PLMGS 4E 5NA AM P2 Prelim 2017 Worked Solutions

Paya Lebar Methodist Girls’ School (Secondary) Department of Mathematics 2017 Preliminary Examination Additional Mathematics Paper 2 (4047/2) Worked Solutions 1.(i) p = – 2 q = 4 r = 1 1.(ii) hk  2  ___________________________________________________________________________ 2.(i) When t = 0, M = 50, 012050 ek k = 120 – 50 = 70 2.(ii) t eM 2 1 70120  When M = 0, t e 2 1 701200  70 1202 1  t e      70 120ln2 1 t      70 120ln2t 08.1t 2.(iii) O 50 1.08 t t eM 2 1 70120  M t

2 3.(i) Graph shows only one x-intercept. 3.(ii) Let )(f x = 123 23  xxx )4(f = 124)4(3)4( 23  = 0 Therefore, x – 4 is a factor of )(f x 3.(iii) 123 23  xxx = )3()4( 2  xxx )4(5123 23  xxxx )4(5)3()4( 2  xxxx 0)53()4( 2  xxx 0)2()4( 2  xxx 0)2()1()4(  xxx 4x or 1x or 2x ___________________________________________________________________________ 4.(i)   BC 312 1 =  1927 BC = 31 )1927(2   = 31 )387(2   = 31 31 31 )31614     = 31 )3(1631431614   = 317  cm 4.(ii) (AC)2 =     22 31731  = 33342893321  = 323296  cm2 ___________________________________________________________________________

3 5.(a) p 22  , 422   2)( 222  = )4(2 p = 4 p  4 p 5.(b) 522   , 422  1 + 1 =   = 45 = 1  2)( 222  = 5 + 2(2) = 9 3  )1()1(   = 1  = 3 – 2 – 1 = 0 A quadratic Eqn is 02  xx ___________________________________________________________________________ 6.(a)(i) 5lg 2lg32log 5 5  2 10lg 2lg5 2lg10lg 2lg5  m m  1 5 6.(a)(i) 210 m 1010 2)10( m m1010 102 

4 6.(b) )4(log2loglog2 333  xx )4(log2log 3 2 3       xx 42 2  xx 0822  xx )8)(1(4)2(4 22  acb = – 28 < 0 there are no real solutions ___________________________________________________________________________ 7.(i) 2)4(9d d xx y  For stationary point, 0d d x y 0)4(9 2  x x = 4 y = 5 stationary point = (4, 5) 7.(ii) For x < 4, x y d d < 0 For x > 4, x y d d < 0 As x increases through 4, the sign of x y d d does not change. The stationary point is a point of inflexion. 7.(iii) O x 5)4(3 3  xy y (4, 5) x 197 5.19

5 8.(i) BCE DE = AD – AE = (  sin2cos5  ) m 8.(ii) )(cos   RDE R = 22 25  = 29 5 2tan   8.21 )8.21(cos29  DE m 8.(iii) CE = AB + CF  cos2sin5  )8.21(sin29   8.(iv) area of triangle CDE = )(sin29)(cos292 1   = )(sin)(cos2 29   = ])(sin)(cos2[4 29   = ])(2sin[4 29   = )22(sin4 29   ___________________________________________________________________________  5 m 2 m A B C D E F

6 9.(i) In  FDB and  FAD DFB = AFD (common angle) FDB FAD (alt seg theorem)  FDB is similar to  FAD ( all corr s are equal) 9.(ii)  FEB is similar to  FAE FE FB FA FE  (  FEB is similar to  FAE ) GBFAFE 2 FD FB FA FD  (  FDB is similar to  FAD ) GBFAFD 2 22 FEFD  FD = FE 9.(iii)  90ABD ( ABF  BE ) AD is a diameter. (  in semicircle)  90ABE ( ABF  BE ) AE is a diameter. (  in semicircle)  90ADF (tan.  rad.)  90AEF (tan.  rad.) AF is a diameter. (  in semicircle) a circle with AF as a diameter passes through D and E. ________________________________________________________________________ 10.(i) midpt of PQ =       2 21,2 43 =       2 3,2 1 Gradient of PQ = 43 21   = 7 1 Gradient of perpendicular bisector = – 7 Eqn of perpendicular bisector is )2 1(72 3  xy 57  xy ------------ (1)

7 10.(ii) y = – 2x ----------------- (2) Subst (1) into (2): 572  xx x = – 1 From (2), y = 2 Centre of C1 = ( – 1, 2) radius = 22 )21()13(  5 = 5 units the equation of C1 is 25)2()1( 22  yx 10.(iii) dist between R and centre = 22 )25()12(  = 4.2426 units < 5 units R lies inside the circle C1. 10.(iv) Centre of C2 = (1, 2) radius = 5 units the equation of C2 is 25)2()1( 22  yx ___________________________________________________________________________ 11.(i) p = 7 11.(ii) when the scooter changes its direction of motion, v = 0 03 1sin87  t 8 7 3 1sin t basic angle = 1.0654 0654.13 t , 0654.1 1963.3t , 2284.6  3.20 , 6.23 11.(iii) tts d)3 1sin87(  Ctt  3 1cos)3(87 Ctt  3 1cos247

8 when t = 0, s = 0, C 0cos240 C = – 24 243 1cos247  tts when t = 2, 243 2cos24)2(7 s = 8.8612 when t = 3, 241cos24)3(7 s = 9.9672 distance moved in 3rd second = 9.9672 – 8.8612 = 1.1059  1.11 m ___________________________________________________________________________ 12.(a)   xxxxxxx ln613ln3d d 22      = xxx ln63  12.(b)   Cxxxxxx  ln3dln63 2 Cxxxxxxx   d3ln3dln6 2 '2 3ln3 2 2 Cxxx  ''4 1ln2 1dln 22 Cxxxxxx  12.(c)(i) At A, y = 0, 0ln xx x = 0 or ln x = 0 (rej) x = 1 12.(c)(ii) 1 2 1 221 2 1 4 1ln2 1dln      xxxxxx                            22 22 2 1 4 1 2 1ln2 1 2 114 11ln12 1      16 12ln8 1 4 1 16 32ln8 1 

9 2 1 222 1 4 1ln2 1dln      xxxxxx                2222 14 11ln)1(2 124 12ln22 1     4 112ln2 4 32ln2  total shaded area =   2 1 1 2 1 dlndln xxxxxx =       16 32ln8 1 + 4 32ln2  = 0.737 sq unit