JPJC 2024 J2 H2 FM Prelim P1 Solutions

1 2024 J2 H2 FM Prelims P1 (Solutions) Qn Solution 1 Let Pn be the statement that 2 1 1 1 2 n r r n   for all integers 2n . To show P2 is true: LHS = 2 2 1 1 5 4r r  . RHS = 1 3 52 2 2 4   . Hence P2 is true. Suppose Pk is true for some integer 2k  , i.e. 2 1 1 1 2 k r r k   . We want to show Pk + 1 is also true, i.e. 1 2 1 1 1 2 1 k r r k      .         1 22 2 1 1 2 2 2 2 2 1 1 1 1 1 12 1 12 1 2 1 12 1 k k r r r r k k k k k k k k k k k k                       Hence Pk + 1 is true. Since P2 is true and Pk implies Pk + 1, by the Principle of Mathematical Induction, Pn is true for all integers 2n . 2(i) Using GC,   1 0 2 2 rref 0 1 0 2 0 0 0 0        A or   7 201 2 3 3 ref 0 1 0 2 0 0 0 0              A TDim( ) Rank( ) 2R A  By the Rank-Nullity Theorem, TDim( ) 4 2 2K   

2 2(ii) Basis for TR 1 2 1 , 1 3 7                       Let 1 0 2 2 0 0 1 0 2 0 0 0 0 0 0 w x y z                       . Then 2 2 0 2 0 w y z x z       2 2 2 w y z x z    2 2 2 2 2 0 2 1 0 0 1 w y z x z y zy y z z                                                     Basis for TK 2 2 0 2,1 0 0 1                              As T 3 1 11 R        , 3 1 11        Ax for some 4x  . Observe 3 1 2 1 1 1 2 1 11 3 7                             , so a possible solution 1 2 0 0            x The general solution of 3 T( ) 1 11        x is 1 2 2 2 0 2 , ,0 1 0 0 0 1                                            x 

3 3(i) Since the daily removal of concentration level is 7%, i.e. the concentration remaining end of each day is 93%. The recurrence relation is   7 1 00.93 2.5, 0n nu u u    i.e. 1 00.6017 2.5, 0n nu u u    Hence, 7 7 7 0 2.5(0.93 ) 1 0.93 (0.93 ) 6.2767, 0 n n n u c c u       When 0, 0 6.2767 6.2767n c c     Therefore,   6.28 1 0.602 (to 3 s.f.), 0, n nu n n     3(i) Alternatively:             7 1 7 7 2 27 7 2 27 7 7 3 3 27 7 7 3 0.93 2.5 0.93 0.93 2.5 2.5 0.93 (0.93 )2.5 2.5 0.93 0.93 2.5 (0.93 )2.5 2.5 0.93 0.93 2.5 (0.93 )2.5 2.5 n n n n n n u u u u u u                      Therefore,          1 27 7 7 7 0 7 07 0.93 0.93 2.5 0.93 2.5 .... (0.93 )2.5 2.5 1 0.93 2.5 since 01 0.93 6.28 1 0.602 (to 3 s.f.), 0, n nn n n n u u u n n                    3(ii) Since    76.2767 1 0.93 6.2767 n nu    for all 0n , the concentration level will always be less than 7 mg/l. So the Local Authority should allow the factory to go ahead with the discharge 4 Since 2 is an eigenvalue of A,     2 2 det 2 0 2 3 3 2 7 3 0 2 8 4 ( 2) ( 7)( 4) ( 3)(8) (2)[( 3)( 4) ( 3)(8)] 2[( 3)( 3) ( 3)( 7)] 0 ( 2)( 3 28 24) 6 24 6 24 0 ( 2)( 3 4) 0 ( 2)( 1)( 4) 0 1, 2 or 4                                                    a a a a a a a a a a a a a a a a a a a a A I

4 4(i)  det 0  A I   2 1 3 3 2 6 3 0 2 8 5 ( 1 ) ( 6 )(5 ) ( 3)(8) (2)[( 3)(5 ) ( 3)(8)] 2[( 3)( 3) ( 3)( 6 )] 0 ( 1)( 6) 0 3, 1 or 2                                                     Using a GC, For 3  , 2 3 3 2 3 3 2 8 8           x 0 . An eigenvector is 0 1 1         . For 1,  0 3 3 2 5 3 2 8 6           x 0 . An eigenvector is 1 1 1         For 2,  3 3 3 2 8 3 2 8 3            x 0 . An eigenvector is 1 1 2         . 4(ii) If  is an eigenvalue of A with corresponding eigenvector x, then  2 3 2 3 2 3 (2 3) , i.e.        A I x Ax x x x x 2 3 is an eigenvalue of A with corresponding eigenvector x. Therefore the eigenvalues of A are –9, –5, 1, with corresponding eigenvectors 0 1 , 1         1 1 , 1         and 1 1 2         respectively. 5(i) 2 2 2 d d 0d d x xm k m xt t    Characteristic equation: 2 2 2 2 2 0 4( )( ) i can be ignored22 ks m m m ks k m ms k k m           ∵ General solution of x is  2e cos( ) sin( ) k tmx A t B t      . At t = 0, x = 0  A = 0. Since 2e sin( ) k tmx B t    , 2 2d e sin( ) e cos( )d 2 k kt tm mx Bk t B tt m       

