RI 9758/02 2017

1 | Page   Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i) Curve given by parametric equations 3,2xy tt To find points on the curve that lie on the line y = 2x, we have for some t, 2yx 322t t   2 33 33 , 2 3 33 , 2 3 tt tx y tx y           3 , 23a n d 3 , 23AB  Alternatively, express the curve in Cartesian form, 33 6 ,2xt y ttx x   And solve the simultaneous equations 6 and 2yy xx .     22 623 3 , 23a n d 3 , 23 L e n g t h 23 43 6 0 21 5 xx x AB AB       (ii) 2 2 36,2 d3 d d2 ,2dd d 3 xy ttx xy y ttt t x       Equation of tangent at point P: 2232 3yp p x p     2 2 222 3 2 43 yp p xp yp x p      Raffles Institution H2 Mathematics (9758) Solution for 2017 A-Level Paper 2

2 | Page    2 When 0 : 4 26When 0 : 4 3 6Thus, is ,0 and is 0, 4 x yp yp x p x p DE p p       Mid-point F is 3 ,2 pp   which is the point P. 36,2xy pp x  . That is, xy = 6. Question 2 No. Suggested Solution Remarks for Student (i)   11 3 133, 2 3 13 1 1562 3 2 uS d d       (ii)    13 11 3 13 13 13 31 3, 156 1 3 1 156 1 15 2 5 2 52 51 0...(1) r uS r rr rr rr            Note that r = 1 satisfies (1). But if r = 1, then S13 = 3(13) = 39 ≠ 156. So common difference cannot be 1. Using GC, r = 1.210024 = 1.21 (3 s.f.) or –1.451067 = –1.45 (3 s.f.) (iii)   1 33 1.2100 100 3 1 2 Use GC to get inequality or table of values, Smallest 42 n n n    

3 | Page   Question 3 No. Suggested Solution Remarks for Student (a)(i) f( )yx f( 2 )yx   Scaling parallel to the x-axis by factor of 1 2 .  Curve f(2 )yx cuts the axes at 1 ,0 , 0 ,2 ab  (ii) f( )yx f( 1 )yx   Translation of 1 unit in the positive x-direction.  Curve f( 1)yx cuts the x-axis at  1, 0a (iii) f( )yx f ( 1)yx                                                      f( 2 1 )yx   Translation of 1 unit in the positive x-direction followed by scaling parallel to the x-axis by factor of 1 2 .  Curve f(2 1)yx cuts the x-axis at 1,02 a  (iv) We reflect the graph of f( )yx about the line yx to obtain the graph of 1f( )yx  .  Curve 1f( )yx  cuts the axes at  0, , , 0ab (b)(i) a = 1 g is undefined when x = 1. (ii)      2gg g 1g1 1 g 1 11 1 1 11 1 1 xx x x x x x x x              Note that if g has an inverse and hg x x then 1gh  . Here,   gg x x So immediately we actually know 1gg  Replace x with 2x Replace x with x - 1 Replace x with x - 1 Replace x with 2x

4 | Page     1 11Let 1 1 11 11 1 11 1 1g1 g 1 yy x x x y x y xx x