RI 2018 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i)    1 3 1 3 2 3 2 3 2 3 2 3 3 2 3 2 d 1 15d 3 1 13 15 d d3 3 9 1 152 3 9Curve passes through 0,69 8 182 9 1 15 182 3 1 215 43 9 1 215 43 9 2f 3 4 45 9 y yx y y x y x C C y x y x y x x y x                                                   (ii)   1 3 2 3 d 1 15 4d 3 237 9 1 237 15 18 542 3 y yx y x               Coordinates are (54, 237) Raffles Institution H2 Mathematics (9758) Solution for 2018 A-Level Paper 2

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (a)        4 3 2 2 4 3 2 2 2 4 20 56 0 Since the coefficients are real, 2 3i is also a root. 2 3i 2 3i 4 13 4 20 56 4 13 4 comparing coefficients: 20 16 4 56 13 4 1 13 13 4 52 69 x x sx x t x x x x x x sx x t x x x ax b a a a b b t b s b a                                         224 4 1 0 2 1 0 1Thus, the other roots are 2 3i and .2 x x x       (b) 3 27w  (i) 3 2 2 27 3 Comparing constant terms, 27 3 9 Comparing coefficients of , 0 3 3 3 9 0 3 9 36 3 3 3 3 3 3i or i2 2 2 2 2 w w w cw d d d w c d c w w w                        (ii) i0 1 2i 32 2i 33 3 3e 3 3 3i 3e2 2 3 3 3i 3e2 2 w w w          

3 | P a g e (iii) Sum of roots = 0 Product of roots = 27

4 | P a g e Question 3 No. Suggested Solution Remarks for Student (i) 5 5 5 4 4 4 1 2 0 10 5 5 0 4 4 2 1 3 AD BC OD OD                                                                      D is (–5, –4, 3) (ii) 10 0 2 0 5 5 0 4 4 10 0 10 10 5 8 0 4 90 2 10 40 8 0 8 : . 90 0 . 90 400 40 10 40 8 90 BC BE BC BE BCE r x y                                                                                                         40 400 4 45 20 200z x y z    

5 | P a g e (iii) 2 2 2 2 2 10 0 2 5 5 0 4 4 8 1 0 1 10 0 16 8 0 8 10 2 5 2 1 80 40 8 4 5 . 45 40 20cos 8 5 40 4 45 BC BA BC BA                                                                                                             220 58.630 58.6    (iv)   Let be the mid-point of 5 5 01 1 4 4 42 2 1 3 2 Required distance ˆ. 0 0 4 0 4 . 45 10 2 20 2441 34 M AD OM OA OD ME n                                                                                  0 6.882441

6 | P a g e