CJC H2 MATH P2 SOL

Page 1 of 25 CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS JC2 PRELIMINARY EXAMINATION PAPER 2 SOLUTIONS 2015 Section A: Pure Mathematics 1 No. Solution Finding table values in SGD, Cheese/kg Chocolate/kg Candy/kg Price/SGD 4 6 6 Price/SGD 8 10 4 Price/SGD 8 5 7 Let x, y, z be the number of three kg packs bought from Denmark, England and Russia respectively. 4 8 8 84 6 10 5 85 6 4 7 77 x y z x y z x y z          5,x 3,y 5z She should buy 13 packs in total.

Page 2 of 25 2 No. Solution (i)(a) 2f ( ) f ( ) f ( )y x y x y x     (b) f '( )yx x y O (1, 0) 1x x y O (2, 0) (–2, 0)

Page 3 of 25 (ii) Ans: 5 (iii) Method :   f ( ) I : translation of 1 units in the negative -direction f ( 1) 1II :scaling parallel to -axis by a scale factor of 2 f 2( ) 1 III : translation of 3 units in the negative -di rection f (2 1) 3 yx x yx x yx y yx          Method : x y O 3x 1x 1 2y Alternative x y O 1 2y 1 2y  fyx  fyx

Page 4 of 25 f ( ) 1I :scaling parallel to -axis by a scale f actor of 2 f (2 ) 1II : translation of units in the negative -dire ction2 1f2 2 III : translation of 3 units in the negative -di rection f (2 1) 3 yx x yx x yx y yx          

Page 5 of 25 3 No. Solution (i) 1i1 i 2 11 i 1 i 2 a     2 2 2 11 i 1 i 2 b       1iarg arg 1ia       arg 1 i arg 1 i    44 ππ    2 π   2arg arg 1ib       arg 2 arg 1 i   π0 4    4 π

Page 6 of 25 (ii) π 24 28 π πBOC       π π 3πarg 8 4 8ab    2 2 1 i 1 i 1 2i i 2i i1 i 1 i 1 i 2a              2 2 1 i2 1 i 2 1i1 i 1 i 1 i 2b         2 2 2i 1 i 1 i2 2 2ab          2 132 2tan 1 1 2 22 2 π 8       (shown) Re Im O A B C

Page 7 of 25 (iii) (a)  02ii 2zz      Locus is a circle, centre (0, 1) and radius 2 (b)   πarg 2zb 22arg i 2 2 2 πz        Locus is a half-line from the point 22,22   (excluding the angle itself) making an angle of π 2 to the positive Re-axis direction. Equation of circle:   222 12x y   When 2 2x ,   2 2 22 122 y        2 2 7 2 771 or 1 (rej. is positive)2 1 142 71 2 2 1 y y y yy         27 1i22z      Alternative for finding y From diagram, using Pythagoras’ theorem, 2 2 271 2 1 22y      4 Re Im O 1 z y

Page 8 of 25 No. Solution (i) d 0.64 1d 10 PP Pt    d 0.064 10d P PPt    1d 0.06410 d P P P t    1 d 0.064d10 PtPP     1 1 1 + d 0.064d10 10 PtPP    11+ d 0.64d10 PtPP    ln ln 10 0.64P P t C    ln 0.6410 P tCP  0.64e10 tCP P   0.64e , where 10 et cP AP A  0.6410 1 e , where tP B BPA   0.6410 1e t P B  0.64 10 1e tP B   When 0t , 1P , 0 101 1e B  9B 0.649 10 1e tP    The population will approach 10 asymptotically in the long run. (ii) d 10 0.4 lnd P PtP   1d 0.4(ln10 ln ) d P P P t  1 0.4d(ln10 ln ) dP tPP    10 1

Page 9 of 25 ln ln10 ln 0.4P t C   0.4ln10 ln e tCP  0.4ln10 ln , whereee t CPA A   0.4ln ln10 e tAP  0.4e10e tAP 

Page 10 of 25 When 0t , 1P , 0e1 10e A ln10A 0.4(ln10)e10e t P  (Similarity) The population approaches 10 asymptotically in the long run in both models. (Difference)  The population in the second model would increase faster first then slower compared to the first model. OR  The Gompertz function will take a longer time to approach 10 as compared to the Logistic function. 1 10 Gompertz Logistic