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 3 3 1 17Tangent: 13 54 17 1 54 54 121 d 17 9At 1, : 2 163 d 54 3 3 Ta yx y y y y P Q yP x y x x y yQ x                                        1 17ngent: 13 54 17 1 54 54 1Solving coordinates of is ,017 y x x y N         

8 | P a g e Question 8 No. Suggested Solution Remarks for Student (i) 1 2 3 2 5 and 15 15 2 5 5 2 2 40 u u A A u u A          (ii)  1 2 3 2 5 : 2 5 ...(1) 15 : 4 2 15 ...(2) 40 : 8 3 40 ...(3) 15Using GC: , 5, 52 n nu a bn c u a b c u a b c u a b c a b c                   (iii)       1 1 15 2 5 52 2 2 115 1 5 1 52 2 1 2 515 2 1 1 52 n n r r r r n n u r n n n n n n                               

9 | P a g e Question 9 No. Suggested Solution Remarks for Student 22 sin 2 , 2sin for 0x y         (i)  2 2 d d2 2 cos 2 , 4sin cosd d d 4sin cos d 2 2cos 2 4sin cos 2 2 1 2sin 4sin cos 4sin cos sin cot x y y x                            Ok to have 0 and  included though cot is not defined. (ii)    2 2 2 Point where is 2 sin 2 , 2sin Equation of normal: 2sin tan 2 sin 2 At point , 0, sin2sin 2 sin 2cos 2sin cos 2 sin 2 2 , that is, 2 y x A y x x x k                                   (iii)        2 2 0 2 2 0 0 2 0 0 0 Total length of 2 2 cos 2 4sin 2 d 4 8cos 2 4cos 2 4sin 2 d 8 8cos 2 d 8 8 1 2sin d 4sin d 4 cos 4 1 1 8 C                                           

10 | P a g e Question 10 No. Suggested Solution Remarks for Student d d , where d d I q qL RI V It C t    (i) Differentiate with respect to t, 2 2 2 2 d d 1 d d d d d d d d d d , since d d d d I I q VL Rt t C t t I I I V qL R It t C t t        2 2 Therefore, d d d 0, when 0 that is, is a constant.d d d I I I VL R Vt t C t    (ii) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 e d e e e ed 2 2 d e e e e ed 2 2 4 4 d dSub into 0,d d e e e e4 2 Rt L Rt Rt Rt Rt L L L L Rt Rt Rt Rt Rt L L L L L Rt Rt Rt Rt L L L I At I R ARA At A tt L L I AR AR AR AR AR t tt L L L L L I I IL Rt t C AR ARAR t AR tL L                                    2 2 2 2 2 2 2 2 2 2 e 0 e e e e e 04 2 1 404 Rt L L Rt Rt Rt Rt Rt L L L L L AtC AR AR AAR AR t t t L L C R L CL C R                    (iii) Note that the given values satisfy (ii), when 2 4 34 and 3, from (ii), 4 LR L C R    By right, we should sub R = 4, L = 3 and C = 0.75 into I and check that it still a solution of the DE.