RI 9758/02 2018

1 | Page 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 d1 15d3 113 15 d d33 91 1523 9Curve passes through 0,69 8 18 2 91 15 1823 12 15 439 12 15 439 2f 3 4 45 9 y yx y yx y xC C yx yx yx xy x − = −  −=   −= +  ⇒= =    −= +  −= +  −= +   = = ++  ∫∫ (ii) ( ) 1 3 2 3 d1 15 4d3 237 91 237 15 18 5423 y yx y x =−= = = − −=  Coordinates are (54, 237) Raffles Institution H2 Mathematics (9758) Solution for 2018 A-Level Paper 2

2 | Page Question 2 No. Suggested Solution Remarks for Student (a) ( )( ) ( )( ) ( )( ) 4 32 2 4 32 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 xx x x s x xt x x x a xb aa abb tb sb a − + − += + −− −+ =−+ − + − += − + + + −=− ⇒= − − = − ⇒= = = = −+= ( ) 224 4 10 2 1 0 1Thus, the other roots are 2 3i and . 2 xx x− +=⇒ − = + (b) 3 27w = (i) ( ) ( ) 32 2 27 3 Comparing constant terms, 27 3 9 Comparing coefficients of , 0 3 3 3 90 3 9 3 6 3 33 3 33 i or i2 22 22 w w w cw d dd w cd c ww w −=− ++ − = −⇒= = −+⇒= + += −± −= = −− −+ (ii) i0 1 2i 3 2 2i 3 3 3 3e 3 33i 3e22 3 33i 3e22 w w w π π− = = = −+ = = −− =

3 | Page (iii) Sum of roots = 0 Product of roots = 27

4 | Page Question 3 No. Suggested Solution Remarks for Student (i) 5 55 4 44 1 20 10 5 5 0 44 213 AD BC OD OD = −     −− = −       −−       = +− =−                  D is (–5, –4, 3) (ii) 10 0 2 05 5 04 4 10 0 10 10 5 8 0 4 90 2 10 40 8 08 : . 90 0 . 90 400 40 10 40 8 90 BC BE BC BE BCE r xy − = −     = −= −       −−       × = ×− =             = =    ++          40 400 4 45 20 200z xyz= ⇒+ + =

5 | Page (iii) 2 2 22 2 10 0 2 550 44 8 101 10 0 16 8 0 8 10 2 5 2 1 80 40 84 5 . 45 40 20cos 8 5 40 4 45 BC BA BC BA θ − =      = −− = −       −           × = ×− = =                       = ++ + +          220 58.630 58.6θ = =  (iv) ( ) Let be the mid-point of 550 11 44 422 13 2 Required distance ˆ. 004 0 4 . 45 10 2 20 2441 34 M AD OM OA OD ME n  −   = + = − +− =−     =        −−          = =        0 6.88 2441 ≈

6 | Page Question 4 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) ( ) 246 46 2 246 46 22 346 2 46 6 6 24 222ln cos 2 ln 1 ...2! 4! 6! 24ln 1 2 ...3 45 24 1 242 ... 2 ...3 45 2 3 45 1 242 ...3 3 45 241 8 824 3 45 2 3 3 xxxx xxx xx xxxx xxx xx x xxx  = −+−+