2014 H2 Mathematics C2 Preliminary Examination Solution Qn Solutions 1 1(i) 1
k 2x k
k
k
1 2
1 2
1 2
2 1 x k
2
1 2
1 3 1 2 2 2 2 2 1 x x ... 2 k 2 k 3 2 1 1 x 2 x ... 2k k
3 52 2 k k x k x ... 2 1(ii) 2 2 k k x 1 1 x 1 x k k 2 2 1(iii) 1 Let x The value of x is chosen 5 such that 1 x 1 and 1 1 5 will lead to producing 5 8 2 2 1 2 2 and 2 . 5 5
1 2
k 2x
3 2
5 1 1 1 3 1 1 2 2 2 2 2 5 2 4 2 25 1 1 3 2 10 2 200 2 223 200 2 223 5 100 1 NOTE: Another suitable value to be used is x and using this value, 4
1
5
256 . 115
2(i)
2 5 1 7 3 2 10 7 6 3 cos = = 14 78 14 78 14 78 84.8o
2 1 From GC, : r 1 1 , 0 1 2(iii) Since and 1 are perpendicular is not parallel to 1. Note that 1 is parallel to p1 and A is not on p1 and 1 do not intersect. 1 is not on p1. Hence, and 1 are skew lines. 2(ii)
2(iv)
2 1 4 Direction vector of 1 1 1 5 3 1 1 0 4 1 : r 0 5 , 2 1
2(v)
Method 1 (Compute length of project vector)
Choose A 0, 0, 2 on p3 and B 2, 1, 0 on p1 . p1 l
B
p3 A
N
l1
Required perpendicular distance is
2 1 1 AB nˆ 14 2
2 1 11 1 . 4 1 6 14 14 3
2
Method 2 (Use of foot of perpendicular) Choose A 0, 0, 2 on p3 .
0 2 Obtain equation of line through A and perpendicular to p1 r 0 1 2 3 Find foot of perpendicular N through intersection of line and p1
0 2 2 11 0 1 1 5 14 3 3 2 2 2 11 Thus NA ON OA 1 1 and required perpendicular distance is 3 14 3 2 11 11 14 . NA 1 14 14 3
Perpendicular distance between the two planes
and
1
equals to the perpendicular distance between
11 . 14
(Refer to same diagram above)
3
3(a) (i)
z 5 32 i z 5 32e
z 5 25 e
i 2 k
Therefore z 2e 3(a) (ii) 3(b) (i)
2 k i 5 5
, k 0, 1, 2 .
Since the two complex numbers are in the 1st and 2nd quadrants corresponding to
k 0 and k 1 , thus w e
2 i 5
.
Im
4
D (3, 3)
A
6
F(2, 2)
(1, 1)
6
B
O
3(b) (ii)
C 4
Re
Method 1 (Using ½ x base x height)
OF 22 22 8 FA tan FOA tan 12 OF FA 8 tan 12 1 Area of triangle AB OF 2 1 8 2 8 tan 8 tan 2.14 (to 3 s.f.) 2 12 12
4
Method 2 (Using sum of areas of triangles)
Let OA OB x Area OAB area OAC area OAD area OCD
1 2 1 2 1 x sin 2 x 4 sin 4 2 6 6 2 2
x 2 8 x 32 0 x 4 4 3 x 4 4 3 since x 0 Thus, area of shaded region = 3(b) (iii)
1 2 x sin 4 2 6
2
3 1 16 8 3
Note that the point (5, –1) lies on the perpendicular bisector. Therefore
3 1 arg z 5 i tan 1 . 4 5 If correct to 3 s.f., answer is 2.36 arg z 5 i 2.94 .
5
4(i)
Method 1 (Horizontal line test) y (4, 4)
x
O
From the graph above, a horizontal line drawn cuts the curve y f x at 3 times. Thus function is not one-to-one and inverse does not exist. Method 2 (Use a counter example)
f 0 f 3 0
4(ii)
y f ( x) is not one-to-one, therefore f 1 does not exist. Largest value of m 1.5 .
4(iii)
2
3 9 y x 2 4
Let y x x 3
x Since x 1.5, f 1 x 4(iv)
3 9 x , x 2.25 . 2 4
Method 1 (Taking f –1 on both sides)
9 x2 4 9 f 1fg x f 1 x 2 4 3 g x x2 2 3 Since x 0 , g x x . 2 From fg x
6
3 9 y 2 4
Method 2 (Making use of composition of functions)
From fg x
9 x2 4
Treating the equation as a quadratic equation
g x 3 g x
9 x2 4 2 9 g x 3g x x 2 0 4
in g x .
9 3 9 4 x2 4 g x 2 2 3 2 x 3 4x since x 0 . g x 2 2 3 2x As domain of f is x 1.5 , thus g x . 2 Method 3 (By comparison of functions)
Since f f 1 x x
4(v) (a)
9 9 f f 1 x 2 x 2 4 4 9 f 1 x 2 g x 4 3 g x x2 2 3 Since x 0 , g x x . 2 3 y xa Translate 10 units in positive y -direction.
y 10
( y3 y 10 )
xa Stretch with scale factor
1 parallel to x -axis. 2
( x 2x )
y 10
3 k x 2x a
7
4(v) (b)
For hf to exist, Rf Dh
Since Rf , 2.25 , therefore Dh , a .
