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 vertical asymptote 2x , 2 2 0b 1b Method : To find equation of vertical asymptote, equate the denominator to 0, 220 2 21 bx x b bb        Substituting  3,13 into 1 2 axy x   , 3113 32 a  13 3 1a 4a (ii) By long division, 4 1 9 422 xy xx    Method : 1y x [replace with 2] translate 2 units in the positive -direc tion xx x  1 2y x  [replace with ] 9 scale parallel to -axis by a scale facto r of 9 yy y  1 92 9 2 y x y x     [replace with 4] translate 4 units in the positive -direc tion yy y  94 2 94 2 y x y x    

Page 7 of 23 Method : 1y x [replace with 2] translate 2 units in the positive -direc tion xx x  1 2y x  4[replace with ] 9 4translate units in the positive -direct ion9 yy y   41 92y x  [replace with ] 9 scale parallel to -axis by a scale facto r of 9 yy y  41 9 9 2 94 2 y x y x     Method : 1y x [replace with ] 9 scale parallel to -axis by a scale facto r of 9 xx x  1 9 9 y x y x   [replace with 2] translate 2 units in the positive -direc tion xx x  9 2y x  [replace with 4] translate 4 units in the positive -direc tion yy y  94 2 94 2 y x y x    

Page 8 of 23 Q4 Solution (i) 1 16 : 6 -----(1)4u p q r    2 10 : 2 0 -----(2)16u p q r    3 15 1 15: 3 -----(3)4 64 4u p q r    Solving (1), (2) and (3), 16, 3, 5p q r   116 3 54 n nun     (ii) Method :     11 1 1 1 1 111 G.P.: 4 , 4 ,number of terms A.P.: 1, 1,number of terms 116 3 54 16 4 3 5 16 4 4 4 4 3 1 2 1 5 5 5 5 n r r n n n r r r r n n a r n a d n r r nn                                                               times 1 1 1 1 11 2 4 1 4 4 1 4 16 3 25 OR 16 3 51 4 1 4 16 31 4 1 532 16 16 7 343 3 2 2 16 16 7 3where , , and .3 3 2 2 1 2 1 1 12 n nn n n nn nn n n n nn AB n D n C                                              Method :           11 1 11 112 G.P.: 4 , 4 ,number of terms A.P.: 2, 3,number of terms 116 3 54 16 4 3 5 16 4 4 4 4 2 1 3 1 5 3 5 n r r nn r rr n n a r n a d n r r nn                                                         1 1 1 1 11 2 2 4 1 4 4 1 4 16 OR 161 4 1 4 16 3 7143 2 2 16 16 7 343 3 2 2 16 16 7 3where , , a 2 3 5 2 2 1 322 . nd 3 3 2 2 nn n n nn nn nn nn A B C D                                                

Page 9 of 23 Q5 Solution (a)(i)         22 1 2 3 3 lnd ln 2 3 3 ln d 2 3 2 3 2 3 xxx x x x x xx x x x      (ii)   2 13 32 d 3 2sin ( 2 )d 1 ( 2 ) xxxx xx    (b) Differentiating   2221xy y y   implicitly with respect to x,     d d d2 2 2 2 1d d d d d d2 2 2 2 2d d d d2 d 4 2 2 2 1 y y yy x y y x x x y y yy y x yx x x y y y x y x y x               Where tangent is parallel to the x-axis, d 00d y yx    When 0y ,     2 2 L.H.S. 2 (0) 0 0 R.H.S. 1 0 1 x      Hence there are no points on the curve where the tangent is parallel to the x-axis.

Page 10 of 23 Q6 Solution (i) (ii) (iii)