RI 2019 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i)      3 3 2 2 3 5 2 2 2 21 1 d3 3 2 41 13 15 I x x x x x x x c           Don’t forget to include the arbitrary constant of integration. (ii)        2 2 2 2 2 4 2 5 3 5 3 2 2 d1 2 d 1 2 d 2 1 d 2 d 2 2 5 3 2 21 15 3 xx u u u I u u u u u u u u u u u u d x x d                         2 1 2 1 1 x u u x     Final answer in terms of x. (iii)                  5 3 3 5 2 2 2 2 3 3 5 5 2 2 2 2 3 5 2 2 3 5 2 2 5 5 2 2 2 2 2 41 1 1 15 3 3 15 2 2 4 21 1 1 13 3 15 5 2 101 1 13 15 2 21 1 13 3 2 21 13 3 x x d x x x c x x x x x d c x x x d c x x x d c x x d c d c                                              Show difference in answer in parts (i) and (ii) do not depend on x. Raffles Institution H2 Mathematics (9758) Solution for 2019 A-Level Paper 2

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (i)    2 2 3 5 8 2 3 8 1 xy x x x x x       (ii) 8 or 1 23x x   (iii)      2 03 8 1 2 03 8 1 8 1 or 23 x x x x x x x x          

3 | P a g e Question 3 No. Suggested Solution Remarks for Student (i) 2 2 2 900 2 2 450 450 r rh r rh rh r            2 2 2 3 2 2 2 2 2 2 2 450 450 d 450 3 0 150d 150 ( 0) d 1506 0 for all 0, that is, gives max .d max 450150 450 150150 150 300 150 450 rV r h r r r r V r rr r r V r r r Vr V r h r r rh                                                        2 2 2 1 450 150 1 450 450 150 2 : 1: 2 r r h r r r h r r h            

4 | P a g e Question 4 No. Suggested Solution Remarks for Student (i)          2 2 3 2 f 2sec 2 tan 2 f 2 2sec 2 tan 2 tan 2 2 2sec 2 sec 2 4sec 2 tan 2 4sec 2 f 0 1 f 0 0 f 0 4 f 1 2 x x x x x x x x x x x x x x               (ii)  0.02 2 0 1 2 d 0.0200053333 0.02001 (5 d.p.)x x   (iii) 0.02 0 sec2 d 0.0200053355 0.02001 (5 d.p.)x x   (iv) The approximation is good for small values of x. Part (ii) uses Maclaurin series with polynomial of degree 2 to approximate the integral with x = 0.02 which only differ from actual value from 9th d.p. This already provides a good approximation. (v) We would need g and all derivatives of g to be defined in order to apply Macluarin series. However 1cosec2 sin 2x x is undefined at x = 0.

5 | P a g e Question 5 No. Suggested Solution Remarks for Student (i)        5 2 4 5 1 4 5 1 4 14 1 4 5 15 1 4 4 5 4 OX OB BX OX OA AX b BD a AC b OD OB a OC OA b b a b a a b a b a a b b a a b OX b                                                      