Thus the smallest value of a is 2.25. y
y=0
O
x
x = 2.25
f h Df Rf , 0 Rhf , 0
8
Qn 5
Section B: Statistics Solution No. of ways that all the girls are seated together = 6! 5! 86400 No. of ways that not all the girls are seated together = 10! 86400 3542400 No. of ways for the two particular girls to be seated directly opposite each other = 8! = 40320 No. of ways that two particular girls are seated directly opposite each other and 4 of the boys are seated together = 5 C4 4! 4! 2 = 5760 Probability =
6(i)
5760 1 40320 7
Method 1 No. of students who study Physics and Biology = 32 70 (100 19) 21 No. of students who study Physics only = 32 21 11 Probability =
11 0.11 100
Method 2 Biology
Physics
70 19 No. of students who study Physics only = 100 70 19 11 Probability = (ii)
No. of male students who study both Physics and Biology = 24 33 (50 8) 15 Probability =
(iii)
11 100
15 0.3 50
Method 1
8 100 1 32 16 8 P( X ) P(Y ) 2 100 100 100
P( X Y )
Therefore the two events are not independent. Method 2
P( X | Y )
8 1 1 P( X ) 32 4 2
Therefore the two events are not independent 9
7(i)
Let S denote the speed of a randomly chosen passenger car. Let X S1 S2 S3 2S .
X
N(3 85 2 85,3 202 +22 202 ) = N 85, 2800
P S1 S2 S3 2S 50 P 50 X 50 (ii)
0.249 (3 s.f.) 5 5 5 P S1 S 2 ... S80 550 60 60 60 5 5 5 Let T S1 S 2 ... S80 . 60 60 60 2 5 5 T N 80 85 , 80 202 60 60 1700 2000 T N , 9 3 P T 550 0.132 (3 s.f.) Method 2:
5 5 5 P S1 S 2 ... S80 550 60 60 60 P S1 S 2 ... S80 6600 Let T S1 S 2 ... S80 . T
N 80 85,80 202 N 6800,32000
P T 6600 0.132 (3 s.f.) Assume that the cars are travelling independently.
10
8(i)
115 P S 115 0.95 P Z 0.95 80 P S 80 0.75 P Z 0.25 115 1.6448536 (1) 80 0.6744897 (2) 35 2.319343
15.1 (3 s.f.) 90.2 (3 s.f.) (ii)
Let Y denote the no. of pupils with height > 115 cm.
Y B(60, 0.05) Since n 60 50 and np 3 5, Y
Po(3) approx
P Y 3 P Y 2 0.423 (iii)
Quota sampling. The university student can set a quota of 30 kindergarten boys and 30 kindergarten girls to be sampled. She is free to choose the boys and girls to fulfill her quota.
11
9(i) I 20
7
(ii)
(iii)
t 1 6 Since the scatter diagram shows a curvilinear relationship, Model (A) will not be suitable. The scatter diagram shows a decreasing concave upwards trend. Hence Model (B) is not suitable as it has a decreasing concave downwards shape. Model (C) will be the most suitable model as it has a decreasing concave upwards shape.
I aebt ln I ln a bt From GC, the regression line is
ln I 2.980120727 0.2013264777t
Hence
ln a 2.980120727 a 19.7 b 0.201
The value of a is the initial radiation intensity of the radioactive source. (iv)
ln(8.0) 2.980120727 0.2013264777t t 4.47
Since the value of I is within the data range and | r |= 0.911 which is close to 1, the prediction will be reliable.
12
10(i)
( x 4) 1 2 s ( x 4) 8 9
2
2
(ii)
1 82 7.38 0.0336 8 9 H0 : 5 against
H1 : 5 X s n Since p-value = 0.0532 0.05 , hence H 0 is not rejected at the 5% level of Using t-test, test statistics t =
significance, i.e. there is insufficient evidence to claim that the company overstated the mean mass of coffee in a capsule. The assumption is that X, the mass of coffee in one capsule, is normally distributed. (iii)
H0 : 5 H1 : 5
(iv)
(v)
1.959963986
x 5 1.959963986 0.25 100
4.95 x 5.05 4.94 x 5.06 x 5 2.4 2.4 0.25 100 Probability = 1 P(2.4 Z 2.4) = 0.0164
13
11(i)
Let X denote the number of customers wanting to rent a camera.
X
Po(2)
P X 0 0.135335 0.135 (3 s.f.) (ii)
P X 3 1 P X 3
(iii)
Let Y denote the number of days out of 7 days where no cameras are rented.
0.1428765 0.143 Y
B 7, 0.1353352832
P Y 3 0.9916589 0.992 (iv)
Required probability 2 3! = P X 0 P X 3 2!
(v)
0.00785 (3 s.f.) Let W denote the number of customers renting a camera in a week. W Po 14 Since 14 10,W
N 14,14 approx
P W 8 P W 8.5 0.0708 cc
(vi)
Let A denote the number of days in a week with no rental of camera. A ~ B(7, 0.135335) Mean = 0.94735, variance = 0.819137
A As n = 52 is large, by CLT
0.819137 N 0.94735, approx 52
P A 1 0.663 (3 s.f.)
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