2014 H2 Maths Prelim Papers - HCI P2 solution.pdf

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 xa Translate 10 units in positive y -direction.

y  10 

( y3  y  10 )

xa 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.)

14