MJC JC2 H2 Maths 2012 Paper 2 Solutions

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 2/Math Dep Page 1 of 10 2012 MJC H2 MATH (9740) JC 2 PRELIMINARY EXAM PAPER 2 – SOLUTIONS Qn Solution 1 Recurrence Relations (i) 101.05 40, 600, 1nnu u u n             10 20 2 0 2 30 32 0 1.05 40 1.05 1.05 40 40 1.05 40 1.05 1 1.05 1.05 40 1.05 1 40 1.05 40 1.05 1.05 1 uu uu u uu u                            1 0Thus, 1.05 40 1.05 1.05 1 40 1.05 1 1.05 60 1.05 1 1.05 600 800 1.05 1 800 200 1.05 nn n n n nn n uu             (ii) At the start of 2020, n = 8.     8 8 800 200 1.05 505 3s.f.u    The predicted population is 505 000. The population will eventually be extinct in this lake. There will be no more fish in this lake. (iii) The proposed number to be harvested should be 0.05 600 000 30 000 Qn Solution 2 Functions (i) Least value of a = – 0.5 Greatest value of b = 2 (ii) Domain of -1f is the range of f under the restricted domain.  1fD 0, 25  Under the restricted domain     f 2 1 2 9x x x   Let 24 16 9y x x    y x 4.5 – 0.5 2   2 1 2 9y x x  

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 2/Math Dep Page 2 of 10 24 4 9y x x      2 4 2 4 9yx       2 4 2 25yx     2 252 4 yx   252 4 yx  Therefore 252 2 xx  or  252 reject 0.5 22 xxx      Hence   -1 25f2 2 xx  (iii)  fD 0.5, 2  fR 0, 25  gfR ln2, ln27 Qn Solution 3 Curve Sketching & Differentiation (Parametric, Tan/Norm) (i) 32 2 43 dd 4 3 2 3dd x t t y t t xy tttt         2 d 2 3 d 4 3 yt xt   When 1t ,         2 4 1 3 1 3 1 2 2 1 3d 1d 4 3 1 x y y x           Alternative method Using GC, When d1, 3, 2, 1 d yt x y x    Therefore, the equation of the tangent at 1t is    23 1 yx yx       (ii) If the normal at P is perpendicular to the tangent at 1t , then the tangent at P will be parallel to the tangent at 1t . 2 d 2 3 1d 4 3 yt xt     g f

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 2/Math Dep Page 3 of 10    2 2 2 3 4 3 3 2 1 0 3 1 1 0 1 or 1 (rejected)3 tt tt tt t                Therefore, at point P, the value of 1 3t  . (iii) At 1t , from part (i) 3 2 x y   At 3t ,     3 2 4 3 3 15 3 3 3 18 x y       From part (i), 2 d 2 3 d 4 3 yt xt   For the gradient to be undefined, denominator = 0.   2 444 3 0 ( is rejected 1,3 )33t t t        At 4 2 3 33t , 3 2 2 3 2 3 16 34 3 3 9 2 3 2 3 4 6 333 3 3 x y                                 15,18 O  3, 2 y x 16 3 4 6 3,93  

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 2/Math Dep Page 4 of 10 Qn Solution 4 Area & Volume (i) 2 2 0 2 0 2 0 2 Area of e 1 d e +2e 1 d e +4e e +4e 5 xa xa x ax x a a Rx x x a              (ii)   1 2f e 1 x x        2 22 21 2 00 2 2 f d e 1 d e +4e 2 5 from e 4e 3 y y y y               (i) It is the volume of revolution formed when the region bounded by the curve  2ln 1yx , the x-axis, the y-axis, and the line y = 2 is rotated completely about the y-axis. Alternatively It is the volume of revolution formed when the region bounded by the curve 2e1 x y , the x-axis, the y-axis, and the line x = 2 is rotated completely about the x-axis. Qn Solution 5 Vectors (i) l1: 41 5 2 , 63                      r p1: 2 3 4x y z   p1: 1 24 3    r Since 11 2 2 , 33                    the line l1 is parallel to n1. The line l1 is perpendicular to the plane p1.

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 2/Math Dep Page 5 of 10 41 5 2 2 4 6 3 3 4 10 4 18 9 4 28 14 2 Coordinates of foot of perpendicular is (2,1, 0).                                    (ii) l2: 11 0 1 , 11                      r 2 1 1 1 2 1 4 3 1 3                                n p2: 1 1 1 r 4 0 4 3 1 3                            p2: 4 3 2x y z   (shown) Since 11 2 1 , 31 k                    l1 and l2 are not parallel and since 41 5 4 6 2, 63                   l1 is not on p2, the two lines are on different planes, Hence, they are skew lines. (iii) p1: 2 3 4x y z   p2: 4 3 2x y z   Using GC, l3: 69 1 3 , 01                       r (iv) p3:  2 7 2 2 1 0.x y z x y z         p3: 12 r 2 2 7            Given that the three planes have no point in common, 1 2 9 2 2 3 0 71 9 18 6 6 7 0 11 22 2                                     6 1 2 1 2 2 07 6 12 2 2 8 11 14                                         2 and 14  

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 2/Math Dep Page 6 of 10 Qn Solution 6 Sampling Methods (i) Quota Sampling (ii) The sample obtained is not representative of the population because the proportion of students selected from each stratum is different. OR The sample obtained is non-random because every student does not have an equal chance of being selected. (iii) Stratified sampling Divide the population of 100 into four non-overlapping strata according to levels and gender. The students from each stratum are randomly selected using simple random sampling with the sample size being proportional to the relative size of each stratum, Males Females JC 1 480 100 301600 560 100 351600 JC 2 240 100 151600 320 100 201600 Qn Solution 7 Hypothesis Testing (i) For a one-tailed test, we are testing whether there is a definite increase or definite decrease in a population parameter. As for a two-tailed test, we are merely testing for a change in a population parameter. (ii) x = 1279 50 + 30 = 55.58; 55.58 50 2779x   s2 = 1 49 [155233 – 22779 50 ] = 15.84040816 Let  denote the population mean weight of the boys. H0 :  = m H1 :  > m Since n is large, by CLT: X ~ N(, 2 )n  approximately. Test statistic: XZ S n  ~ N(0,1) Level of Significance: 5% Critical Region: Reject H0 if z-value  1.6449 i.e. 55.58 15.840 / 50 m  1.6449 m  54.654 m  54.7 (3s.f.)

MJC/2012 JC2 Preliminary Exam Suggested Solutions/H2 Math (9740)/Paper 2/Math Dep Page 7 of 10 Qn Solution 8 PnC, Probability (i) Method 1 Method 2 Required Probability 2! 6! 2 7! 7  Required Probability 5! 6 2 2 7! 7  (ii) Let be the event that no two women are seated next to each other.A Let be the event that couple is seated together.B Method 1 Method 2     Required probability n n 2! 3! 3! 2! 6! 1 20 AB B     Method 1   1From (ii) P | . 20AB    3! 4!P 7! 1 35 A      Since P | P , the events are not independent. A B A Qn Solution 9 Correlation & Regression (i) (ii) Using GC, r = 0.989. Using regression line of y on t, 0.041899 0.94916yt When 7t , 6.6860 6.69y cm2 (3s.f.) Since 7t is within the data range and r = 0.989 is close to 1, the answer is reliable. (iii) When 80t , 75.974 76.0y cm2 (3s.f) t y 10 0 8.8 0.3