RI 2018 P1 A-Level H2 Math Solution

1 | P a g e Question 1 No. Suggested Solution Remarks for Student (i) 2 ln d 1 ln=d xy x y x x x   (ii) e e 2 21 1 e e 2 21 1 e e 1 1 ln ln 1 1d d 1 1 lnd d 1 ln 1 11 e e 21 e x xx xx x xx xx x x x x                          Raffles Institution H2 Mathematics (9758) Solution for 2018 A-Level Paper 1

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (i)    2 3Sub into 2 7 3 2 7 2 7 3 0 2 1 3 1, 32 6, 1 y y xx xx x x x x x x y y                  (ii)    3 2 1 2 2 3 3 3 1 1 2 2 9Required volume 7 2 d 1 97 26 1 36 3 186 125 6 x x x x x                                Note that question said exact and did not mention answer in terms of 

3 | P a g e Question 3 No. Suggested Solution Remarks for Student (i) 2 2 2 2 3 3 d 2 6 ...(1)d ...(2) d d 2 ...(3)d d Sub (2) and (3) into (1): d 2 2 6d d 6d d 6 d yx yx y ux y ux xux x ux x xu uxx ux x u x x               (ii) 2 2 2 2 2 3 3 3 2 when 1: 2 3 1 3 u Cx y C y Cxx x y x C C y x                

4 | P a g e Question 4 No. Suggested Solution Remarks for Student (i)           2 2 22 2 22 2 2 2 2 2 2 3 2 2 2 3 2 2 2 3 2 2 0 2 3 2 2 2 3 2 2 0 2 2 2 4 4 0 1 0 or 2 2 0 2 120, 1, 2 1 3 x x x x x x x x x x x x x x x x x x x x x x x x x x                                      (ii) 1 3 1 or 0 1 3x x       22 3 2y x x   2y x  2 0 122 x y

5 | P a g e Question 5 No. Suggested Solution Remarks for Student f : for , , 1 g : for x ax x x b ax b x x x          Given ff = g,            2 2 2 2 2 1 1 1 1 1 1 Comparing coefficients of : 1 x a ax b xx a bx b x a ax ab xx a bx b x a a b xx b a b x a a b x b x a b x b                            1 ff f f 1 x x x ax x x      ff = g, we assume question is just referring to the rule.

6 | P a g e Question 6 No. Suggested Solution Remarks for Student Given 3 2a b a c      (i)   3 2 3 2 2 2 0 / / 3 2 Thus, 3 2 , where is a constant. a b c a b a c a c a c a b c b c a                                   (ii)     2 2 2 2 2 . cos 60 2 3 2 . 3 2 . 9 12 . 4 144 24 4 124 2 31 b c b c b c b c a a b b c c a                                   

7 | P a g e Question 7 No. Suggested Solution Remarks for Student (i)   2 2 2 2 2 2 2 2 2 2 2 4 1 2 2 8 Differentiate with respect to : d d4 16 2 2d d d 2 16 2d d 2 d 2 16 x y x xy x y x xy x y yx y x xy yx x y xy y x yx y x y x xy y               (ii)   2 2 2 2 1 4 1When 1, 21 2 8 1 1 3 1 1Let and be 1, and 1, respectively, 3 3 121 d 17 9At 1, : 2 163 d 54