AJC_H2Maths_2012Prelim_P1_Solution

AJC Preliminary Examination 2012 H2 Mathematics Paper 1 (9740/01) Solution Qn Soln 1  100 2 99 100002 nS a d   2 99 200ad   a, a+d, a+4d are consecutive terms in GP: 4a d a d a a d     (a+d)2 = a(a+4d)  d2 = 2ad  d = 2a since d  0 . Sub d = 2a into 2a + 99 d = 200, get d = 2 and a = 1. 2 22 8Ax By Cy   (2,1)  4A + B + C = 8 -----(1) Diff (*) wrt x : dd2 2 0 dd yyAx By C xx    d2 d2 y Ax x By c   Tangent at (2,1) // y-axis : 2B + C = 0 ------(2) [1] Diff again wrt x : 222 22 d d d2 2 0 d d d y y yA B y C x x x       When y = 0 , d3 d2 y x  and 2 2 d9 d4 y x   392 2 0 24A B C            (3) Solve the 3 eqns : get A = 3, B = 4 and C = -8 3  6 4 (2 1)( 3) 03 x x x x       221 03 xx x    (2 1)( 1) 03 xx x     3 (2 1)( 1) 0x x x     1 or 1 32xx   6 4cosec 1 2sin1 3cosec      6sin 4 1 2sinsin 3     Replace x by sin , 1sin , 1 sin 32    7 11 or6 6 2      4 (i) Reversing the transformations: a. Stretch parallel to y-axis by factor ½ gives 2 1 24 y x   b. Translate 1 unit to the right gives 2 1 2 4 ( 1) y x   c. Reflection in y-axis gives 22 11 = 2 4 ( 1) 2 4 ( 1) y xx       = f(x) 4 (ii) The graphs of y = g(x) and y = g-1(x): y = g(x) y = g-1(x) -2 -2 ½ ½ x y B 1 3 1 A 2 4 6 −1.5 −1 −0.5 0.5 1 1.5 x y

4 (iii) area of the region bounded by y = g-1(x), the x-axis and the line x = 1 = region A = region B = Rectangle - 0 3 dyx   =    0 2 3 11 3 d 4 x x   = 0 1 3 3 sin 2 x      = 30 3       = 3 3  . 5 2 2 2 2 .....1 2 3 2 3 4 3 4 5 ( 1)( 2) nS n n n            1 2 3 1 1 1 5 1 1 9 1 1 = , = , =3 2 2 3 12 2 3 4 20 2 4 5S S S        . (ii) nS = 11 2 ( 1)( 2)nn  by observation. (iii) Let nP be the statement “ nS = 11 2 ( 1)( 2)nn  ” for nZ  P1 is true from (i) Assume that kP is true for some kZ  ie. kS = 11 2 ( 1)( 2)kk  We need to show that 1kP  is true, ie to prove that 1kS  = 11 2 ( 2)( 3)kk  LHS = 1kS  = kS + (k+1)th term = 11 2 ( 1)( 2)kk  + 2 ( 1)( 2)( 3)k k k   = 1 3 2 2 ( 1)( 2)( 3) k k k k     = 11 2 ( 2)( 3)kk  = RHS Therefore 1kP  is true. Since 1P is true and kP is true 1kP  is true,  by MI, nP is true for nZ  6 (i) By pythagoras’ theorem: 24lr and 2 2 2 2 (2 ) 4 4R r R r R      A = πrl  A = π 4 4 4RR 24πA R R   (ii) dA dA dR dV dt dR dV dt   22 2 (2 1) 1 84 dA R dt R RR     2 (4 1) 1 3 3 2 84 (4) 24 2 2 dA dt       7 (i) Since the sequence converges to L, ie as 1, and nnn x L x L    3 22 1 1 12 3 2 1 13L L L L L LLL 