2012 NYJC H2 Math Prelim P1 solutions

1 2012 NYJC H2 Math Preliminary exam Paper 1 solutions 1 Let the equation of C be 3 2 y ax bx cx d = + + + Subst (0, –1) into equation, 1d = − Subst ( –1, 1) into equation, 1 2 (1) a b c d a b c − + − + = ⇒ − + − = − − − − − At ( –1, 1), d 0 3 2 0 (2) d y a b c x = ⇒ − + = − − − − ( ) 3 3 2 2 31 1 d 4ax bx cx x + + − = ∫ 3 4 3 2 2 31 4 3 2 4 a b c x x x x   + + − =     81 9 8 31 9 3 4 2 2 4 2 3 4 a c b b a c     + + − − + + − =         65 19 5 35 (3) 4 3 2 4 a b c + + = − − − − Solving eqn (1), (2) and (3), 1, 0, 3 a b c = = = − 3 3 1 y x x ∴ = − − 2 1 1 sin sin cos cos a b θ φ θ φ         × = ×             sin cos cos sin cos cos sin sin θ φ θ φ θ φ φ θ −    = −     −  ( )sin 2sin sin 2 2 2cos sin 2 2 φ θ θ φ φ θ θ φ φ θ     − −   + −   =     + −     sin 2 2sin sin 2 2cos sin 2 δ θ φ δ θ φ δ     −   +  =     +   

2 a b × = 2 2 2 2 sin 2 4sin (sin cos ) δ δ β β + + , where 2 θ φ β += 2 2 sin 2 4sin δ δ = + 2 2 2 2 4sin cos 4sin 2sin 1 cos δ δ δ δ δ = + = + Using sin a b a b α× = 22sin 1 cos δ δ + =( )( ) 2 2 2 2 1 sin cos 1 sin cos sin θ θ φ φ α + + + + ( )( )2 2 sin α= . 2sin sin 1 cos α δ δ = + 3 ( ) 3 2 3 2 2 2 2 2 2 (i) 4 3 2 Differentiating wr.t. : d d 12 6 3 3 d d d3 3 12 6 d x x y y x y y x xy x y x x yx y x xy x + = − + + = − = − − ( ) ( ) ( ) 2 2 2 d 4 2 d 2 2 y x xy x y x x x y y x y x += − += − + ( ) ( ) 3 2 3 3 Curve meets when: 4 3 2 2 2 1 and 1 Thus, coordinates of is 1,1 y x x x x x x x y P = − + − = − − = − ⇒ = − = − ( ) d(ii) At 1,1 , is undefined. d y x− Equation of tangent at : 1 is a square. P x OQPR = −

3 4 (a) 11 2sin 0 sin 2x x − > ⇒ < π 0 6x⇒ ≤ < . π π π 2 6 2 π0 0 6 1 2sin d 1 2sin d 1 2sin d x x x x x x − = − + − ∫ ∫ ∫ = ( ) ( ) π π 6 2 π0 6 1 2sin d 1 2sin d x x x x − + − − ∫ ∫ = [ ] [ ] π π 6 2 π 0 6 2cos 2cos x x x x + − + = ( ) π2 3 1 6− − . (b) ( ) 2 cos cos d d 2cos 2 1 2 1 2sin 1 θ θ θ θ θ θ = − − − ∫ ∫ = ( )2 1 cos d 1 4sin θ θ θ−∫ = 2 1 d 1 4 x x−∫ using x = sin θ sin 20 0 cos 1 d d 2cos 2 1 1 4 x x α α θ θ θ = − −∫ ∫ = sin 0 2 1 d 12 4 x x α − ∫ = ( ) sin 1 0 1 sin 2 2 x α −      = ( ) 11 sin 2sin 2 α− 0 cos π d 42cos 2 1 α θ θ θ = −∫ ⇒ ( ) 11 sin 2sin 2 α− = π 4 ⇒ ( ) 1 πsin 2sin 2α− = ⇒ 2sin 1 α = ⇒ 1 πsin = 2 6 α α = ⇒ .

