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