RI 9758/02 2023 - new, pls ignore previous version due to NRIC in metadata

1 | Page Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (a)            22 22 32 5 2 92 5 2 92 5 2 0 36 5 236 5 2 0 15 1 1 0 111 5 xx xx xx xx xx xx x                 The required set is 11:1 5xx   Alternatively, From the graphs, 111 5x The required set is 11:1 5xx   (b)        2 2 22 2 25 3 4 5 35 225 3 11 10345 45 45 5 1 xx x xxxx x xx xx xx x x            Raffles Institution H2 Mathematics (9758) Solution for 2023 A-Level Paper 2 5 2

2 | Page           2 2 25 345 25 3045 35 2 051 13 5 2 5 0 51o r < 2o r 53 x xx x xx xx xx xx x x xx x               Question 2 No. Suggested Solution Remarks for Student (a)  2 2 2 3 3 2 2 2 ds e c t a nln(sec ) tan ds e c d secd d 2sec sec tand 2sec tan dd2( S h o w n )dd yx xyx x xx y xx y xx xx xx yy xx           (b) 232 43 2 32 43 2 dd d dd dd22 2dd d dd dd yy y yy y y xx x xx xx            When 0,x ln sec 0 0,y d tan 0 0,d y x  2 2 2 d sec 0 1,d y x    3 3 d 21 0 0 ,d y x     4 2 4 d 20 0 21 2d y x   Therefore, 24 2 4 3 20 0 0 ... ...2! 4! 2 12 xx x xyx x     (c) When 1 π,4x   24 24 π 1ln sec ln 2 ln 242 ππ 44 ...21 2 ππ 32 3072 y        Therefore,  24ππln 2 16 1536 (d)  11 24ππ10 10 00 ln sec d d 21 2 0.0052187161 0.005219 (to 4 s.f.) xxxx x    Use GC to evaluate 5 3 1 2 5 + + + - -

3 | Page Question 3 No. Suggested Solution Remarks for Student (a)  2 2 32 11 5 32 2 2 1 0 85 1 4 BC AB OC OB OB OA OC OB OA                                      (b) 1 2OD d     , 11 2 1 1 0 22 0 , 2 2 4 55 88 AD BD dd dd                                      2222 22 20 04 58 25 48 41 0 2 5 1 61 6 6 4 65 1 17 2 AD BD dd dd dd dd d d                        (c) Let angle ADB =  1 20 0.4 71 .7 22cos 6549 141 644 7cos 83.81769576 83.8 (to 1 d.p.)65 AD BD AD BD                               Angle ADB 83.8 (to 1 d.p.) 

4 | Page (d) Since AD BD     , so ABD is an isosceles triangle. Let M be the midpoint of of AB. Then, 11 22 085 02 6.5 OM            Equation of the line of DM is 01 0 02 0 6.5 8.5 6.5 01 02 , 6.5 2                           r  Since P lies on line DM, then 01 022 , f o r s o m e 6.5 2 6.5 2 OP                