JPJC_9758_2022_Promo_Solutions.pdf

Jurong Pioneer Junior College H2 Mathematics JC1-2022 Year-End Exam Solution Q1 (i) (ii) 0.349 2 x or 2.15x Q2 (a) (i)   33 3 33 d 9e 3eln 9 3ed 9 3e 3 e xx x xxx     (ii)   4 2 3 2 2 3 2 2d sin 4sin cos 2 8 sin cosd x x x x x x xx  (iii) x = 2 y x 0.349

2 1d sec3 sin 2d xxx      1 2 1sec3 2 sin 2 3sec3 tan 3 12 x x x x x   1 2 2sec3 3tan 3 sin 2 14 x x x x    (b) 32 4 9 10 0y y x x     2 dd3 4 2 9 0dd yyyx xx    2 d 9 2 d 3 4 yx xy   Q3 32 d13 d uu t t t    When 0t  , 1u . When 2t  , 9u          22 325 3333 0 0 9 3 1 9 33 1 9 23 1 912 1 9 2 1 dd 11 11 d3 11 d3 1 d3 1 3 1 2 1 1 1 1 1 1 1 32 1 3 2 3 9 2 81 2 243 ttt tt tt u uu u uuu u u u uu uu                                    Q4 (i)

3 2 4 1 x kxy x          2 2 2 1 4d d 1 x k x x kxy x x            2 2 24 1 x x k x     d 0d y x     2 2 4 0x x k    For two stationary points:   22 4 4 0k   5k  Alternative: 2 45 ( 1)11 x kx ky x k xx            2 5d 1d 1 ky x x   d 0d y x     2 15xk   15xk   For two stationary points: 50 k 5k  (ii) When 4k  , 2 44 1 xxy x   (iii) 2 4 ( 3)( 1) 0x kx mx x      x = –1 y x y = x + 3 (0,4)

4 2 4 ( 3)( 1)x kx mx x     2 4 31 x kx mxx   , where 4k  . For graphs of 2 4 1 x kxy x   and 3y mx to have no intersection and hence 2 4 ( 3)( 1) 0x kx mx x      to have no real root, 01 m . Q5 (i) Using similar triangles, 0.6 0.7 6 7 x y xy   Let the volume of water in the tank be V.   21 6 12422 7 7V xy y y y     d 24 d7 V yy  Since d d d ,d d d V V y t y t     24 d0.0025 0.47d d 0.0018 m/s (4 d.p)d y t y t    (ii) Time taken =     21 120.6 (0.7)(4) 0.4 0.0025 226s27     (nearest second) Or Time taken =   2212 12(0.7) 0.4 0.0025 226s77     (nearest second) Q6 (i) 0.7 y x 0.6

5 Using GC, from graphs, the intersection points of C and l are (2, 4) and (5,1) . Alternative 2 2 or 6 8 16 7 10 0 (or use GC polynomial rootfinder) ( 5)( 2) 0 5 2 x x x xx xx xx               When 2x , 2 6 4y   When 5x , 5 6 1y   The intersection points of C and l are (2, 4) and (5,1) . (ii) Area of the region R   5 2 54 6 d 4 d a x x x x     or 5 2 54 6 d 8 16 d a x x x x x        532 5 4 4623 a xx x         or 52 3 2 54 8