2017 SCGS Sec 4 AM Prelim P2 Solutions

Paper 2 1. (i) Express 592 97 2   xx x in partial fractions. [3] (ii) Hence find   7 6 2 d 592 97 x xx x [3] 1. (i) )5)(12( 97 592 97 2     xx x xx x Let 125592 97 2    x B x A xx x )5()12(97  xBxAx Let ,5x A1144 4A Let ,2 1x           52 192 17 B B2 11 2 11  1B 12 1 5 4 592 97 2    xxxx x (ii)   7 6 2 d 592 97 x xx x        7 6 d12 1 5 4 xxx 7 6 )12ln(2 1)5ln(4      xx            13ln2 11ln415ln2 12ln4 70.2 (6 marks)

2 2. Given that 22332)(f 23  xxxx is exactly divisible by ,2 2 cbxx  (i) find the value of b and of c, [4] (ii) show that 022332 23  xxx has only one real root, [3] (iii) find the remainder when )(f x is divided by )32( x . [2] 2. (i) 22332)(f 23  xxxx )2(f  22)2(3)2(3)2(2 23  0 )2( x is a factor. Let )2)(2(22332)(f 223 cbxxxxxxx  Comparing coefficient of 2x , 34 b 7b 11c (ii) 0)(f x 0)1172)(2( 2  xxx ,2x 01172 2  xx Discriminant )11)(2(4)7( 2 039 01172 2  xx has no real roots. Hence, 0)(f x has only one real roots. (iii)      2 3f 222 332 332 32 23              13 (9 marks)

3 [Turn over 3. The roots of the quadratic equation 0432 2  xx are  and . (i) Without using a calculator, show that 32 99 22      . [5] (ii) Hence find the quadratic equation whose roots are 2  and 2  . [2] 3. (i) 0432 2  xx 2 3 2 22      22 33    22 22 ))((    22 2 ]3))[((    2 2 )2( )2(32 3 2 3              4 64 9 2 3      32 99 (ii)            22      1 2 1 Quadratic equation, 02 1 32 992  xx or 0169932 2  xx (7 marks)

4 4. The diagram below shows a quadrilateral BCDF whose vertices lie on the circumference of a circle. The tangent to the circle at the point F meets CB extended at A and CD extended at E. The lines CF and BD intersect at G and BG = GF. (i) State an angle which is equal to angle BFA. [1] (ii) Prove that angle FDE = angle FAB + angle FCB. [2] (iii) Prove that quadrilateral BCDF is a trapezium. [3] 4. (i) FCB = BFA or BDF (Alternate Segment Theorem) (i) CBF = FAB +BFA (Exterior angle of ) = FDE (ext  of a cyclic quadrilateral) FDE = FAB + FCB. (iii) FBG = BFG (Base s of an isos ) FBG = FCD (s in the same segment) FCD = BFG Since the alternate ang