SAJC_H2_Math_P1_ANS

1 2015 H2 Math Prelim Paper 1 Solutions 1 2 2 14 32 3 5 x x x x + − − ≤− 2 2 2 2 2 2 2 2 14 3(2 3) 3 0, , 1 2 14 6 3 9 0 4 5 5 (2 3)( 1) 5 (2 3) 0 4 ) 5 0 ) 5 ( 1) (2 3)( 1) ( (2 3)( 1) ( 2 4 (2 3)( 1) ( 2) (2 3 ) 1 ) 0( 1 0 x xx x x x x x x x x x x x x x x x x x x x x x x x x x − − − + + − + − − + − + − + − + + − + − ≤ ≠ − ≠ − − − − + ≤ − − ≤ + + + + − ≥ ≥ ≥ Since 2( 2) 1 0x + + > for all x ∈ ℝ Therefore, (2 3)( ) 01x x + − ≥ 1

2 Hence, 2 3x ≤ − or 1x ≥ . Since and 1 2 3x x ≠ − ≠ , 2 3x < − or 1x > . 2 Let Pn be the statement ( )cos 4 cos 10 6 sin sin11 sin 21 ... sin(10 1) 2sin 5 x n x x x x n x x − + + + + + + = for 0,1,2,3,... n = When 0n = , LHS sin x= ( ) cos 4 cos 6 RHS 2sin 5 2sin 5 sin 2sin 5 sin 2sin 5 2sin 5 sin LHS x x x x x x x x x x −= − − = = = = Hence 0P is true. Assume Pk is true for some { }0,1, 2,3,... k ∈ , i.e. ( )cos 4 cos 10 6 sin sin11 sin 21 ... sin(10 1) 2sin 5 x k x x x x k x x − + + + + + + = . To prove 1Pk + is true, i.e. ( )cos 4 cos 10 16 sin sin11 ... sin(10( 1) 1) 2sin 5 x k x x x k x x − + + + + + + = .

3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) LHS sin sin11 ... sin(10 1) sin(10 11) cos 4 cos 10 6 sin 10 11 2sin 5 cos 4 cos 10 6 2sin 10 11 sin 5 2sin 5 cos 4 cos 10 6 cos 10 6 cos 10 16 2sin 5 cos 4 cos 10 16 2sin 5 x x k x k x x k x k x x x k x k x x x x k x k x k x x x k x RHS x = + + + + + + − + = + + − + + + = − + + + − + = − + = = Hence Pk is true implies 1Pk + is true. Since 0P is true, and Pk is true implies 1Pk + is true, by Mathematical induction, Pn is true for all {0,1, 2,3,..} n ∈ 3(i) ( ) 3 5 1 f 3 2 2 xx x x −= = + − − . ( ) ( ) 2 1f ' 2 x x = − − ( )f ' 0 x < for all , 2 x x∈ ≠ R since ( ) 2 2 0 x − > for all , 2 x x∈ ≠ R . Hence, f is decreasing on any interval in the domain. 3(ii) From graph of ( )fy x = , 1 ffD {R \ 3} − = = R . Let ( )fy x =

4 13 2 12 3 1 2 5 2 3 3 y x x y yx y y − = − − = − −= + = − − Hence, 1 2 5 , for , 3 3f : x x x xx− − ∈ ≠ −֏ R . 4(i) 3 2 2 3 1x y x y + = Differentiate with respect to x : ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 2 2 3 3 2 2 3 2 2 2 2 d d 2 3 3 2 0 d d d 2 3 2 3 d 2 3 d d 2 3 y y x y y x x y y x x x y x y x y y x x y x xy y x y x x y x y     + + + =         + = − + += − + For stationary point, d 0d y x = . Since 0, 0 x y≠ ≠ : 0 22 3 3y x y x ⇒ = − + = or 2 3x y = − Substitute back into equation of curve: 2 3 3 2 3 3 12 2 x x x x     − + − =         5 5 9 27 14 8 x x − = 59 18 x− =

5 5 8 9x = − 5 5 3 8 3 8 2 9 2 9 y   = − − =       Hence, the coordinates of A is 5 5 8 3 8 ,9 2 9   −      or 5 5 2 27 27 ,3 4 4   −      (ii) Since B is the reflection of A in y x = , the coordinates of B is 5 5 3 8 8 ,2 9 9   −      5(i) Transformation 1: stretch with scale factor k parallel to x-axis Transformation 2: m units