SAJC 9758 2022 Prelim P1 Solutions

1 St Andrew’s Junior College 2022 Preliminary Examination H2 Mathematics Paper 1 (9758/01) Q Solution Mark scheme 1(i) ( ) ( )224 20x y x y+ + − = ---- (1) Differentiate (1) with respect to x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) dd8 1 2 1 0 dd dd8 8 2 2 0 dd d10 6 8 8 2 2 0 d 2 5 3d 10 6 d 6 10 2 3 5 yyx y x y xx yyx y x y x y x y xx yx y x y x y x xyy x y x x y x y    + + + − − =       + + + + − − − = + + + − + = ++=− =−++ d 5 3 (Shown)d 3 5 y x y x x y +=− + (ii) Since the tangents are perpendicular to the line yx= , hence the gradient of tangents = - 1 d 5 3 1d 3 5 y x y x x y +=− =−+ 5 3 3 5 22 (*) x y x y xy xy + = + = = −−− Substituting into (1);

2 Q Solution Mark scheme ( ) ( ) ( ) 2 2 2 2 24 20 4 20 4 5 2 2 5 5 4 xx x x xx x x = −= = = = ++ Given y = x from (*) When 55,22xy== When 55,22xy=− =− Hence, the points are and 22 5 5 5 5, 2 , 2     −−          22 2 55 5 5 yx x yx − =− −  =− + =− +

3 Q Solution Mark scheme 55 5 22 5 2 yx x yx  − − =− − −    =− − =− − The equation of the tangents are 5yx=− + and 5yx=− − .

4 Q Solution Mark scheme 2(i) ( )( ) 2 66 4 2 2y x x x== − − + Asymptotes are 2, 2, 0x x y= =− = Intersections with axes: When x = 0, ( ) 63 2 2 2y== (Also the stationary point) 26yx=− Intersections with axes: When x = 0, 6y= = > (0 , 6) When y = 0, 2 2 0 6 or 6 6 6 x x x = = = − − ( ) ( )0 or6, 0 6,−

5 Q Solution Mark scheme 2 (i) (ii) The x-coordinates of the intersection points between the graphs 2 6 4y x= − and 26yx=− are 2.77− , 1.53− , 1.53 and 2.77 (to 3 sig. fig.) For 2 2 6 64 xx −− From the graph above, 2.77 2 or 1.53 1.53 or 2 2.77x x x−  − −     (iii) Replace x with –x, y x O y = 0 (0 , 6)

6 Q Solution Mark scheme after the reflection about the y-axis, the solution is: 2 2.77 or 1.53 1.53 or 2.77 2xxx   −   −  − Replace x with 4x+ , after the translation of 4 units in the negative x direction, 2 4 2.77 or 1.53 4 1.53 or 2.77 4 2 2 1.23 or 5.53 2.47 or 6.77 6 xxx x x x   +  −  +  −  + − −  − −  − −  − the solution set is therefore  : 6.77 6 or 5.53 2.47 or 2 1.23x x x x −  − −  − −  −

7 Q Solution Mark Scheme 3(i) Given ( ) ( )f 1 f 2 0− = = but 21−  and fD1, 2−  , f is not a one -to- one function. Hence, f does not have an inverse. (ii) Let 32f ( )y x ax bx cx d= = + + + Curve passes through ( )0, 3− 3d =− 32f ( ) 3y x ax bx cx= = + + − Curve passes through ( )1,0− 3a b c− + − = ----------- (1) Curve passes through ( )2,0 8 4 2 3abc+ + = ----------- (2) 2d 32d y ax bx cx = + + Tangent to the curve at 1x= is a horizontal line, d d y x = 0, 3 2 0a b c+ + = ------------ (3) Solving (1), (2) and (3) using GC, 39, 0, ,22a b c= = =− The equation of the curve is 339f ( ) 3 22y x x x= = − −

8 Q Solution Mark Scheme (iii) (iv) Smallest k = 1 (v) The graphs of f ( )yx= and 1f ( )yx −= are reflections of each other about the line y = x. (vi) ( ) ( ) ( ) ( ) ( ) 1 1 3 3 Since the graphs f , f and intersec t at the same point, the solution of f f is the sa me as the solution of f . 39 322 3 11 6 0 (shown) Solving the equation using GC, 2.1 y x y x y x xx xx x x x xx x − − = = = = =  − − =  − − = = 4(3 s.f.) since 1x (vii)  )fR6 ,=−  ( )gD 5 ,=−  Hence gf does not exist y x (2,0) (-1,0) (1,-6) (0,-3) f 0

9 Q Solution Mark Scheme 4 (i) From the graph above, 11 or 2xx−  y x O

10 Q Solution Mark Scheme (ii) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 0 2 10 21 10 21 1 d21 13 d2 2 2 1 1 3 1 3 dd2 2 2 1 2 2 2 1 1 3 1 3 ln 2 1 ln 2 12 4 2 4 1 3 3 1 3ln 3 1 ln 5 0 ln 32 4 4 2 4 1 3 3 ln 3 1 ln 52 4 4 x xx xx xxxx x x x x − − − −− − −− + − =+ −    = + + − +       −−       = + − − + −              = − + − − + − − − +              =− + − − +     13 ln 324 331 ln 3 1 ln 524 33ln 3 ln 5 (Shown)24 −+  =− + + − =−