EJC 9649 2023 Prelim P1 Solutions

2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over Q1 Solution 3d 4 d4 , 8dd xy ttt t t       22 3 6 2 2 62 2 62 2 2 3 24 2 2 2 2 d d 4 48dd 1616 32 64 1616 32 116 2 116 16 1 xy ttt t t t t t t tt t tt t t t t t                                   Surface area       1 5 4 2 62 1 62 62 1 4 2d 32 d 32 1232 23 34 1 π4 π 11π 62 1π 6 16 π3 k k k k t t tt k k t tt tt k                 Solving   6216 π 384π33 4k k   , (or 62 3 76 0k k   ) we get 2k  . Since 1k  , 2k  . Q2 Solution Multiply throughout by 1nn  1 14( 1) 2nn nnn x n x     Let n nnn x y , 142nnyy  where 1 1y  142ny   2 24 4 2 2ny     32 34 4 2 4 2 2ny        1 2 3 14 4 2 4 2 4 2 2n n ny          

2 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 Q2 Solution 1 1 2(1 4 )4 14 n n       11 24 1 4 3 nn   12 5 4 3 n   Hence 12 5 4 3 n n nx n    . Alternative 1 142nnyy  where 1 1y  The general solution is of the form 4n n By A   . 1 1 4 1y A B    2 6 16 6y A B    Using GC, 52,12 3AB  Hence 52 412 3 n ny    . Alternative 2 142nnyy  where 1 1y  1 22 433 nnyy     1 1 22 433 n nyy      152433 n ny    Q3 Solution (a) Let y ux  d d dd y uuxxx 2 3 2 3d d yx y x x y yx           32 3 2 d d ux ux u x x x ux uxx       2 3 4 3 3 3 3 d d uu x ux x ux u xx     23d 1d uux u u ux      32 d1 d1 uu xxu u u     2 1dd( 1)( 1) u ux xuu     Let 22 11( 1)( 1) ( 1) u A B C uuu u u     

3 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over Q3 Solution  2( 1) ( 1)( 1) ( 1)u A u B u u C u       Let 1u : 14 A  1 4A Let 1u : 12 C   1 2C  Let 0u : 110 42 B    1 4B  2 1 1 1 d ln4( 1) 4( 1) 2( 1) u x cuu u       1 1 1ln 1 ln 1 ln4 4 2( 1)u u x c u        1 1 1ln ln4 1 2( 1) u xcuu      4( 1)11ln4 1 2( 1) xu cuu    4( 1) 2ln 11 xu cuu   (where 4cc ) Sub. yu x : 4 1 2ln 11 yx x cyy xx      4() 2ln x y x x cy x y x   Since xy , 0yx , so the general solution is 4() 2ln x x y x cy x y x   (b) 3 2 3 2 d d y x x y y x xy  Let   3 2 3 2 22f , 1 x x y y y xxy yx y x     , 0 2x  , 0 1y  . Using Euler method with step size 0.5,  1 0 0 0 2 2 0.5f , 211 0.5 1 1 2 2.375 y y x y     

4 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 Q3 Solution (c) The approximation is an over-estimate. Q4 Solution (i) Volume of revolution about y-axis, V 3π 12 0 2π d a xy x      3π 2 0 d2π ( sin ) (1 cos ) dd xa t t a t t t      3π 22 0 2π ( sin ) (1 cos ) da t t a t t   3π 32 2 0 2π ( sin )(1 cos ) da t t t t   (Shown) (ii) V 3π 32 2 0 2π ( sin )(1 cos ) da t t t t   3π 3 2 0 22π ( sin )(1 2cos cos )da t t t t t    3π 3 2 0 cos2 12π ( sin )(1 2cos )d 2 ta t t t t     3π 3 2 0 3 cos22π ( sin )( 2cos )d22 ta t t t t    3π 3 2 0 3 3 cos 2 sin cos 22π sin 2cos sin 2cos d2 2 2 2 t t ta t t t t t t t        3π 3 2 0 3 3 sin 3 sin cos 22π sin sin 2 2cos d2 2 4 2 t t ta t t t t t t         3π 3 2 0 3 5sin sin 3 cos 22π sin 2 2cos d2 4 4 2 t t t ta t t t t        x y (2, 1) (2.5, 2.375)

