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.          1 22 1 2 11 1 1 1 1 1 24 1 1 2 3 44 k r r r k k k k k k k k                  For 1,n k                   1 2 1 22 1 2 2 3 2 2 3 2 2 1 1 1 1 1 1 1 2 1 2 24 1 1 2 4 8 84 1 1 5 10 84 1 1 2 3 44 k r k r r r r r k k k k k k k k k k k k k k k k k k k k k k k                                      Therefore 1kkP P is true. Since 1P is also true, nP is true for n  by Mathematical Induction.

7 | P a g e (ii) 3 1 0 1 2 3 2 4 14 44 n nu u n n u u u u        (iii)  1 1 0 1 2 1 3 2 1 2 1 0 ... n r r r n n n n n u u u u u u u u u u u u u u                          0 1 1 3 1 2 1 2 2 2 1 12 1 2 from (i)4 n n r r r n r n r u u u u r r r r n n n n                      

8 | P a g e Question 7 No. Suggested Solution Remarks for Student (a)                2 2 1 8i 17 7i 0 ...(1) 1 5i 1 8i 1 5i 17 7i 24 10i 41 3i 17 7i 24 41 17 10 3 7 i 0 Thus, 1 5i is a root of (1) w w                                                2 Let i be the second root, then 1 8i 17 7i 1 5i i 17 7i 1 5i i 17 7i 5 5 i Comparing Re and Im parts, 5 17 5 7 Thus, 2, 3 Second root is 2 3i x y w w w w x y x y x y x y x y x y x y                                            2 2 2 2 2 OR Let i be the second root, then i 1 8i i 17 7i 0 2 i 8 8 i i 17 7i 0 Comparing Re and Im parts: 8 17 0 ...(2) 2 8 7 0 ...(3) Solving to get the same answers... x y x y x y x y xy x y x y x y x y xy x y                               (b) Since coefficients are real, both 1 + ai and 1 – ai are roots.       3 2 3 2 2 2 2 2 5 16 1 i 1 i 5 16 2 1 Comparing coefficients and constant term, 2 5 3 2 1 16 3 since 0 1 30 z z z k z a z a z b z z z k z z a z b b b b a a a k a b                               

9 | P a g e Question 8 No. Suggested Solution Remarks for Student f tany x ax b   (i) 2 2 2f sec 1 tanx a ax b a ax b a ay          2 2 2 3f 2 f 2 2 2x ay x ay a ay a y a y          2 2 2 2 2 2 2 2 3 3 2 3 2 3 4 3 3 2 3 4 f 2 f 6 f 2 6 2 2 6 6 2 8 6 x a x a y x a a ay a y a ay a a y a y a y a a y a y                (ii)  f tan 4y x ax         f 0 tan 14          2f 0 1 2a a a    2 2 2f 0 2 2 4a a a    3f 0 16a    3 2 2 38f 1 2 2 ... 3 ax ax a x x    

10 | P a g e (iii) Using part (i) with a = 2 and b = 0, f tan 2y x x  f 0 0     2f 0 2 2 0 2    f 0 0  f 0 16   38f 2 ... 3x x x   OR Using part (ii) 2 3 2 3 2 3 2 3 64tan 2 1 4 8 ...4 3 tan 2 tan 644 1 4 8 ... 31 tan 2 tan4 tan 2 1 641 4 8 ...1 tan 2 3 tan 2 1 tan 2 1 2 2We can rewrite 11 tan 2 1 tan 2 1 tan 2 2 64 2 4 8 ...1 tan 2 3 1 1 tan 2 x x x x x x x x x x x x xx x x x x x x x xx                                       2 3 1 2 3 2 2 3 2 3 3 2 3 3 3 321 2 4 ... 3 321 tan 2 1 2 4 ... 3 32 321 2 4 2 43 3 322 4 ... 3 81 2 ...3 8tan 2 2 ...3 x x xx x x x x x x x x x x x x x x x x x x                                             