RI 2017 P1 A-Level H2 Math Solution

1 | P a g e Question 1 No. Suggested Solution Remarks for Student          2 2 3 2 3 2 3 2 22 2 3 2 2 3 e ln 1 2 21 2 ... ... , 1 12! 3! 2 3 22 ...2 3 2 4 3 6 ...2 3 x ax x x ax axx ax ax ax ax ax xax ax x ax a a a a aax x x                                           2No term in x , we have 24 0 0 (rejected since 0) or 4.a a a a a      Raffles Institution H2 Mathematics (9758) Solution for 2017 A-Level Paper 1

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (i) (ii) To solve 1 b x ax a  , we consider the graphs 1y x a  and y b x a  drawn in (i), noting they intersect at a point where x > a.    2 1 , . 1 1 1 , reject since b x a x ax a x a x a x a x ab b b                1 b x ax a  1x a b  or x a You need to refer to (i) if you are using “Hence”. ab x=a y=b|x – a|

3 | P a g e Question 3 No. Suggested Solution Remarks for Student (i) 2 22 5 10 0y xy x    … (1) Differentiate (1) with respect to x: d d2 2 2 10 0d d y yy x y xx x    d d 5 0d d y yy x y xx x    … (2) When d 0d y x , 5y x Sub into (1): 2 2 2 2 1 125 10 5 10 0 20 10. or 2 2x x x x x         (ii) Differentiate (2) with respect to x: 22 2 2 2 d d d d d 5 0d d d d d y y y y yy xx x x x x          …(3) When 1 5 d, and 0 d2 2 yx y x   . Sub into (3): 2 2 2 2 2 2 5 d 1 d d 5 25 0 0d d d 42 2 y y y x x x      Thus, 1 5, 2 2     is a maximum turning point. Use 2nd Derivative Test here. 1st derivative test cannot be used as 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 | P a g e Question 4 No. Suggested Solution Remarks for Student (i) 4 2 14 9 1 42 2 2 xxy x x x        Note that C is defined on all values of x except –2.   2 d 1 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, 4x y  (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 | P a g e Question 5 No. Suggested Solution Remarks for Student (i)     3 2f 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 a b c a b c a b c a b c a b c a b c                                   Solving (1), (2) and (3), 3 3, , 72 2a b c   (ii)  3 23 3f 7 2 2x x x x    We know that in general a cubic function has 1, 2 or 3 x-intercepts.    2 2 2 2 2 3f ' 3 3 2 33 2 1 1 33 4 4 2 1 3 13 0 for all since 0 for all 2 4 2 x x x x x x x x x x x          