Tampines Prelim 2009 Am 2 Solutions

Answer scheme for Prelim 4E 2009 Add Math Paper 2 1(i) (ii)

M = 24g M = 24 x ( ½ )0.03472 x 45 = 8.13 g (iii) 10 = 24 x ( ½ )0.03472t 5 1 lg  lg( ) 0.03472t 12 2 5 lg 12  0.03472t 1 lg 2 t = 36.4 years

A1 M1 A1 M1 M1

   2.

2 3

B2

4   3

2 2      2   3(   )  3( )  2 B1 3 (2   )(  2  )

 2 2  4    2  2 2

A1

B1

2

 5  2(   ) 4 2 4  5( )  2[( ) 2  2( )] 3 3 3 20  9

B1 B1

20 0 9 or 9x2 – 18x + 20 = 0 x2  2x 

B1

3(i)

LHS 

(cos A  sin A)(cos A  sin A)  (cos A  sin A)(cos A  sin A) (cos A  sin A)(cos A  sin A)

2

2

2

 cos  2 sin A cos A  sin A  cos A  2 sin A cos A  sin A cos 2 A  sin 2 A 2 cos 2 A  2 sin 2 A  cos 2 A 2  cos 2 A  2 sec 2 A

(ii) 2sec2A = 5 2 B1 5 B1  = 66.40 2A = 66.4, 293.6, 426.4, 653.6 A = 33.2o , 146.8o , 213.2o , 326.8o

B1

2

cos 2A =

M1 A1

M1 B1 A1

4a) f(m) = 0 m2 – (m-2)m – m2 + 4m – 8 = 0 B1 m2 – m2 + 2m – m2 + 4m – 8 = 0 M1 6m – m2 – 8 = 0 m2 – 6m + 8 = 0 (m – 4)(m – 2) = 0 m = 4 or 2 A1, A1 4b) subst x = 1, -2 = D subst x = -2, 3 = -3C C = -1 subst x = 0, -2 = -2B B = -1 Subst x = 2, 7 = 2A + 1 A=3 Remainder = - x – 1

5a) Area = (3  4 3 )(5  = 15 

18 3

6 3

)

 20 3  24

=  9  20 3 

18



3 =  9  20 3  6 3 = 14 3  9

B1

3

M1

3

A1

B1 B1 B1 B1 A1

6ai) FDE = FCE (s in same segment) B1 FCE = NAF (alternate s, EC // AB) B1 NDA =  NAF B1 AND = FNA (common angles) B1 NFA and NAD are similar as all corresponding angles are equal. NA ND  aii) NF NA NA 2  ND  NF b) By Tangent-secant theorem, AB2 = AF X FC (2NB)2 = AF X FC 4NB2 = AF X FC NB2 = ¼ AF X FC

5b) log4(3x+5) – log4(y-2) = 2 3x  5 log 4 ( )2 M1 y2 3x  5  16 B1 y2 3x + 5 = 16y – 32 3 x  37 y= A1 16 log 2 8 5c)  2 log 2 x  7 M1 log 2 x 3 M1  2w  7 w 3 + 2w2 – 7 = 0 (2w -1)(w – 3) = 0 w = ½ or 3 B1 x = 1.41 or 8 A1 A1 dy M1  x( 2)( x  3)  ( x  3) 2 dx 7a)  ( x  3)( 2 x  x  3)  ( x  3)(3 x  3) B1  3( x  3)( x  1) dy 0 x = 1 or 3 dx P(1,4) Q(3,0) A1 A1

B1 7b) Area of triangle = ½ x 1 x 4 = 2 units2 Area under curve = B1 B1 B1



3

1

B1

x( x  3) 2 dx

B1

3

=

  ( x 3  6 x 2  9 x)dx 1

x 4 6 x3 9x 2 3 [   ]1 4 3 2 = 4 units2 Shaded area = 6 units2

B1 B1 A1

AO  cos 3 AO = 3 cos  COB = 90o -  OCB =  OB  sin  7 OB = 7sin  AB = AO + OB = 3cos  + 7 sin 8ii) R = 9  49 8i)

=

B1

A1

B1

B1

M1

58

3 tan  = 7  = 23.2 o AB = 58 sin ( + 23.2o)

9)(i) 3 A1 (ii) 180o o (iii) max pt (45 , 2) A1 min pt (135 o, -4) A1 (iv)

(v) M1

A1

58 sin ( + 23.2 o) = 6.5 sin ( + 23.2o) = 0.85349 B1 basic angle = 58.6o B1  + 23.2 o = 58.6o  = 35.4 o A1 1 10i) 1  2  3 M1 2x x2 = ¼ B1 x = ½ or – ½ P( ½ , 2 ½ ) A1 1 2 10ii) y =  (1  x )dx 2 1 x 1 = x c M1 2 1 1 =x+c 2x subst x = ½ , y = 2 ½ c=3 M1 1 y= x+3 A1 2x 10iii) y – 2 ½ = - 13 (x – ½ ) M1 1 2 y= - 3x+2 3 A1 dx 10iv) 0.15 = [ 1 + ½ (0.5)-2 ] x M1 dt dx 0.15 = 3 x B1 dt dx = 0.05 A1 dt

8iii)

range : 0  y  4 (vi) 5

A1

A1

8  4  6  10 , ) 2 2 = (2, 2)

11i) C = (

M1

length AD = 16 2  (12) 2 = 20 units radius = 10 units (x-2)2 + (y-2)2 = 100 x2 + y2 – 4x – 4y – 92 = 0

M1

11ii) grad AD = 

4 3

B1 A1 B1

grad BE = ¾ eqn BE: y – 2 = ¾ (x -2) M1 y= ¾x+ ½ B1 solving eqn BE and eqn of circle: (x – 2)2 + ( ¾ x + ½ - 2)2 = 100 M1 9 9 9 x 2  4 x  4  x 2  x   100 M1 16 4 4 2 2 16x – 64x + 64 + 9x – 36x + 36 = 100 M1 x2 – 4x – 60 = 0 x = -6 or 10 y = -4 or 8 E(-6, -4) B(10, 3) A1 A1