Bowen 2009 Prelim Am P2 Solutions

2009 prelims A-Math 4EX/5NA P2 Solutons 1ai

  60  2 x  30   60  ,120  ,420  ,480  x  15  ,45  ,195  ,225 .

1aii

1  3(1  sin 2 y )  4 sin y 3 sin 2 y  4 sin y  4  0 (3 sin y  2)(sin y  2)  0 2 sin y  or sin y  2 (NA) 3   41.8  x  41.8  ,138.2 

1bi

sec (90o + A) 

1 cos(90   A)

1 cos 90 cos A  sin 90  sin A 1   sin A = - cosecA 1bii sin 5 B  sin 3B  sin B 2 sin 3B cos 2 B  sin 3B  cos 5B  cos 3B  cos B 2 cos 3B cos 2 B  cos 3B sin 3B (2 cos 2 B  1)  cos 3B (2 cos 2 B  1)  tan 3B 2a AB  7 sin  , BC  4 cos  P  2(7 sin   4 cos  )  14 sin   8 cos 2b P  260 sin(  29.7  )  10 10 sin(  29.7  )  260    8.63 

2c



Max = 260 cm = 16.1 cm when   29.7   90    60.3 .

3 3a

dy  ln 2 x  1 dx ln 2 x  1  0

1 2e  0.184 d2y 1   2e > 0 dx 2 x  min pt. dy  0.2 dt dx 1  0.2  dt ln 2  1 = 0.118 units2. dy  2 sec 2 2 x dx 2   cos 2 4 4 x

3b

4a

grad of normal = 

4b

1   1   c 4 8   c  1 32 x   y   1 4 32 dy 2 x  3  2 x  dx (2 x  3) 2 3  (2 x  3) 2 0  no turning pt.

1 4

1 2

1

5a 5b

5c

YPZ  PXY (angles in alt seg) PZY  XZP (common angle) PZ YZ  (similar ∆s) PX PY  PZ  PY  PX  YZ YZ  XZ  PZ 2 (tan-sec thm)

 PX  YZ     PY  2 YZ  XZ  PX    YZ 2  PY 

2

2

6a

XZ  PX     YZ  PY  Let f ( x)  x 3  3x 2  2 x  k 2 f (3)  f (2) 2(k  6)  k  16

6b

k  4 x  0, s  3.

x  1, p  6. x 3 : q  5. x2 : 6  5  r  r  1 7a

6 2 3



 2  2 3 2  3  2  2 3 3 2 2   23  2 3  3 2  2

2 3

=

6

3 2 3  2  3    3  2  3  2  3  2 2 3 2 33 23 2 2 3  2 3 3

 2 4 3  2 3 3

7b

 

3(3 x ) 2  9 3 x  6  0 (3 x ) 2  3(3 x )  2  0 (3 x  1)(3 x  2)  0

3 x = 1 or 3 x =2 lg 2 x = 0 or x = lg 3 = 0.631

8a

8b

9a

2

2

Perpendicular height of isos ∆ = 13x   5 x  = 12x. 1  10 x  12 x  h  3840 2 64 h 2 x 1 A = 2h(13x )  h(10 x )  10 x 12 x  2 2304 =  60 x 2 x dA 2304   2  120 x  0 dx x 3 x  19.2 x  2.68 d 2 A 4608  3  120 > 0 dx 2 x  min A = 1290 cm2  1  Tr 1 10 C r (2 x 2 )10 r   3   3x 

r

r

 1  C r (2)     x 205r  3 20  5r  5 r 5 5  1 coeff. = 10 C 5 (2) 5     3 896 = 27 6 (1  ax)  1  6ax  6C 2 a 2 x 2  ... 10  r

10

9b

(1  bx )(1  ax) 6  (1  bx)(1  6ax  6C 2 a 2 x 2  ...) x : 6a  b  0 x 2 : 15a 2  6ab   a2 

1 9

1 a   , b  2 3

7 3

10a

Centre = mid-pt = (0, -1) 32 Radius = 2 2 2 ( x  0) 2  ( y  1) 2  8 x2  y2  2y  7  0

10b

11a

( x  4) 2  ( y  1) 2  5 y   x 2  8 x  11  5 dy x4  dx  x 2  8 x  11 dy 1 1 x  3,   or dx 4 2 1 3  (3)  c 2 3 c 2 x 3 y  2 2 y  a(1  x) n lg y  lg a  n lg(1  x ) plot lg y against lg(1 + x) lg(1 + x) lg y

0.301 0.5416

Gradient = normal ( x  4) 2  ( y  1) 2  5 centre = (4, 1) to circ = (3, 3) gradient = -2 dy 1   dx 2 1 3  (3)  c 2 3 c 2 x 3 y  2 2

0.477 0.683

0.602 0.782

0.699 0.860

0.778 0.924

lg y

lg y  lg a  n lg(1  x )

lg(1 + x)

From graph, gradient = n = 0.8, Y-intercept = lg a = 0.3, a = 2.

11b

12

2 1  xy    2   2 5x  2 2 y 3  x 5x 2 ( x  k )  2( x  k )  1  4 x  0 x 2  ( 2k  6) x  k 2  2 k  1  0

13 13ai

(2k  6) 2  4(k 2  2k  1) < 0  16k  32 < 0 2
  (   ) 2  2  = 2 2  2 36  4  4 8 13aii (   ) 3   3  3 2   3 2   3  3   3  (   ) 3  3 (   )  6 3  3(2)(6)  180 13b    New sum = 2 2  =8  1  new product = 4 4 1 new eqn: x 2  8 x   0 4