2 + β 2 = 3 and α 2 β 2 =

1 4

1 α + β = 3 or α β = 4 α + β = 2 or αβ = 1/2 2

2

2

2


3 4 2 1

Number 6

Solution and marking scheme a) − 3( x − 1) + 2 2

Sub Marks

Full Marks

2

2 2 ⎡ ⎛2⎞ ⎛2⎞ ⎤ − 3⎢ x + 2 x + ⎜ ⎟ − ⎜ ⎟ ⎥ − 1 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎢⎣

1

b) x = 1

1

5 x < − ,x >1 3

3

3

7

5 − 3

3

2

1

(3x + 5)(x − 1) > 0 8

1

629(5 3n −3 )

2

5 3n (5 + 5 −2 − 5 −3 )

1

2

9 y=

3

1 3

2

32 y = 3 9 y (9 − 1) = 24

3

1

10 k=

m+3 81

3

2

81k 2 = m + 3

1


3

Number

Solution and marking scheme 2 + log 3 k =

11

log 3 (m + 3) log 3 9

Sub Marks 1

x=3

3

x 2 − 2x − 3 = 0

2

log 3 x =

log 3 (2 x + 3) log 3 9

Full Marks

3

1

12 − 72

S12 =

3

12 [2(16) + 11(− 4)] 2

2 3

a = 16 or d = −4

1

2187 5

4

13

a = 729 and r = −

2 3

Solve simultaneous equation a + ar + ar 2 = 567 or ar 3 + ar 4 + ar 5 = −168

14

15

1 3 a + ar + ar 2 = 13ar 2

3 4

2

1

2

4

1

n = 1 and k = 16

4

n = 1 or k = 16

3

2n = 3 or log 2 k = 4

2 1

2n log 2 x − log 2 k


4

Number

Solution and marking scheme

Sub Marks

Full Marks

16 x 2 + y 2 − 4x − 3y = 0

3

2

(x − 2) + ⎛⎜ y − 3 ⎞⎟ = 5 2⎠ 2 ⎝ ⎛ 3⎞ midpo int = ⎜ 2, ⎟ ⎝ 2⎠ 2

17

2

3

1

p = −7

3

p +1+ 6 = 0

2

⎛ 2 ⎞ ⎛ − 3⎞ ⎛ x ⎞ ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ p + 1⎠ ⎝ 6 ⎠ ⎝ 0 ⎠

1

ab + 1 − a 2 1 − b 2

3

cos 20 0 = 1 − a 2 or sin 30 0 = 1 − b 2

2

sin 20 0 cos 30 0 + cos 20 0 sin 30 0

1

4

18

19

46.7483

3

1 (12)2 (0.92) − 1 (7 )(7 )sin 52.710 2 2

2

1 (12)2 (0.92) or 2

1

1 (7 )(7 )sin 52.710 2

3

3

20 2

3

1 ⎡ 2(1) − 1 ⎡ 2(− 1) − 1⎤ ⎤ −⎢ ⎢ 2 ⎥⎥ 2 ⎣⎢ 12 ⎣ (− 1) ⎦ ⎦⎥

2

1 ⎡ 2 x − 1⎤ 2 ⎢⎣ x 2 ⎥⎦ −1 1

1


3

Number 21

Solution and marking scheme 5 11 and k = 3 6

h=−

Solve simultaneous equation h + 2k = 2 or − h + k =

22

7 2

2

A

24

3

= 390 and

∑x

2

B

= 388

2 1

1 6

3

⎛1 1 4⎞ ⎛1 3 1⎞ ⎛ 2 1 1⎞ ⎛1 1 1⎞ ⎜ × × ⎟+⎜ × × ⎟+⎜ × × ⎟+⎜ × × ⎟ ⎝3 4 5⎠ ⎝3 4 5⎠ ⎝ 3 4 5⎠ ⎝3 4 5⎠

2

Either 2 operation above correct

1

a)

b)

25

2

3

x=8

23

Full Marks

1

Player B

∑x

Sub Marks 3

1

1 15

2

1× 4 P4 × 2 6 P6

1

b)

3

2

1 6 1× 5 P5 6 P6

a)

3

0.1587

4

2

1540 − 1500 40

B1

1562

2

( x − 1500) = 1.55 40


4

B1


Modul Perfect Score Sbp 2009

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