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 angles, FCD and BFG are equal, BF is parallel to CD. Hence, quadrilateral BCDF is a trapezium. (6 marks) A B C D E F G

5 [Turn over 5. Solve the following equations. (i) 32)3( )2( 123  xx [2] (ii) 23)3(7 11   xx [4] (iii) 2log)1(log)5(log 333  xx [4] 5. (i) 3232 123  xx 4 332323 xx 2424 x 1x (ii) 23)3(7 11   xx Let xt 3 2321  tt 02123 2  tt 0)3)(73(  tt )( 3 ,3 73 rejt x  x 3lg 3 7lg     771.0 (iii) 2log 3log )1(log)5(log 3 3 3 3  xx 2log 3log2 1 )1(log)5(log 3 3 3 3  xx 2log )1( 5log 323    x x 2)1(25  xx 2425 2  xxx 0352 2  xx 0)3)(12(  xx 3 ),rej( 2 1x (10 marks)

6 6. A curve has the equation xe xy 2 . The point (p, q) is a stationary point on the curve. Determine (i) the exact value of p and of q, [4] (ii) the nature of the stationary point (p, q). [3] Hence (iii) write down the values of x for which y is increasing. [1] 6. (i) x y d d x xx e exe 4 22 )2( xe x 2 21 At (p, q), 021 2  pe p 2 1p , eq 2 1 (ii) 2 2 d d x y 22 22 )( )2)(21(2 x xx e exe  xe x 2 44 At (p, q), 02 d d 2 2  ex y Hence (p, q) is a maximum point. Alternative Method x 0.5 0.5 0.5+ x y d d + 0  Slope /  \ Hence (p, q) is a maximum point. (iii) y is increasing when 2 1x . (8 marks)

7 [Turn over 7. The population, P, of a certain species of fish after t years is given by ).21(1750 t keP  When t = 2, the population is 4600. (i) Find the value of k. [3] Any species is considered as an “Endangered Species” if its population falls below 2500. (ii) Determine, with working, whether this species of fish will become an “Endangered Species” after 10 years. [2] (iii) Hence sketch the population-time graph. [2] 7. (i) When t = 2, P = 4600, )21(17504600 2 ke ke 22  11750 4600 70 572  ke      70 57ln2k k      70 57ln2 1 10272.0 103.0 (ii) When t = 10, P ]21[1750 1070 57ln2 1     e 3003 > 2500 Since the population of the fish is greater than 2500, the species of fish will not be considered as an “Endangered Species”. Alternative Method When P = 2500, )21(17502500 .102720 te te .102720 211750 2500  14 3.102720  te 14 3ln.102720  t t .102720 14 3ln  0.15 > 10 Since the population of the fish takes 15 years to fall below 2500, the species of fish will not be considered as an “Endangered Species”.

8 (iii) (7 marks) 8. The depth of water, d metres, at a pier, t hours after low tide, can be modelled by the formula )cos(btacd  , where a, b and c are positive constants. (i) If low tides occur every 12 hours, find the value of b. [1] Given that the depth of water at the pier was 2 metres during low tide and 8 metres during high tide, (ii) find the value of a and of c. [2] The pier will be open when the depth of the water is more than 4 m. (iii) For how long will the pier be open in a 12-hour period after low tide. [3] 8. (i) 612 2  b (ii) 32 28 a 5c (iii)      td 6cos35  When d = 4, 46cos35      t 3 1 6cos      t Basic angle = 1.2309 5.0522 ,2309.16 t .64899 ,3508.2t Length of time for which the pier is open 3508.26489.9  h 30.7 (6 marks)

9 [Turn over 9. The diagram shows a solid which consists of a cube fixed on top of a cuboid. The cube has sides x cm. The cuboid has a square base of side 2x cm and a height of y cm. Given that the volume of the solid is 270 cm3, (i) show that the total surface area, A cm2, of the solid is given by xxA 54010 2 . [5] Given that x can vary, (ii) find the value of x for which A has a stationary value and determine whether this value of A is a maximum or a minimum. [5] 9. (i) 2704 23  yxx 32 2704 xyx  2 3 4 270 x xy  A )2(4)2(24 22 xyxx         2 3 2 4 270812 x xxx 22 254012 xxx  xx 54010 2 (ii) 2 54020d d xxx A  When 0d d x A , 054020 2  xx 54020 3 x 273 x 3x 32 2 108020d d xx A  When 3x , 060d d 2 2 x A Hence, A has a minimum value when 3x . (10 marks) x cm x cm x cm 2x cm 2x cm y cm

10 10. The diagram shows part of the curve 43 8   x y . The curve intersects the y-axis a