CJC_9758_2022_Promo_Solutions

Page 1 of 23 CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 2022 JC1 PROMOTIONAL EXAMINATION SOLUTION Q1 Solution (i) x y 3 -3 -1.5

Page 2 of 23 (ii) Method : Hence (Graphical Method) For exact points of intersection, let 2 323 28xx x    22 10 6 0xx   2 5 3 0xx        25 5 4 1 3 21x    5 13 2x  or 5 13 2x  5 13 2x  (reject 5 13 2x  since from graph, 1.5x )   22 8 323x xx    22 8 323x xx    22 6 0xx  30xx  0 or 3xx  3x  (reject 0x , since from graph, 1.5x ) Therefore, for 22 3 2 8 3x x x    , 5 13 or 32xx   Method : Otherwise (Definition of Modulus) 22 3 2 8 3x x x    22 3 2 8 3x x x     (for 1.5x ) OR   22 3 2 8 3x x x     (for 1.5x ) 22 3 2 8 3 0x x x      22 3 2 8 3x x x    22 10 6 0xx    20 2 8 3 2 3x x x     2 5 3 0xx    20 2 6xx 203 xx  03 xx x y 3 -3 -1.5

Page 3 of 23 2Let 5 3 0xx   ,  Let 3 0,xx       25 5 4 1 3 21x    Critical Points: 0, 3xx  Critical Points: 5 13 2x  , 5 13 2x  Test Point Method: 5 13 5 13 (rejected) or 22xx       3 or 0 (rejected)xx   x +  + x 0 3 +  +

Page 4 of 23 Q2 Solution (i) Let a and d be the first term and common difference of the arithmetic series respectively. Let b and r be the first term and common ratio of the geometric series respectively. A.P. G.P. 2 5 10 4 9 T a d T a d T a d    2 3 2 4 ub u br u br    Method :        2 2 2 2 2 2 94 4 94 10 9 8 16 2 7 0 2 7 0 0 or 3.5 a d a d a d a d a d a d a d a ad d a ad d ad d d a d d a d                rejected 0 d  3.5 4 3.5 5 3 ddr dd r    Method :         2 4 1 3 ------(1) 94 1 5 ------(2) br b a d a d b r d br br a d a d br r d             (2) (1) :     1 5 13 5 3 br r d b r d r    (since 1r  )

Page 5 of 23 Method :     2 2 4 3 ------(1)3 94 5 ------(2) br b a d a d br b d br bd br br a d a d br br d              Substitute (1) into (2):      2 2 2 5 3 3 3 5 5 (since 0) 3 8 5 0 3 5 1 0 51 or 3 br br br b r r r b rr rr rr                rejected 1 r  (ii) Using 73.5 2 3.5a d d    Using 7 2 9b a d            1 1 1 1 1 1000 9 7 1 2 1000 59 7 2 1 10003 59 2 5 10003 n n n n br a n d rn n n                       n 1 59 2 53 n n    10 868.06 < 1000 11 1461.4 > 1000 12 2451.7 > 1000 Least n = 11

Page 6 of 23 Q3 Solution (i) Method : Since C has a ver