RVHS_H2_MATHS_P2_ANS

RVHS 2015 Y6 H2 MA Prelim Paper 2 (Solutions) Section A: Pure Mathematics [40 marks] Question 1 [8 Marks] (i) Vertical asymptote when x = 2 2 + b = 0  b = –2 Substitute x = –1, y = 0 into the equation for C:       b a   1 1110 2  a = 2 (ii) (iii)        2 2 2 12 1 1 , 22 x x k x x xx xk        Therefore, draw the graph   2 1 kx y  

2 By observation, the graphs intersect twice when k < –2. Furthermore, when k = –2, equation becomes:       22 2 1 2 2 ( 2)[ 1 ( 2) 1] 0 2 or 2.10380 2 is an answer as well. x x x x x x xx k             Ans: 2, kk  Question 2 [9 Marks] (i) Any horizontal line y = k will cut the graph of y = f(x) at most once. Therefore, f is one-one. (ii) Let y = f(x). Then:   )1 negative (Rej 72 72 34 2 22 2    xyx xy xxy  Thus, 2 ,72:f 2  xxx  (iii) fg exists iff fg DR  i.e. ),1[g R So,  g 12,R   g 1,1D   (1, 2 ) y = f(x) (0, 12) x = –1 x = 1

3 (iv)   gf g 1,1 [12, ) [ 189, ) (OR [13.7, ))D          Therefore, fg [13.7, )R  Alternatively:      31 1241 12 1 12f gf fg 2 2 2 2                   xx x x x From the graph, fg [13.7, )R  y = fg(x) x = 1 x = –1 (0,13.7)

4 Question 3 [11 Marks] (i) 1|2|  iz is a circle centred on (0, 2) with radius 1. 4)arg( w is the half line from the origin (excluding the origin), which makes an angle of 4  with the positive real-axis. (ii) Intersect at one point implies the half line is a tangent to the circle. 62 1sin   Therefore, 362  k or 3 2 62  k Re Im 1 3 2 1 2 3

5 (iii) 2 34 2 1 2 1 2 32  iiiz 22 3 iw  1 2 3

6 Question 4 [12 Marks] (a) 2 dy duy ux x u dx dx     Substituting into DE: 2 2 2 4 ( 2) 0 4 ( 4 4) 0 4 (shown) dy ydx dy yydx dy yydx                 2 2 1 1 1 4 1 1 42 2sin 2 2 2sin 2sin dy dx yy dy dx y y xc y x c xcu x              (b) (i) From part (i),  2sin 2s t c   when 5 6t  , 1s : 51sin 62 5 66 c c c           2sin 2st     (ii) The object oscillates about the starting point which is 2m from O, with an amplitude of 2m. The motion assumes the absence of resistance whereby the amplitude remains constant which is unrealistic in real-life. 4 t s 2 0 2π 4π

7 Section B: Statistics [60 marks] Question 5 [3 Marks] (i) No. of ways =  ( ) 1 !i j k   (ii) No. of ways = 3! ! ! ! 6 ! ! !i j k i j k Question 6 [6 Marks] (a) A population is the entire collection of data that we want to study while a sample is a subset of the units in the population having the same characteristics that we want to measure. (b) (i)