RI 2016 P1 A-Level H2 Math Solution

1 | P a g e Question 1 No. Suggested Solution Remarks for Student      22 2 2 2 4 4 14 3 44 4 14 34 4 4 4 14 12 4 3 5 2 4 3 1 2 4 x x x xx x xx x x x x x x x x x x x x                           24 4 14 34 3 1 2 04 x x xx x x x           3 1 2 4 0x x x    12 or 43x x   Detailed working needed. We can use GC to check our answers, though question states “without using a calculator”. Raffles Institution H2 Mathematics Solution for 2016 A-Level Paper 1

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (i) cos2 xy 0 d 0d x y x   2 d 0.6931471 0.693 (3 s.f.)d x y x    Can obtain answer from GC directly. No working needed. (ii) Tangent at x = 0: y = 2 Tangent at :2x  2 1 0.69315y x    0.693 2.09(3sf)y x   2 2 1When 2, 0.69315 0.12811y x x      Required coordinates (0.128, 2).

3 | P a g e Question 3 No. Suggested Solution Remarks for Student           4 3 3 f f 4 f 0 4 0 since 0 x k x l m x k x l a k a l l a k                   4Thus fx k x a m       4f a b k a a m b m b            4Thus fx k x a b       4 4 f 0 0c k a b c c bk a        (a,1/b) (0,1/c) y = 0 0

4 | P a g e Question 4 No. Suggested Solution Remarks for Student 4 7 14 3 ...(1) 8 ...(2) 11 ...(3) a d br a d br a d br       (i) (3) – (1): 14 4 4 108 1 ...(4)d br br br r    (2) – (1): 7 4 4 35 1 ...(5)d br br br r    10 10 3 3 10 3 (4) 8 1: 5 5 8 8(5) 5 1 5 8 3 0 Since 1, using GC, 0.74045 0.74(2 d.p.) r r rr r r r r             (ii)  3.85 0.741 n nbr br

5 | P a g e Question 5 No. Suggested Solution Remarks for Student 2 1 , 0 2 a u v b                    (i)   2 2 2 0 1 2 2 2 2 2 4 4 2 u v u v u u v u u v v v v u a b b b a bi b a j ak a                                                            Note that 0u u v v u v v u                 (ii)        Given that 1 2 4 1 11 2 18 1 2 18 b a u v u v a u v u v a a                             (iii)   2 2 22 2 . 0 . . 0 0 2 1 2 3 u v u v u u v v u v v u                            

6 | P a g e Question 6 No. Suggested Solution Remarks for Student (i) Let nP be the statement    2 2 1 11 1 24 n r r r n n n n       for .n  For 1,n        1 2 2 1 2 1 1 1 1 2 1 81 1 1 1 1 2 24 4 r LHS r r RHS LHS               Therefore, P1 is true. Assume kP is true for some k  , i.e.    2 2 1 11 1 24 k r r r k k k k       We need to show1kP is true, i.e.    