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,