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 inf