5 At t = 0, d .d x vt vB    2e sin( ) k tmvx t     5(ii) Period of vibrations, 2 .T   Let 0 ,2t   Amplitude at time t0 after the nth period,  02e k t nTm n vA     Amplitude at time t0 after the (n+1)th  0 ( 1)2 1 e k t n Tm n vA             0 0 ( 1)2 1 2 2 e e e k t n Tm k Tn m k t nTn m v A A v           (constant) Thus, amplitude of successive vibrations follows a geometric progression. 5(iii) Roots to the characteristic eqn will be real, distinct and negative. Thus the general solution will be   1 2e em t m tx A    where m1 and m2 are positive constants. x will initially increase, then gradually decrease and tend to zero (overdamped system). A possible graph is − 0 . 5 0 . 5 1 1 . 5 2 −0 . 1 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 x y x t

6 6(i) 2 1n n nx x kx   6(ii) (ii) = = 0 The auxiliary equation is = 0 m =        2 1 1 4 1 0.11 2       = 1 1.44 2  m =1.1 or = Sub. , 20 = Sub. , 25 = Solving, c = 22.5 d = = 6(iii)   2 1 1 11 n n n n n n y y ky ry r y ky           7(i)     2d1 2 1 0d yx y x yx       2 22 1d 2 d 1 1 y x yy y yx x x      Using Euler’s method,   2 1 1 2f , (0.5) 1 n n n n n n n n n yy y h x y y y y x            , where 0 0 0, 1x y  . 20 1 0 0 0 2(0.5) 1 21 (0.5) 1 1 1.5 yy y y x                21 2 1 1 1 2 2(0.5) 1 2 1.51.5 (0.5) 1.51 0.5 1.375 yy y y x              2nx  1 0.11n nx x  2 1 0.11n n nx x x   2 0.11m m  0.1  nx    1.1 0.1 nnc d   0 20u  c d 1 25u  1.1 0.1c d 2.5  nx    22.5 1.1 2.5 0.1 nn  

7 7(ii) 2 1 ... (1) d 1 d ... (2)d d y z y z x z x   Subst. (1) and (2) into     2d1 2 1 0d yx y x yx     ,         2 2 1 d 1 11 2 1 0 d d1 2 1 0d d 2 1 (shown)d 1 zx x z x z z zx z xx z zx x                                 2 d1 22ln 1 I.F. e e 1 xx x x                         2 2 2 2 2 2 2 3 2 d 21 1 1 d 1 d 1 1d 1 1 d 1 3 1 3 1 zx x z xx x x z xx x z x x x c x cz x                               3 2 2 3 1 31 3 1 3 1 , where 3 1 x c y x xy A c x A        When x = 0, y = 1, 3 1 21 AA    .     2 3 3 1 1 2 xy x     7(iii) When x = 1,     2 3 3 1 1 6 51 1 2 y     . 1.2 1.375% error 100 1.2 14.6    To improve, use a smaller step size h

8 8(i)      d dsin 1 cos 1 cos ,, sind d xx r y r r y r            Surface area           2π 2 2 0 2π 22 2 2 0 2π 2 0 2π 2 0 2π 2 0 2π 2 2 0 2 2 2 2 2π d 2π d 2π d d d d d 1 cos 1 cos sin 1 cos 1 2cos cos sin 1 cos 2 2cos 1 cos 2 1 1 2sin 2 1 cos sin2 2 si 2π d 2π d 8π d 8π n 2 y r r r r r r r r x y                                                                2π 0 2π 2 0 2 2 2 3 cos sin2 2 2 co d 8π 2 cos 2 2 6 s2 3 2 4π8π 2 23 3 3 r rr                        2 2 64πGiven 432 π.3 81 9 4 2 r r r    

9 8(ii) At  3π 2, 2 ,   3π2 1 cos cos 02 2       Equation of L: 2 3π 2y x  Volume of solid generated                 2 2 2 1.5π 0 1.5π 0 4πd 3π 2 3 π2 sin 1 cos 1 cos d π 3π2π