ASRJC_FM_2023_Prelim_P1 (Solutions)

1 2023 ASRJC H2 FM Prelim Paper 1 Solutions 1 031cos 60 223     031cos 54.7363 3     0 0 060 54.736 5.264      0.09193 3 cos0.0919 sin 0.0919 2.99 0.275ia e i i      2 Using the reflective property of ellipse,             2 2 2 2 ' 2 ' cos 2    c FP F P FP F P          2 2 2 2 ' 2 ' cos 2   c FP F P FP F P         22224 ' 2 ' 1 2sin    c FP F P FP F P           22224 ' 2 ' 4 ' sin    c FP F P FP F P FP F P      2224 ' 4 ' sin   c FP F P FP F P      2224 2 4 ' sin c a FP F P    2 2 2' sin  FP F P a c    2 2' sin  bFP F P , shown

2 3(i) 1 nuy      ddd 1 d 1d d d d 1 nn yyuu n y yx x x x n          23 1 d 1 d1 1 nn yuyy xn x n x  1 23 d 1 1 d nun yx xx  23 d 1 1 d un ux xx    , shown (ii) 23 d 2 1 d u ux xx Integrating factor = 2 2 2 d ee x xx   2 2 2 32 1 1 2e e d e d 2 x x xu x x xxx      22 2 11e e d2 2 xx xx x   22 11e + e24 xx Cx  2 11+e24 xuC x    2 2 11+e24 xyC x      1 2 2 11+ e , 024 xy C y x      4(i) 1sin cos 22r    21sin 1 2sin2   2 31 sin42     maximum value of r is 3 4 . Alternatively,  d cos sin 2 cos 2sin cos cos 1 2sind r               d 0 cos 1 2sin 0d r      cos 0 2    or 1sin 2  5,66 

3 2   , 1 2r  5,66   , 3 4r  maximum value of r is 3 4 . (ii) (iii) 1sin 26    Area of triangle 1 3 3 1 3 3 3tan 32 4 4 6 2 4 4 32                   Area of sector 26 0 1 d2 r    2 6 0 11 sin cos 2 d22      226 0 11 sin cos 2 sin cos 2 d24              6 0 1 1 1 11 cos 2 1 cos 4 sin 3 sin d2 2 8 2            6 0 1 5 4cos 2 cos 4 4sin 3 4sin d16           6 0 1 1 45 2sin 2 sin 4 cos3 4cos16 4 3           1 5 1 2 4 42sin sin cos 4cos 416 6 3 4 3 3 2 6 3                      5 9 1 396 128 6    Required area 3 5 9 13332 96 128 6     3 5 13128 96 6    where 3 5 1,,128 96 6a b c   1 2

4 5 Since   11 2 2 0 1d xx is the area of the quarter circle 21yx with centre at origin and radius 1, therefore     11 22 2 0 111 d 1 . 44I x x     Using Simpson’s Rule,   0 1 2 3 4 222 2 1 4 ( 4 2 4 )3 1 1 1 31 4 1 2 1 4 1 1 112 4 2 4 1 15 3 71 4 2 4 012 16 4 16 0.7708988 0.7709 (4 ) I y y y y y dp                                       From symmetry , area under the curve from 1 2 to 1 is 1 1 1 2 4 2 8 4