CJC 2026 Prelim P1 Ans

1 CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 2026 JC2 PRELIM EXAMINATION SOLUTION 1 Suggested Solution     f : translate 1 unit in positive -direction f 1 : scale parallel to x-axis by a scale factor of 2 f 12 : reflect in the -axis f 1 2 2 2 y x A x y x B xy C y xy y x                   So reversing the order:     2 2 ': reflect in the -axis 2 2 1,: scale parallel to x-axis by a scale factor of 2 2 1 2 2 1 ': translate 1 unit in negative -direction 1 1 1 1 2 y x C y y x B y x x A x y x x                

2 2 Suggested Solution (a)         1 2 2 d tan 2cos3d 1 2 sin3 3 1 2cos3 6sin 3 1 4cos 3 xx x x x x       (b)         9 2 18 9 2 18 1 9 18 1 1 sin 3 d 2 1 4cos 3 1 1 6sin3 d2 6 1 4cos 3 1 tan 2cos312 1 tan 1 tan 312 1 12 4 3 144 x x x x xx x                                     

3 3 Suggested Solution (a) (b) Lines of symmetry: x = 1, y = 0 (c) For C to exist, 2 4 x  

4 4 Suggested Solution (a) (b) (c)

5 5 Suggested Solution (a) Method 1 (wrt x-axis) Area of region P              4 4 4 4 14 2 4 4 3 2 4 43 2 4 13 2 2 5 9 4 5 9 4 d 2 9 4 d 6 4 d 46 3 2 4 4 2 64 2 units where 64 x x x x x x x x x k                                           Method 2 (translated graph) Area of region P               5 72 2 5 72 5 72 2 5 5 723 5 2 5 8 d9 52 8 d 9 5 89 3 72 72 8 5 72 0 8 527 16 2 48 2 64 2 units where 64 y y y y y y k                                                         x y  4,5  4,5 72

6 Method 3 (wrt y-axis) Area of region P                       5 72 11 2 2 11 5 5 72 11 2 2 11 5 5 72 113 3 11 5 5 52 4 d 2 4 d 9 9 5 548 2 4 72 24 2 4 d 2 4 d 9 9 5 2 58 72 2 4 2 4 4 5 7 11 9 3 9 12 3 4 y y y y y y y y y y y y                                                                                                     3 3 2 72 72 6 648 2 2 4 5 72 4 11 2 4 11 0 4 527 27 27 48 2 2 16 2 20 24 2 36 2 8 44 20 64 2 units where 64k                                                   (b) Volume generated       14 22 2 11 3 55 18 2 4 d 9 1271.72 units (2 d.p.) y y                x y  4,5  5,14  5, 4 11 1

7 6 Suggested Solution (a)  2 2 12 1 m mx xy x m x m x m          VA: x = m OA: y = x + m-1 (b) Subst (1,1) into 2 2x xy x m    21 1 21 1 m    1− m = −2  m = 3 OR         2 2 2 1 1 2d d x m x x xy x x m        Given tp at (1,1), subst f’(1)=0       20 1 2 1 1 1 1 2m       m = 3 (c)       2 2 2 2 92 4 12 3 34 12 3 3 4 xx x x x x x x x x x              Since x < 3,   2 2 1 33 4x x xx   

8 Insert the graph of  1 34y x  , showing downward sloping and -ve y-intercept. Thus 2.29 x  3 or 1.49x

9 7 Suggested Solution (a) Using similar triangles,         8 8 9 8 9 8 8 9 8 9 8 H r H h H r H r H r rH H rH r rH r              Volume of the container                                   2 2 2 2 2 2 2 2 2 2 2 2 1 1 3 3 8 9 8 91 1 8 93 8 3 8 1 512 893 8 8 512 8 81 93 8 8 64 81 93 8 8 8 8 81 93 8 8 81 9 83 R H r H h r r r rr r r r r r r r r rr r r r r r r r r r rr r r rr r                                                                                   2 8 1 9 8 64 (shown)3 r r r r            (b)   21 9 8 643V r r r              2 2 2 2 d 1 9 2 8 8 64 1d 3 1 2 10 72 8 643 1 3 2 83 V r r r rr r r r r r r                       

10 At stationary point,    2 2 d 0d 3 2 8 0 3 2 8 0 3 4 2 0 4 or 23 V r r r r r r r r r              Method 1: r Sign of d d V r 0 Tangent / __ \ V is a maximum when . Method 2:     2 2 2 d 1 3 2 8d 3 d 1 6 2d 3 V r rr V rr          2 2 2 d 10 0d 3 rVr    V is a maximum when 2r . Maximum volume, V      2 3 1 9 2 2 8 2 643 196 cm        2 2 2 0 0 2r