RI 9758/01 2017

1 | Page Question 1 No. Suggested Solution Remarks for Student ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 23 2 3 23 2 22 2 32 23 e ln 1 221 2 ... ... , 1 12! 3! 2 3 22 ...23 2 4 36 ...23 x ax x x ax axx ax ax ax ax ax xax ax x ax aa a a aax x x +       = + + + + − + − −< ≤     = − ++ − + +   − −+= ++ +     2No term in x , we have 24 0 0 (rejected since 0) or 4.aa a a a− =⇒= ≠ = Raffles Institution H2 Mathematics (9758) Solution for 2017 A-Level Paper 1

2 | Page Question 2 No. Suggested Solution Remarks for Student (i) (ii) To solve 1 bx axa <−− , we consider the graphs 1y xa= − and y bx a= − drawn in (i), noting they intersect at a point where x > a. ( ) ( ) 2 1 ,. 11 1 , reject since bx a x axa x a xa xa xabb b = −>− − =⇒=+ =− >  1 bx axa <−− 1xa b>+ or xa< You need to refer to (i) if you are using “Hence”. x y ab x=a y=b|x – a|

3 | Page Question 3 No. Suggested Solution Remarks for Student (i) 22 2 5 10 0y xy x− + −= … (1) Differentiate (1) with respect to x: dd2 2 2 10 0dd yyy x yxxx− −+ = dd 50dd yyy x yxxx− −+ = … (2) When d 0d y x = , 5yx= Sub into (1): 2 22 2 1125 10 5 10 0 20 10. or 22 x xx x x− + − =⇒ = ∴= − (ii) Differentiate (2) with respect to x: 222 22 d d d dd 50d d d dd y y yyyyx x x x xx + − −−+ = …(3) When 1 5d, and 0 d22 yxy x= = = . Sub into (3): 22 2 22 2 5d 1d d 52 50 0dd d422 yy y xx x− +=⇒ = − < Thus, 15, 22   is a maximum turning point. Use 2 nd Derivative Test here. 1st derivative test cannot be used a s d d y x depends on both x and y. Note that exact values with working are required as no calculator is allowed for this question.

4 | Page Question 4 No. Suggested Solution Remarks for Student (i) ( )4 2149 1422 2 xxy xx x +++= = = +++ + Note that C is defined on all values of x except –2. ( ) 2 d1 0 for all 2d 2 y xx x =− < ≠− + Thus, gradient of C is negative for all points on C. (ii) 14 2y x= + + Asymptotes are 2, 4xy= −= (iii) A translation of 2 units in the positive x direction follow by a translation of 4 units in the negative y direction. Replace x by (x – 2) follow by replacing y by (y + 4) is not acceptable.

5 | Page Question 5 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) ( ) 32f f 1 8 1 8 7 ...(1) f 2 12 8 4 2 12 4 2 4 ...(2) f 3 25 27 9 3 25 9 3 2 ...(3) x x ax bx c abc abc a bc a bc a bc a bc =+ ++ = ⇒+++= ⇒++= = ⇒+ + += ⇒ + += = ⇒ + + += ⇒ + += − Solving (1), (2) and (3), 33, , 722a bc= −== (ii) ( ) 32 33f7 22xx x x= − ++ We know that in general a cubic function has 1, 2 or 3 x-intercepts. ( ) ( ) 2 2 2 22 3f ' 3 3 2 33 2 11 33 44 2 13 13 0 for all since 0 for all 24 2 x xx xx xx x xx x = −+ = −+ = −+− +  = − +> − ≥   ( ) ( )f as and f as . Since gradient of curve is always positive, curve is always increasin