2024 H2 MATH P1 with Solutions

1 1 The graph of 2 1xy ax bx c += ++ , where a, b and c are non–zero constants, has an asymptote at 1 2x=− . The graph also has a turning point at 12, 9 −− . Find the values of a, b and c. [4] [Solution] Asymptote at x = 1 2−  2 11 022a b c   − + − + =       2 4 0a b c− + = …Eq(1) When x = –2 and y = 1 9− , 1 2 1 9 4 2a b c −+−= −+ 4 2 9a b c− + = …Eq(2) y = 2 1x ax bx c + ++  ( ) 2 1ax bx c y x+ + = + Differentiating wrt x: ( ) ( ) 2 d 21d yax bx c ax b yx+ + + + = When x = –2, y = 1 9− and d d y x = 0, ( ) 10 4 1 9ab + − + − = 49ab− + =− …Eq(3) From GC, a = 2, b = –1 and c = –1 2 Two vectors a and b are such that |a  b| = 3. It is given that a is a unit vector and b.b = 9. Show that a and b are perpendicular. [3] [Solution] 3=ab  sin 3 =ab where  is the angle between vectors a and b. ( )( )1 3 sin 3 = since |a| = 1 and |b| = 93= |sin  | = 1  sin  = 1 or –1   = 90 or 270  a and b are perpendicular. (Shown)

2 3 The equation 32 0,z az bz c+ + + = where a, b and c are constants, has roots 3 and w is a complex number. (a) State a condition on a, b and c for the third root to be w*. [1] (b) Given that the condition in part (a) holds, and that w = – 1 + 2i, find the values of a, b and c. [3] [Solution] (a) All three numbers, i.e. a, b, and c, are real numbers. (b) Let f(z) = 32z az bz c+ + + f(3) = 27 + 9a + 3b + c = 0  9a + 3b + c = –27 …Eq(1) f(–1 + 2i) = ( ) ( ) ( ) 32 1 2i 1 2i 1 2ia b c− + + − + + − + + = 0 From GC, (11 – 2i) + a(–3 – 4i) – b + 2bi + c = 0 (11 – 3a – b + c) + i(–2 – 4a + 2b) = 0 By comparing real and imaginary parts, 3a + b – c = 11 …Eq(2) 4a – 2b = –2 …Eq(3) From GC, a = –1, b = –1, c = –15 4 (a) Without using a calculator, solve the inequality 43 2 x xx −+ . [4] (b) Hence, solve the inequality 34 2 x xx −+ . [2] [Solution] (a) 43 2 x xx −+ 43 02 x xx −−+ ( )( ) ( ) 4 2 3 02 x x x xx − + − + ( ) 246 02 x x x xx − + + +

3 ( ) 2 56 02 xx xx −− + ( )( )( )2 6 1 0x x x x+ − +   –2 < x  –1 or 0 < x  6 (b) Replacing x with |x|, we have –2 < |x|  –1 (reject since |x|  0) or 0 < |x|  6  –6 < x < 6 Since x  0, the solution is –6 < x < 0 or 0 < x < 6 5 A geometric series has first term a and common ratio r, where r < 0. The sum to infinity of the series is 18 and its third term is 8 .3 (a) Show that 32 27 27 4 0.rr− + = [3] (b) Find the values of r and a. [1] The nth term of the series is .nu (c) It is given that 11 80 . k rr r k r uu  = + = = Find the value of k. [3] [Solution] (a) S = 1 a r− = 18  a = 18 – 18r …Eq(1) ar2 = 8 3  3ar2 = 8 …Eq(2) Subst Eq(1) into Eq(2): 3(18 – 18r)r2 = 8 54r2 – 54r3 = 8 27r3 – 27r2 + 4 = 0 (Shown) (b) From GC, r = 1 3− , 2 3 , 2 3 (reject 2 3