2012 NYJC H2 Math Prelim_ P2_solutions

1 2012 NYJC H2 Math Preliminary exam Paper 2 solutions 1 (a) ( )( )1 i i i i 6 11i zw a b b ab a = + − − = − − − + = − 6 1 11 a b ab − = + = ( )1 6 11 a a ⇒ + − = 2 6 10 0 a a − − = 6 36 40 6 76 3 19 2 2 a ± + ± = = = ± Since a > 0, 3 19 a = + , 3 19 b = − + (b) 22, arg 3u u π= = − 35, arg 4v v π= = 2 5 4 v u = 2 3 4 25 arg arg 2 arg 4 3 12 v v u u π π π   = − = + =     2 1arg 12 v u π  ∴ =     2 5 cos isin 4 12 12 n n v n n u π π       = +             Since 2 n v u       is imaginary, cos 0 12 nπ = 3 5 , , ,... 12 2 2 2 nπ π π π ⇒ = Since Im part of 2 n v u       is negative, sin 0 12 nπ < 3 7 , ,... 12 2 2 nπ π π ⇒ = Hence, smallest n = 18 2 (i) Clearly f is 1-1 on the domain ( )1,1 − since every horizontal line y = k ( k ∈ /Rbb) cuts the graph at most once. So f –1 exists. Alternatively, Since 2 n v u       is imaginary and negative, 2 , 12 2 n k k π π π= − + ∈ /Zbb 6 24 n k = − + To get smallest positive n, let k =1. Hence, smallest n = 18.

2 Let y = x 2 – 4 x. Rearranging, we have x2 – 4 x – y = 0. Solving for x, we obtain x = ( ) ( ) 2 444 2 y−−−± = 2 4 y± + . Since 1x < , x = 2 4 y− + as 2 4 2 y+ + ≥ . So f –1 (y) = x = 2 4 y− + . Hence, f –1 : x → 2 4 x− + , x ∈ /Rbb, 3 5 x− < < . (ii) Since R f = ( )3,5 − ⊂ /Rbb\{–3} = D g, gf exists. (iii) ( ) ( ) ( ) 2 2 1gf g 4 1 4 3 x x x x x = − = + − + = 2 11 4 3 x x + − + (iv) ( ) ( ) ( ) 2 2 2 2 2 4 4 2 h 4 3 4 3 x x x x x x x − − ′ = − = − + − + Given 1 1 2 2 2 x x − < < ⇒ − < − < 2 4 2 6 x⇒ < − < . Since ( ) 224 2 0 4 3 0 x and x x − > − + > , ( )h 0 x′ > and so h is increasing on D h. Note that ( ) 1 9 h 1 1 1 4 3 8 − = + = + + and h( x) → ∞ as x → 1-. Hence h 9R , or (1.125, ) 8   = ∞ ∞     . x 5 1 -1 y -3 y = f( x) O

3 3 2 2 1 4x y + = . Let V be the volume of the silo. 2V x y π= 21( ) 4 y y π= − 31= ( ) 4 y y π − km 3 21= ( 3 ) 0 4