DHS_H2_MATH_P2_Mark_Scheme

1 2015 Year 6 H2 Math Prelim Exam Paper 2 Mark Scheme Qn Suggested Solution 1(i) Given 1 1 4u  and 1 22 4 ,1 .(1 ) (2 ) nnuu n nn n     Let P(n) be the proposition that 22 1 (1 ) nu nn  , 1.n Consider P(1): LHS = 1 1 (given)4u  RHS = 22 11 LHS1( 1 1 ) 4 Hence P(1) is true. Assume P(k) is true for some ,k  i.e. 22 1 (1 ) ku kk  Consider P(k+1). 22 22 2 2 2 22 2 2 22 2 22 1LHS 4 (1 ) (2 ) 14 (1 ) (1 ) (2 ) (2 ) 4 (1 ) (1 ) (2 ) (1 ) (2 ) 1 = RHS(1 ) (2 ) k k u u kk k kk kk k kk kk k k kk k kk             Hence P(k) is true  P(k+1) is true. Since P(1) is true & P(k) is true  P(k+1) is true, by mathematical induction, 22 1 (1 ) nu nn  for all 1n .

2 (ii) 122 11 11 ()(1 ) (2 ) 4 NN n nn nuunn n            1 12 23 1 1 11 22 22 1 1 ()4 1 [4 () () ] 1 4 11 1 44 (1 ) (2 ) 11 16 4( 1) ( 2) n N NN N n NN n uu uu uu uu uu uu NN NN                      (iii) As 22 1,04( 1) ( 2)NNN   So 2 1 2 11 1 0( 1)( 2) 16 16 N n nn n   Hence, the series is convergent and 1 22 11 ( 1)( 2) 16n nn n   (iv) 22 2 2 22 22 2 22 1 1 2 11 (1 ) ( 1 ) ( 1 ) 1 (Let 1)(1 ) (2 ) 11 16 4( 1 1) ( 1 2) 11 16 4 ( 1) NN nn N k nn n nn knkk k NN NN                        Qn Suggested Solution

3 2 Surface area of cone 2π(12 )(13 ) 156 πx xx Surface area of side and bottom of model 22π(12 ) 2 π(12 ) 144 π 24πx xy x x y   Cost of making the cone in dollars   22156π 0.05 7.8 πx x Cost of making the side and bottom of model in dollars  22(144π 24π ) 0.02 2.88 π 0.48πx xy x xy  Total cost of making model in dollars = 22 27.8 π 2.88 π 0.48 π 10.68 π 0.48 πx xx yxx y  Given that the model costs $100 to make, 2 2 10.68π 0.48π 100 100 10.68 π ...... (1)0.48π xx y xy x   V, Volume of the model     22 21π 12 π 12 5 48 π (5 3 )3x yx x x x y  Substituting equation (1) into V, 2 2 33 3 100 10.68 π48π 53 0.48π 240π 30000 3204 π 30000 2964 π (shown) xVx x x xx x xx        2d 30000 8892 π 0d 30000 1.0363 (5 s.f.) = 1.04 (3 s.f.)8892π V xx x     30000100 10.68 π 63.9688892π 40.9 (3 s.f.)0.48π 1.0363300000.48π 8892π y    2 2 d 17784π 0 0 d maximum volume occurs when 1.04 and 40.9. V xxx xy      

4 Qn Suggested Solution 3(i) 2 4 * *4 4 2 0 (shown) z z zz z zz      (ii) πi 313 i 2 e  (iii) Let A, B, P, Q be points that represent the complex numbers 2 , 13 i , 1z and 2z respectively. (iv) Method 1 12 12 Since ( ) and ( ) lie on circle (centred at origin) of radius 2, 2 Pz Qz zz   2πi 3 2 πi 3 2 1 1 2π 3 π angle bisector3 2πarg ( ) 3 2e 1 3i πarg ( ) (angle on straight line)3 2e 1 3i AOB AOQ QOB z z z z                Method 2 2zz Im (Qz

5 Cartesian equation o