2017 A Level H2 Maths Paper 2 Suggested Solutions_CWY_edited

1 2017 H2 Mathematics Paper 2 Pure Math (Suggested Solutions) 1(i) Substitute 3, 2 into 2 x y t y x t   2 32 2 3 3 t t t t              when 3, 3, 2 3 3,2 3 when 3, 3, 2 3 3, 2 3 t x y A t x y B           Length of AB =     2 2 3 3 2 3 2 3 60 2 15 units      1(ii) At 3 ,2 P p t p p         2 2 d d d 3 2 2d d d 3 y y x t x t t t            Gradient of tangent at P = 22 3 p Equation of tangent at P:   2 2 2 3 2 3 2 43 py p x p py x p            At D : let y = 0 6 6 coordinate of ,0 x D p p         At E : let x = 0  4 coordinate of 0,4 y p E p   Coordinate of F= 6 0 4 0 3 , ,2 2 2 pp pp                  

2 3Let (1) 2 (2) from (2), (3) 2 substitute (3) into (1) 6 6 x p y p yp x y xy       ⋯ ⋯ ⋯ The Cartesian equation of the curve traced by F as p varies is given by xy = 6 (or any equivalent form). 2(i) For the given arithmetic progression, 13 3, 156 a S    13 6 12 156 2 1.5 d d    (ii) For the given geometric progression, 13 3, 156 a S     13 13 13 3 1 156 1 1 52 52 52 51 0 (shown) r r r r r r         Using GC to sketch the graph of 13 52 51 y x x    and solve for the values of x- intercepts. We obtain 1.21, 1.45 or 1 r  If 1r  , all the terms of the GP will be the same resulting 13 156 S  . Hence, 1r  even though 1r  is a root of the equation. Possible values of the common ratio are 1.21, -1.45.

3 (iii) Since 0, 1.210024 r r           1 1 1 3 1.210024 100 3 1 1.5 3 1.210024 150 150 3 1.210024 150 150 0 n n n n n n                Using GC to solve, n   1 3 1.210024 150 150 n n    41 -149.9 < 0 42 991.74 > 0 43 2404.7 > 0  The smallest possible value of 42 n . 3(a) (i)  f 2 y x  results in a scaling of the graph of y = f( x) with scale factor 1 2 parallel to x-axis   f(2 ) ,0 ,0 2 x aa         coordinates of x intercept: ,0 2 a      coordinates of y intercept:  0, b (ii)  f 1 y x   results in a translation of the graph of y = f( x) by 1 unit in the positive direction of the x-axis         f( 1) f( 1) ,0 1,0 0, 1, x x a a b b        coordinates of y intercept: not possible to find (iii)  f 2 1 y x   results in a translation of the graph of y = f( x) by 1 unit in the positive direction of the x-axis, followed by scaling with scale factor 1 2 parallel to the x-axis.     f(2 1) f(2 1) 1,0 ,0 2 10, , 2 x x aa b b                 coordinates of x intercept: 1,0 2 a      coordinates of y intercept: not possible.

4 (iv)   1fy x   reflection of the graph of  fy x  about the line y x          1 1 f ( ) f ( ) ,0 0, 0, ,