4 5 1 3 5 ay x −= − + 1 1 3 C 5 ay a x − −= + − + : 1 1 3 B 2 5 ay a x − −= + − + + : ( ) 1 1 3 A 2 5 1 3 1 3 3 3 3 1 3 ay a x a a ax a ax x ax x − −= + − − + + − − + + = + = + + += + : ( ) (1 − 3a)/5 (1 − 3a)/5 y = x 1g ( ) y x −= y = 5 x = 5 y = 0 x = 0 1 3 g( ) 5 ay x x −= = − + y x

5 6 (a)(i) tan 3 AC xAB π  = −     tan tan 3 1 tan tan 3 x x π π − = + 1 3 tan 3 tan AB x AC x += − (ii) 1 3 3 AB x AC x +≈ − when x is small ( )( ) 1 1 3 3 x x − = + − ( ) 1 11 3 1 3 3 xx −   = + −     ( ) 1 1 3 1 3 3 xx   = + +     1 1 3 ... 3 3 x x  = + + +     1 4 ... 33 x= + + Hence, 1 4 , 33 a b = = (b)(i) ( ) 2 2 d1 1 d yx xy x x+ + = + ( ) 2 2 2 2 d d d 1 2 d d d 1 y y y x x x x y x x x x + + + + = + ( ) 2 2 2 2 d d 1 3 d d 1 y y x x x y x x x + + + = + When x = 0, y = 1, d 1d y x = , 2 2 d 1d y x = − 2 1 ... 2 xy x ∴ = + − + (ii) 2 2 1 2 2 e e e e x x x x y + − −   ≈ =      

6 22 2 1e 1 2 2 2 x x x x       ≈ + − + −               ( ) 2 21e 1 2 2 xx x     ≈ + − +         ( )e 1 x≈ + 7 (a)(i) 2 6 2 a d a d r a a d + + = = + ( ) ( ) 2 2 6 a d a a d + = + 24 2 d ad = 1 2d a = ( 0d ≠ ) ( )1 22 2a a r a += = (ii) ( ) ( ) ( ) 0.1 2 1 2 2 2 1 2 1 2 2 na n a a n −     > + −     −     ( )2 1 20 5 2 1 n n n − > + −     ( )2 1 20 5 2 1 0 n n n − − + − >     Smallest 11 n = (b)(i) 1st hit 2nd hit 3 rd hit …………… Distance travelled by the ball when it strikes the floor for the third time h 3 5 h 2 3 5 h     

7 2 3 3 2 2 5 5 h h h     = + +         2.92 h= (ii) Total distance travelled S∞< ( ) ( ) ( ) 2 3 2 0.6 2 0.6 2 0.6 h h h h     = + + + +         … 2 3 2 0.6 0.6 0.6 0.6 2 1 0.6 h h h h   = + + + +     = +   −  … 4h= (shown) 8 (a) z5 = 32 i = 32 22i n e π π  +    , where n = 0, ±1, ±2. z = 10 2 52 i n e π π+      , where n = 0, ±1, ±2. z = 9 3 7 10 10 10 10 22 , 2 , 2 , 2 , 2 . i i i i i e e e e e π π π π π − − Area of P1 P2 P3 P4 P5 = 21 2 5( (2) sin ) 2 5 π ≈ 9.51 units 2 (b) | z – 2 – 2 i| ≤ 1⇒ |z – (2 + 2 i)| ≤ 1 P3 P4 P5 P1 P2 π /10 2π /5 0 2π /5 0 2π /5 0 2π /5 0 2π /5 0 2 2 2 2 2 Re (z) Im (z)

8 θ = tan -1(1/3) β = π − 2θ α = ( β − (π /2))/2 ≈ 0.46365 3 arg( 3 2 ) 4 or arg( 3 2 ) ( ) 3 arg( 3 2 ) 4 or arg( 3 2 ) 2.68 z i z i z i z i π π π π α π π π ∴ ≤ − − ≤ − < − − ≤ − − ⇒ ≤ − − ≤ − < − − ≤ − 9 (a) When n → ∞ , nx l → and 1nx l + → , 1 2 l l = − − 2 1 2 l l = − 2 2 1 0 l l + − = ( )2 4 4 1 2l − ± − − = 2 2 2 2 − ± = 1 2 = − ± Since 0nx < , 1 2 l = − − (b)(i) ( ) 2 2 2 A B r r r r = + + + Im (z) Re (z) 1 2 3 1 2 3 • (2,2) 1 • (2,3) locus of P • (3,2) θ β α tan − 13 or equiv

9 By cover-up rule or otherwise, A = 1, B = –1 Hence ( )1 1 1 1 1 1 2 2 2 n n r r r r r r = =   = −   + +   ∑ ∑ 1 1 1 3 1 1 2 4 1 1 3 5 1 2 1 1 2 1 1 1 1 1 1 2 n n n n n n   −      + −       + −     = +     + −   −    + −   − +     + −   +  /vertellipsis ( ) ( ) ( ) ( ) 1 1 1 1 12 2 1 2 1 3 2 1 2 2 1 2 3 2 3 4 2 1 2 n n n n n n n n n   = + − −