5 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over 2 3 2 3 0 3 0 23 5cos cos 2 cos3 sin 2 sin 22π 2sin 2sin d4 4 2 12 4 4 t t t t t ta t t t t                       3 3 2 2 2 00 3 3 5cos cos 2 cos3 sin 2 cos 22π 2sin 2cos4 4 2 12 4 8 t t t t t ta t t t                     3 2 2 0 3 3 3cos 3cos 2 cos3 sin 22π 2sin4 4 8 12 4 t t t t ta t t                    3 23 9 3 3 1 3 12π 0 0 1 1 1 0 1 0 2 14 4 4 8 12 2 4a                       3 2 2 272π3 1 1 6 7 1a     2 3 6 27π6 8 17a       231 π 81 722 44 3a   Alternative V 3π 3 2 22 0 2π ( 2 cos cos ) ( sin )(1 cos ) da t t t t t t t t      Let cos d sin cosI t t t t t t c    ----(1) Let 2cos dJ t t t du 1dut t   2 1 1 1 2 2 4 dv cos (1 cos2 )d sin 2d t v t t t tt       21 1 1 1 2 4 2 2 sin 2 ( sin 2 )dJ t t t t t t     221 1 1 1 1 2 4 2 2 4 sin 2 cos2J t t t t t c       21 1 1 1 2 2 2 4 sin 2 cos2J t t t t c    Let 2( sin )(1 cos ) dK t t t  

6 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 23 1 3( sin )(1 cos ) d (1 cos )K t t t t c      3π 3 2 22 0 2π ( 2 cos cos ) ( sin )(1 cos ) d V a t t t t t t t t      3π 21 2 2 32 1 1 1 1 2 2 2 4 31 3 0 2( sin cos ) 2π sin 2 cos 2 (1 cos ) t t t t a t t t t t          3π 23 2 4 3 11 48 31 3 0 2 sin 2cos 2π sin 2 cos 2 (1 cos ) t t t t a t t t t       227 16 33 11 88 1 3 π 0 3π 0 2 0 2π 0 2π 0 0 aa             32 27 17 16 122π π 3πa   

7 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over Q5 Solution (i)           T3det det det ( 1) det det     A A A A A since A is skew symmetric    2det 0 det 0   AA . (ii) (a) Method 1 ax 11 22 33 ax ax ax                  2 3 3 2 3 1 1 3 1 2 2 1 a x a x a x a x a x a x    3 2 1 3 1 2 2 1 3 0 0 0 a a x a a x a a x             32 31 21 0 0 0 aa aa aa     M Method 2 1 23 32 10 0 0 a aa aa                                ai 13 2 31 0 10 0 aa a aa                               aj 12 21 3 0 0 10 aa aa a                                ak 32 31 21 0 0 0 aa aa aa     M (ii) (b) M is skew symmetric and so by (i)  det 0 M Hence M is not invertible. (ii) (c)    ker |T k k a , the set of vectors parallel to a or line through origin and parallel to a   3R( ) | 0T    v a v , the set of vectors perpendicular to a or plane through origin and perpendicular to a

8 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 Q6 Solution (i) cos cos 2 sin 2 cos sin sin 2 cos 2 sin cos 2 cos sin 2 sin sin 2 cos cos 2 sin cos(2 ) sin(2 ) cos sin                                         T sin cos 2 sin 2 sin cos sin 2 cos 2 cos cos 2 sin sin 2 cos sin 2 sin cos 2 cos sin(2 ) cos(2 ) sin cos                                        T The eigenvalues are 1 and 1 with eigenvector cos sin     and sin cos     respectively. (ii) 110 01     T R R (iii) π 3  13 22 31 22     T 13 3122 22 2 11 131 3 222                   T (iv) TT cos 2 sin 2 cos 2 sin 2 sin 2 cos 2 sin 2 cos 2                 cos 2 cos 2 sin 2 sin 2 cos 2 sin 2 sin 2 cos 2 sin 2 cos 2 cos 2 sin 2 sin 2 sin 2 cos 2 cos 2                   cos 2( ) sin 2( ) sin 2( ) cos 2( )             which is a rotation through an angle of 2( )

9 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 1 [Turn over Q7 Solution (i) 2 1f ( ) 2xx x   2 11f (2) 2 2 024     2 18f (3) 3 2 039     Since f ( )yx is a continuous curve over the interval