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 in positive x-direction Transformation 3: n units in negative y-direction ( ) ( ) ( ) ( ) ( ) 2 2 2 2 Trans 1 1 2 2 2 2 2 2 2 2 Trans 2 Trans 3 2 2 2 2 : 1 1 6 3 6 3 1 1 3 3 6 6 x x y y kC x m x m y n y k k       + =  → + = − − + → + =  → + = Final equation: ( ) ( ) ( ) 2 2 2 2 2 : 1 36 x m y n C k − + + = 5(ii) If 2C is a circle with centre (4, −7), then ( ) ( ) ( ) 2 2 2 2 1 36 x m y n k − + + = to ( ) ( ) 2 2 2 2 4 7 1 (6 ) 3 x y k − + + = means 4, 7 m n = =

6 and 16 3 2k k = ⇒ = 6 (i) 2 d 5 d 1 u t t t= + Integrating both sides with respect to t, ( ) ( ) 2 2 2 5 2 d 2 1 5 ln 1 , since 1 0, where is arbitrary con stant. 2 tu t t t C t C = + = + + + > ∫ Substitute values 0 and 3 t u = = : 3C = Particular solution is ( ) 25 ln 1 3 2u t = + + . 6 (ii) As t → ±∞ , 2 d 5 0d 1 u t t t= → + . The gradient of every solution curve tends towards zero as t → ±∞ . 6 (iii) u t 0 3 1

7 7 (i) 7 (ii) Area ( ) ( ) ( ) ( ) 4 2 2 0 2 2 2 0 0 2 2 0 0 d 4 4 2 4 sin 3cos sin d 8 12 cos sin d 3 cos sin d 8 312 cos sin d sin 2 d 8 4 y x π π π π π θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ = − − = + − = + − = − − + − ∫ ∫ ∫ ∫ ∫ ∫ ( ) ( ) 3 0 0 0 2 cos 3 12 1 cos 4 d 8 3 8 3 sin 4 4 1 1 8 8 4 3 units 8 π π π θ θ θ θ θ π   = − + − −       = − − − + − −     = ∫ y x 0

8 8 (i) ( )( ) ( ) ( ) ( )( ) ( )( ) ( )( ) 2 2 2 2 1 2 3 2 1 2 3 1 1 2 1 2 2 6 4 3 6 1 2 4 1 2 r r r r r r r r r r r r r r r r r r r r r r r r r + + − + + + − + ≡ + + + + + + − − + + ≡ + + +≡ + + (ii) ( )( )1 1 32 1 1 2 3 4 1 2 2 3 1 1 2 ( n n r n r rS r r r r r r = = += + +   = − +   + +   = − + ∑ ∑ 32 2 3 + − 1 4+ 2 3 + 3 4 − 1 5 + 2 2 2 1 n n − − + + ⋮ ⋮ ⋮ 3 1n−− − 3 n 1 n+ + 1 1 2 n n + + 3 1 1 2 ) 2 3 2 1 3 1 1 2 2 1 1 2 3 2 1 2 1 2 n n n n n n n + + − +   = − + + − +   + + +   = − + + +

9 (iii) ( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 3 4 ( 4)( 1) 1 1 ( 2) 1 ( 2) 4 1 2 4 5 1 2 6 3 2 1 5 2 1 2 6 2 2 1 3 1 2 n n r r n r n r r r r r r r r r r r r r r r r r r r r n n n n = = = = + − + − = − + + − + += + + += − + + = − + − + + = − + + + ∑ ∑ ∑ ∑ 9(i) Since ( ) ( )g 2 g 6 5 = = , the function g is not one-to-one and hence does not have an inverse function. (ii) ( ) ( ) ( ) ( ) 3g 3 ggg 3 gg 1 g 4 3 = = = = . Since ( ) 2g 3 4 = and ( ) 3g 3 3 = , n can be 2,5,8,... The set of values of n is { }3 1: k k +− ∈ Z (Also accept answers such as { }2,5,8,11,... , { }3 2 : 0,1, 2,3,... k k + = etc.) (iii) ( ) ( ) ( ) ( )( ) ( )g g 1 g 2 g 1 1 ,for 1 2 x x x = + − − < < ( ) ( )( ) ( ) g 1.5 4 5 4 1.5 1 4 1 0.5 4.5. = + − − = + = ( ) ( ) ( ) ( )( ) ( )g g 2 g 3 g 2 1 ,for 2 3 x x x = + − − < < ( ) ( ) ( ) ( ) g 2.7 5 1 5 2.7 2 5 4 0.7 2.2. = + − − = − =

10 (iv) (v) When ( )g x k = has four real distinct roots, the graph of y k = intersects the graph of ( )gy x = at four distinct points. From (iv), 2 3 k< < . 10(i) AB AB −= − b a b a 10 (ii) By Sine Rule, sin sin 2 2 6 3 1 23 1 3 OA AM AM π π =   =           = a a Hence,