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:     d ln ln ee e k k t ktt t kt     Multiplying the DE by the IF,   1 d d d d d Since 1 1 1 k k k kk kk k k k ykt t y ttt t y tt t y t t k tt y C k ty Ctk             Given that 0y when 1t  ,  101 1 1 1 k Ck C k      Hence,  11 1 1 1 kkty t t tk k k       (b) When 2k  ,   d 2d d 2 2 1 f ,d yt y tt y t y y tyt t t      Step size, 0.1h 1 0.1nntt  ,   2f , 1 n nn n yty t  1 20.1f , 0.1 1 n n n n n n n yy y t y y t         0 1t  , 0 0y  1 1.1t  ,  0 10 0 2020.1 1 0 0.1 1 0.1 1 yyy t          

7 2 1.2t  ,  1 21 1 2 0.120.1 1 0.1 0.1 1 0.181818 1.1 yyy t           Hence, 0.182y (3sf) (c) When 2k  ,   21 3y t t  For 1t , the solution curve is increasing and concave downwards, the line segments used in Euler’s method lies above the curve, this results in an overestimate of the exact value of y when 1.2t  . 1 3yt  1,0 y t O

8 5 (a) 2 2 dd 2 3 4dd yy yxx  = 0 Auxiliary equation is 2 2 3 4mm = 0 m =      2 2 3 2 3 4 1 4 2  = 2 3 4 2  = 2 3 2i 2  = 3i General solution is y =  f x =   3e cos sinx A x B x  f 0 0 0 =   0e cos0 0A  0 = A   f x = 3e sinxBx  f x =   33e cos 3 e sinxxB x B x   f 0 1  1 =   0e cos0 0B  1 = B   f x = 3e sinx x (b) Let nP be the statement “    3 1f 2 e sin 6 n nxx x n   ” for all n  LHS of 1P =  f x = 3d e sind x xx = 33e cos 3e sinxx xx =   3e 3 sin cosx xx = 31 1e 2sin tan 3 x x     (R-formula) = 3 12e sin 6 x x   = RHS of 1P 1P is true. Assume that kP is true for some k  , i.e.,    3 1f 2 e sin 6 k kxx x k   .

9 We want to prove that 1kP is true, i.e.,      1 13 1f 2 e sin 1 6 k kxx x k        . 1LHS of kP  =   d fd k xx = 3d1 2 e sind6 kx xkx     = 33 112 3 e sin e cos 66 k x x x k x k              = 3 112 e 3 sin cos 66 kx x k x k              = 31 112 e 2sin tan 6 3 kx xk      (R-formula) = 3 112 e 2sin 66 kx xk     =   13 12 e sin 1 6 kx xk    = 1RHS of kP  1 is true is truekkPP  Since (1) 1P is true, and (2) kP is true 1kP  is true,  by mathematical induction, nP is true for all n  .

10 (c) 3e sinxyx , 03 x  . (d) (i) Given that the positive roots of the equation   1f x x are denoted by 12, , ..., , ...,n   in increasing order. [Refer to the above graphs of   3f e sin xy x x and  1y x , and look for the intersection points at ,,, 1 2 3 4x α α α α .]  f x = 1 x 3e sinx x = 1 x 3 1e sinx x x = 0 Let  g x = 3 1e sinx x x  g x =   2 1f x x  