2012 VJC H2 MA Prelim P2 Solutions

Victoria Junior College H2 Mathematics (9740) – 2012 Preliminary Examinations Paper 2 This list of common mistakes and comments are compiled to maximise your learning through the mistakes made. Do not miss the opportunity. Go through all the details carefully so that you will not make the same mistakes in the A level exams. 1. (i) The locus of point P is the half-line from (1, 2) (excluding it), inclined at an angle of  with the positive real axis. (ii) 2 6 2i 2w   (iii) 2 6 2i 2w   2. (i)   1 1 1 11r r r r      1 1 1 1 1 122 11 1 1 1212 1 2 2 n n n r r r rr r r r r n n                    1 .3 1.. n 1 1 1(1 ) 1 1 n nn n             1 11 12 1 nn rr rr u u r rr         1u 0 2 u u   1u 1 ... nnuu  0 1( 1) 1 1 11 ( 1) 1 (since 1 as given)1 1( 1) 2 (shown)1 n n nn n u n n u n u n n n                  Locus of P meets locus of Q more than once if 1122 55sin sin 0.412 0.412          2 2 5 |w – 6 – 2i| = 2 (ii)  4 (1, 2) /6  (1, 6) |z – 1 – 6i|min |z – 1 – 6i|min 4sin 26 4sin 3 23       (1, 2)

    1 1 2 1 Let 1. When 1, 2; when 1, 12( 1) ( 1)( 2) 12 1 1122 1 1(1 1) 15( 1) 1 12 13( 1) 12 n r n k n k k r r k r n k n r rr k kk k kk nn n nn n                                         3. i) Since replacing x with –x results in the same equation, the graph is symmetrical about the y-axis. (ii)   22 2 2 2 Differentiate with respect to : dd.2 2dd d 2 1 2d d2 (shown)d 2 1 y y y y e x y x yyex xx y exx yx xe      (iii) The line xk is vertical, so at x = k, d d y x is undefined. 2 112 1 0 ln . 22 yey     When 11ln ,22y     11 22 2 2 2 ln 11ln22 1 1 1 1ln 1 ln 22 2 2 2 ek k      Hence, 1 ln 2 2k  . (iv) When d1, 0, 2. d yxy x   Equation of tangent :  0 2 1yx   2 2.yx   Solving equations of the tangent and the curve simultaneously, we get 4 4 2 2 2.xe x x    From GC, 1x (rejected as this gives the previous point) or 2.73x (3 s.f.) (iii) Alternative    1 1 1 2( 1) 12 n r r rr      = 4 6 8 ... 2 n    1 1 1 ... 2.3 3.4 ( 1) nn         1 2 11 1 2 1.2 n r rr r    15 ( 1) 1 12 13 ( 1) 12 nn n nn n           

4. (i) 2 4 6 solve simultaneously using GC, 6 x y z x y z         1 2 1 2 5 1 0 xz yz zz    So the equation of the line l is 51 1 1 , . 02                  r (ii) Given that the system of equations of the 3 planes has infinitely many solutions, it implies that line l lies on the plane p3. So 5 12 21 a b                     . for all real values of .   5 2 2 2 5 4 2 a a b a a b               This is satisfied when a = 4 and therefore b = 18. OR 3 1 normal of 1 2 0 4 21 a l p a                    . . 3 54 (5,1,0) lies in 1 2 18 01 p b b                   . (iii) plane is perpendicular to p1 and p2  it is perpendicular to l. normal to the plane is 1 1 2    . Equation of the plane is 1 1 1 1 1 1 2 1 2                         r 1 1 4. 2     r (iv) p1, p2 and p4 have no common point: p1 p2 examples of p4 line l line m (3, 4, 1)  m is parallel to l. distance between m and l = distance from (3,4,1) to l. B(3,4,1) m  l  A(5,1,0) 1 1 2    2 1 2 1 5 1 3 1 5 2 1 2 5 661 1 2 5 3 5 units 62 AB d                                           d

5. (i) Let  cm be the population mean increase in heights of orchid plants in the nursery. 0 1 H : 1.34 H : 1.34     As n = 55 is large, by Central Limit Theorem, X is approximately normal. no assumption about the increase in heights is needed. (ii) Level of significance: 5% Test-statistic: When H0 is true, 1.34 ~ N(0,1)0.24 55 XZ  Distribution of Z 0, 1.34To NOT reject H 1.64490.24 55 0.241.34 1.6449 1.3932 55 xz x         : 1.39xx (iii) There is a 5% probability of the test concluding that there is a significant improvement in the mean increase in heights of orchids when there is not. 6 (a) Due to time and resource constraints, the committee would prefer to poll a sample instead of all members. Obtain a numbered name list of the club members. Use a random number generator to choose an integer from 1 to 10. Choose the member corresponding to that number on the name list followed by every 10th member subsequently until 100 members are selected. If there are cyclical patterns in the name list, the survey might be biased. (b) Case 1: Adam & Bernice are together : 9! 2! 725760 Case 2: Adam & Bernice separated by exactly 1 person: 8 18! 2! 645120C   Required number of ways = 10! 725760 645120 2257920   7. Let X be the number of free gifts claimed per week.  ~ Po 8 .X (i) Let n denote the number of gifts kept in stock at the start of the week. For   0.95P X n , from the GC: n P(X ≤ n) 12 0.9362 13 0.96582 0 1.6449 0.05 the least number of free gifts needed is 13. 0.95 Rejection Region:

7. (ii) Let Y be the number of free gifts claimed in m weeks.  ~ Po 8Ym *If 8m > 10,  ~N 8 , 8 approximately.Y m m     P 50 0.1 Using continuity correction, P 50.5 0.1 Y Y   . Standardising, we have 50.5 8P 0.10 8 mZ m   . 50.5 8 1.2816 8 50.5 8 1.2816 8 0 m m mm       largest integer value of m = 5. *for Normal approximation to Poisson distribution, m > 5/4. 8. Let X and Y denote the waiting times, in minutes, at the fitting room for a man and a woman respectively. Given E(X) = 4.2, Var(X) = 1.62 & E(Y) = 8.5, Var(Y) = 2.22 (i) Let 1 2 50 50 Y Y YY    . By Central Limit Theorem, 22.2~ N 8.5, approximately.50Y   For  P 0.10Yk , Y 8.8987k least integer value of k = 9 Given X ~ N(4.2, 1.62), Y ~ N(8.5, 2.22) , (ii)   16 22 ... 3 ~ 6 4.2 3 8.5, 6 1.6 9 2.2 ~ (-0.3, 58.92) T X X Y TN TN                  P 4 P 4 P 4 0.28767 0.31489 0.603 (to 3 s.f.) T T T         Assumption: The waiting times for the 6 men and 1 woman are all independent of one another. (iii) Let W be the number of woman out of 4 with waiting time more than 8.5mins. W ~ B(4, 0.5) P(W ≥ 3) = 1 – P(W ≤ 2) = 0.3125 0 1.2816 0.10 50.5 8 1.2816 8y x x   Z 0.10 8.8987 0.90 0.90

9. (i) (ii) r = 0.971 (to 3 s.f.) The