7/31/2019 Mark Scheme for Mock Trial Exams

1/3

Answers for Mock Trial Exam Papers1 (i) Number of taxis = 493

(ii) n = 6.11, round up to 7

Year = 2000 + 7 = 2007

2 Coordinates of A or B :2 2

2 ,5 5

, ( 3, 4)

Distance AB = 6.49 units (3 s.f.)

3 x = 1 , y = 2 ; x = 1 , y = 3

4 (i) Using completing the square, Least/minimum value =2

4

k and

the corresponding value of x =2

k

(ii) Coordinates of intersection = (0, 0) and ( 1,1 )k k

(iii ) (a ) k = 3 , Coordinates of intersection = (0, 0) and (2, 2)

Quadratic curve with x-intercepts 0 and 3.

Straight line graph with point (0, 0) and (2, 2)

(b ) Range of values of x : 0 x 2

Least value of f( x) =9

4

5 (i) 5 2 3 4 5(2 ) 32 80 80 40 10 p p p p p p

(ii) Sub p with x2, 2 5 2 4 6(2 ) 32 80 80 40 ...... x x x x

2 5 2 2(2 ) (1 ) x x = 2 4 6 2 4(32 80 80 40 ......)(1 2 ) x x x x x

= 6 6 680 160 40 x x x

= 640 x

Coefficient of x6 = 40

6 (i) (a )

(b

)(ii)

(iii )

7/31/2019 Mark Scheme for Mock Trial Exams

2/3

7 (i) a = 16e 2t

(ii) s = 5t + 16e 2t 16

(iii ) Solve v = 0, 0.235t .

Sub 0.235t into s , s = 4.83 m

Distance from O = 4.83 m

(iv) v is a decay function which decreases exponential from 5 m s 1 at t = 0. As t increases, v decreases. Therefore, particle P cannot travel faster than particle Q which travels at constantspeed of 5 m s 1 .

8

Equations Y X m c

y = pq x lg y x lg q lg p

y = Bxk lg y lg x k lg B

9 (i) Sine graph for 1 cycle and cosine graph for 1 cycle for 0 x

Amplitude for cosine graph is 3 and for sine graph is 1.

(ii) Point of intersection = 3

,6 2

(iii ) Area of region bounded =1

2units 2

7/31/2019 Mark Scheme for Mock Trial Exams

3/3

10 EITHER

(i) 2V

h x

(ii) Cost of production , C = Curved area 1 cent + Plane area k cents

= [ 2 xh ](1) + [ 22 x ](k )

= 2 x ( 2

V

x) + 22 kx

22 2 V

C kx x

(iii ) 2d 2

4 dC V

kx x x

0 = 22 4V kx x

2

24

V kx

x

Rearranging, 32V

xk

(iv) From 2V

h

x

, V = 2 x h .

Sub V = 2 x h into 32V

xk

,2

3

2 x h

xk

Simplifying,2h

xk

which fulfils the equationd

0dC x

for C to be stationary.

2

2 3

d 44

dC V

k x x

Sub2h

xk

,2 2 2

2 3

d 32 4

dC Vk kh

x h> 0

Thus, C is a minimum when2h

xk

.