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) This procedure will not result in a random sample as each of the 120 teachers does not have an equal chance of being selected, e.g. P(a Humanities teacher chosen) = 10 21 while P(a Language teacher chosen) = 10 33 . (ii) We calculate the numbe r of teachers from each department based on simple ratio and proportion: Department Number of teachers selected Humanities 21 40 7120 Language 33 40 11120 Mathematics 27 40 9120 Science 39 40 13120 7 Humanities teachers, 11 Language teachers, 9 Mathematics teachers and 13 Science teachers are chosen randomly from each department.

8 Question 7 [7 Marks] Let X denote the number of correct answers gotten by someone totally clueless of Singapore’s history. Then ~ B(8, )Xp , where p is the probability of getting a correct answer by mere guessing. Given P( 1) 0.50331648X  87 8(1 ) (1 ) 0.5031p p p     87(1 ) 8 (1 ) 0.503 0p p p     From the graph, p  0.200 = 1 5 . Therefore, there are 5 options for each of the question. (shown) So ~ B(8,0.2)X P(X = 0) = 0.1678 P(X = 1) = 0.3355 P(X = 2) = 0.2936 The mode is 1. 1.28~ N 1.6, 50X   approx. by CLT, since n = 50 is large. P( 2) 0.00621X   (to 3 s.f.) Question 8 [7 Marks] (i) The average number of calls requiring ambulance assistance is constant throughout the operating hours of the hospital in a day. OR For any time interval within the operating hours of the hospital in a day, the mean number of calls requiring ambulance assistance is proportional to the time interval. (ii) Let X denote the number of calls requiring ambulance assistance in 1 week. Then ~ Po(7)X . Required probability = P( 5) P( 4) 0.172992 0.173 (to 3sf)XX     87(1 ) 8 (1 ) 0.503y x x x    

9 (iii) Let Y denote the number of weeks (out of 52) with more than 4 calls requiring ambulance assistance. Then ~ B(52,1 0.172992)Y  i.e. ~ B(52,0.827008)Y Since n = 52 is large, 43.004416 5 and 8.995584 5np nq    , ~ N(43.004416,7.439419933)Y approx. Required probability P(40 45) P(40.5 45.5) C.C. 0.641 (to 3sf) Y Y        Question 9 [8 Marks] (i) Let X denote the time required by the machine to complete a task and μ be the mean time. Test 0H : 47.0 against 1H : 47.0 (ii) 48.1x 2 2 2 12 (1.9) 3.9381818181 11 n ns n    Test statistic: Under 0H , 47.0 ~ (11) 3.938181818 12 XTt  Using GC, p-value = 0.0405659745 > 0.03 (level of significance)  do not rej. 0H . There is insufficient evidence at 3% level of significance that the mean time required by the machine to complete the task is understated. (iii) Given: σ = 2.1 Test statistic: Under 0H , 47.0 ~ N(0,1)2.1 12 XZ  For 0H to be rejected, cal criticalZZ  i.e. 47.0 1.880793612.1 12 x  48.14017053x 48.1 (or 48.2)xx  

10 Question 10 [8 Marks] (i) P(all 3 balls have different colours) = 3 4 3 1 4P  = 3 8 or 0.375 (ii) 33 4 4 3 3 3 2 P(the 3 balls have diff. colours and nos '0,0,0' or '0,1,1') P(all 3 balls have different colours) 11 88 3 8 1 2 P P C              1P( | ) 2AB  (from part (ii)) 33 3 1 P( ) P(0,0,0) P(0,1,1) 11 22 1 P( | )2 A C AB                A and B are independent. ALTERNATIVE METHOD 3P( ) 8B  (from part (i)) 33 3 1 P( ) P(0,0,0) P(0,1,1) 11 22 1 2 A C               1 3 3P( ) P( ) 2 8 16AB      3In addition, P( ) from (ii)16 P( ) P( ) AB AB   Hence, A and B are independent.