JJC_H2_MATHS_P1_Solution

1 2014 H2 Mathematics Prelim P1 Worked Solutions Qns Solution Notes 1 Let the amount of money for a blue voucher, a yellow voucher and a red voucher be x, y and z respectively. Then 3 4 7 27.40 (1) 5 2 4 20.80 (2) 2 8 5 45.00 (3) xyz xyz xyz       From G.C., 2, 5, 0.20xyz 4 2 43.40 43.40 4(2) 2(0.2) 75 x py z p   2 (i) Let nP be the statement  n nu 2813 1  for all 0n  . LHS of P0 = 3 0 u (Given) RHS of P0 =  32813 1 0   P0 is true. Assume that kP is true for some 0k  , ie  k ku 2813 1  . We want to prove 1kP is true, ie   1 1 2813 1    k ku . 1LHS of kP = kk uu 211  (Given) = 1 –  k 2813 2  =  k 23 16 3 1  =  k 2)2(3 8 3 1  =   1 2813 1   k 1kP is true Since P0 is true and kP is true 1kP is true, by Mathematical Induction, nP is true for all 0n  . (ii) The sequence is divergent as n  ,   n 2 does not converge to a finite number

2 3 when r = 0, 3 = 3b r = 1, 1 = (4  a)b Alternative Method (r + 1)4 + (r + 1)2 + 1  (r2 + ar + 3)(r2 + r + b) (r4 + 4r3 + 6r2 + 4r + 1) + (r2 + 2r + 1) + 1 (r2 + ar + 3)(r2 + r + b) Comparing coefficients: r0: 3 = 3b r3: 4 = 1 + a a = 3, b = 1 22 22 2 2 42 11 ( 3 3 ) ( 1 ) 13 3 ( 1 ) ( 3 3 ) 2( 1) (1 )(1 )1 rr r r r r rr r r rr r rr            42 0 22 0 1 (1 )(1 )1 11 1 21 3 3 N r N r r rr rr r r                 22 22 11 13 11 37 1 2 11 11 1 1 3 1 3 11 13 3 NN N N NN N N                     = 2 11123 3 NN   42 2 1 N r r rr  = 1 42 1 1 (1 )(1 )1 N r r rr       2 2 11 1 2 3 (1 ) 3 (1 ) 3 11 1 23 1 NN NN        

3 4 (i) Area of region R 3 2 0 253 3 d 4 1.37 (to 3 s.f.) x x        (ii) Equation of new curve 2 2 2 35 4 2 2 4 y x y x      Volume of revolution 23 2 0 3 2 2 0 3 2 2 0 3 1 0 2 2 d 4 48 4 d4 4 124 1 d (Shown)4 4 124 l n 2 s i n 2(2) 2 2 12 3 24 l n 3 43 23 x x xx x xx x xx xx                               5 222 22 4 2 2 24 2 24 1 3 3 rl k kl r lrh kk rhr rr krVr r rk rV              24 4 24 d2 d3 3 Vk r r r kr   

4 24 4 24 24 4 42 2 4 dAt stationary point, 0d 2 3 3 2 3 3 V r kr r kr kr r rk kr         6 (a) (i) 2 d1ln d vux x x  d1 1 d u vx xx  21 11 1 ln d 11 1ln d n n n xx x x xxx x           1 ln 1 ln 1 1 ln 1 + 1 n n nx n nn n nn          (a) (ii) 21 1l n 1ln d + 1lim n nxx nnx     = 1 (b) dsec sec tand xxa a    When x = a, sec 1 cos