Bowen 2009 Prelim Am P1 Solutions

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2009 prelims A-Math 4EX/5NA P1 Solutons 1a

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1b

5 11  3( x  1) 3( x  2) dy 5 11   2 dx 3( x  1) 3( x  2) 2 dy 13 x  1,  dx 4 4 AOB  9  BOC  9

2a

y

2b Shaded region =

-(

+

+

)

 ( 4) 2  1 4 1  1 4    (4)(4) sin  (4)(4) sin  (4)(4) sin  2 9 2 9 2 9  2  6.64cm 2 

3a

 9 13  Mid-pt =  ,  2 2  m  3 13 9  3   c 2 2 c  20  y  3x  20 

3b

 20  P   ,0 , Q  0,20   3  2

 20  PQ  20     21.1units  3  20 10 0 9 0 A= 3 2 5 20 8 0 5 2

3c



4a

1  100   160     180   2 3   3 

 100units 2 log 8 x  p 8p  x 23 p  x log 2 x  3 p

4b

log 2 y 2  log 2 2 4  log 2 ( y  5) y 2  16( y  5) y 2  16 y  80  0 ( y  4)( y  20)  0  y  4 or y  20

4c

2 3 x 3 2 2 x 8  2 5 x 5  20(2 5 x 1 )

2 5 (5 x ) 72(5 x )

4 9 2  3 

5

x  2 y  1, log 2 ( y  x)  log 2 4  log 2 ( x  4) yx

 x4 4 y  5 x  16. x  3, y  1 6a 6b

6c

Gradient = 2 

k 4

1 k 2 2 4 k 6 x2 6 y   c, 2 x c  4 x2 6  4 2 x A  (0,7),

y

7

x 2  4x  7  0 x  3. B  (3,4). Area =

+ 3

  x 2  4 x  7 dx  0

3

1  4 4 2

 x3     2 x 2  7 x  8 3 0  20units 2

8

9a

1  4  2   10  3 1   a  1  4  2   5         b  10  3 1  5  a  3, y  1 dy  sin x(sin x  cos x )  cos x(cos x  sin x)  dx (sin x  cos x) 2 A 1 

 sin 2 x  sin x cos x  cos 2 x  sin x cos x (sin x  cos x ) 2 1  (sin x  cos x) 2  k  1 

 3 0



 3 1 3 dx   3 dx 2  0 (sin x  cos x ) (sin x  cos x) 2

 3 1   3 1 23   0  1  2  3   3  1  3 1  3  1 3  9b



4

1

33 3 2

6

6

f ( x )dx   f ( x )dx   f ( x )dx  7 1

4

4

4

4

  4.5  1.8 f ( x )dx   4.5dx  1.8 f ( x)dx 1

1

1

 4.5 x   1.87 4 1

 0.9

10a 10b

ds  3t 2  10t  2 dt a  6t  4

v

v   adt  3t 2  4t  c t  1, v  4  c  3  3t 2  4t  3. s   vdt  t 3  2t 2  3t  6 10c

t 3  2t 2  3t  6  t 3  5t 2  2t  4 3t 2  t  2  0 2 t  s. 3 v P  ive, vQ  ive.  same direction.

11

y

y = 1  3 sin 2 x

y  4

4x 

x From graph, 5 solutions.

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