RI 9758/02 2019

1 | Page   Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i)     33 22 35 22 22 11 d33 24 1131 5 Ix x x x xx xc          Don’t forget to include the arbitrary constant of integration. (ii)       2 22 22 42 53 53 22 d12 d 12 d 21 d 2 d 22 53 221153 yxu u x Iu u u u uu u uu u uu d xx d               2 1 2 1 1 xu ux   Final answer in terms of x. (iii)                 53 3 5 22 2 2 33 55 22 22 35 22 35 22 55 22 22 2 411 1 153 3 1 5 2 24211 1133 1 5 5 21 011 131 5 2211 133 221133 xx d x x x c xx x x xd c xx x d c xx x d c xx d c dc                                   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 | Page   Question 2 No. Suggested Solution Remarks for Student (i)    2 22 35 8 3 8 1 xxy xx x x     Label Asymptotes: 8 ,1 , 03xx y   Label axial intercepts: 10, , 2, 04   (ii) 8 or 1 23xx   (iii)       2 038 1 2 038 1 8 1 or 23 x xx x xx xx        

3 | Page   Question 3 No. Suggested Solution Remarks for Student (i) 2 2 2 900 2 2 450 450 rr h rr h rh r         2 22 3 22 2 2 2 2 2 450 450 d 450 3 0 150d 150 (0 ) d 150 6 0 for all 0, that is, gives max .d max 450150 450 150150 150 300 150 450 rVr h r r r r V rrr rr V rr r Vr Vr h r r rh                                         2 2 2 1 450 150 1 450 450 150 2 :1 : 2 r r hr rr hr rh          

4 | Page   Question 4 No. Suggested Solution Remarks for Student (i)           2 23 2 f2 s e c 2 t a n 2 f 2 2sec 2 tan 2 tan 2 2 2sec 2 sec 2 4sec2 tan 2 4sec 2 f0 1 f0 0 f0 4 f1 2 xx x x xx x x x xx x xx            (ii)  0.02 2 0 1 2 d 0.0200053333 0.02001 (5 d.p.)xx  (iii) 0.02 0 sec 2 d 0.0200053355 0.02001 (5 d.p.)xx  (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 | Page   Question 5 No. Suggested Solution Remarks for Student (i)       52 4 5 14 51 4 141 4 5151 44 5 4 OX OB BX OX OA AX bB D a A C bO D O B aO C O A bb a b a a b a ba ab ba ab OX b                                                                 