JPJC_9649_2023_Prelim_P2_Solutions

H2 Further Mathematics 2023 JPJC JC2 Prelim Examination Paper 2 Solutions 1               1 2 12 1 2 2 1 ln ln 2ln2 ln 2ln 4ln n n nn n n n nn XX X X X X X X X          Let  lnnnYX , we have 1 1 2 2 24 2 4 0 n n nn nn Y Y Y Y Y Y        Auxiliary equation:   2 2 i3 2 4 0 2 2 4 4 2 1 3i 2e mm m        Hence, general solution,   2 cos sin 33 2 cos sin 33 ln 2 cos sin 33 e n n n n n nnAB n nnY A B nnX A B X          

2 2 (a)                        1 2 1 1 1 12 2 2 11 1 122 1 1 122 1 11 112 2 2 11 11 122 11 1 1 1d 1 2 1 d 2 1 d 2 1 1 1 d 2 1 1 d 2 1 1 d 2 1 d 2 1 d 2 2 1 2 2 (Show n n nn n n nn nn n n n nn I x x x x nx x x n x x x n x x x n x x n x x x n x x n x x I n I n I n I n I                                              n) (b) From above,   12 1 2 nnn I n I  1 2 21 nn nII n   31 22 1 2 1 21 11 1 3 4 31 42 31 d8 1 3 sin8 3 8 2 2 3 8 II I x x x                       (c)   31 2 2 3 1 2 1dI x x   Using Simpson’s rule with 5 ordinates, let     3 2 2f1 xx x -1 -0.5 0 0.5 1  f x 0 33 8 1 33 8 0     2 12 d11 d d 21d n n vux x u xn x v xx       

3         31 2 2 3 1 2 1d 1 3 3 3 30.5 0 4 2 1 4 03 8 8 1 2 3 36 I x x                         (d) Using (b) and (c),     31 2 3 386 4 2 3 39    

4 3 Let X, Q and R be the points as shown in the diagram. Note that at any time, OQ = PQ = r x-coordinate of P = OQ – PR =  sin sinr r r      y-coordinate of P = XQ – XR =  cos 1 cosr r r    (a) When 2 ,  2 sin 2 2x r r       1 cosyr  When y is maximum, cos 1     sinx r r      and  1 cos 2y r r    Hence, the coordinate of the maximum point is  ,2rr . (b) d( sin ) (1 cos )d xx r r         d(1 cos ) sin d yy r r      Surface area of revolution 222 0 dd2d dd xyy               R y x O  ,2rr 2 r P r Q y x O  X

5 2 2 2 2 2 0 2 2 2 2 0 2 2 0 32 2 2 0 3 2 222 0 2 23 0 2 (1 cos ) (1 cos ) sin d 2 (1 cos ) 1 2cos cos sin d 2 (1 cos ) 2 2cos d 2 2 (1 cos ) d 2 2 2sin d 2 8 sin d (Shown)2 r r r r r r r r                                                 2 23 0 2 22 0 2 22 0 2 3 2 0 2 22 Exact Area 8 sin d 2 8 sin 1 cos d22 8 sin sin cos d2 2 2 cos 28 2cos 2 23 228 2 2 33 64 units3 r r r r r r                                           

6 4 (a) d d d 1d yt ky tt yk ytt   Integrating Factor