RI 2017 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i) Curve given by parametric equations 3, 2x y tt  To find points on the curve that lie on the line y = 2x, we have for some t, 2y x 32 2t t       2 3 3 3 3, 2 3 3 3, 2 3 t t t x y t x y              3,2 3 and 3, 2 3A B   Alternatively, express the curve in Cartesian form, 3 3 6 , 2x t y tt x x     And solve the simultaneous equations 6 and 2y y xx  .      2 2 62 3 3,2 3 and 3, 2 3 Length 2 3 4 3 60 2 15 x xx A B AB          (ii) 2 2 3 6, 2 d 3 d d 2, 2d d d 3 x y tt x x y y tt t t x        Equation of tangent at point P: 22 32 3y p p x p        2 2 22 2 3 2 43 y p p x p y p x p      Raffles Institution H2 Mathematics (9758) Solution for 2017 A-Level Paper 2

2 | P a g e   2 When 0 : 4 2 6When 0 : 43 6Thus, is ,0 and is 0,4 x y p y p x p x p D E pp           Mid-point F is 3, 2pp     which is the point P. 3 6, 2x y pp x   . That is, xy = 6. Question 2 No. Suggested Solution Remarks for Student (i)    1 13 133, 2 3 13 1 1562 3 2 u S d d          (ii)    13 1 13 13 13 13 3 13, 156 1 3 1 156 1 1 52 52 52 51 0...(1) ru S r r r r r r r                 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 12 Use GC to get inequality or table of values, Smallest 42 n n n        

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