HCI 2022 FM Prelim P1 - modified (Solutions)

Page 1 of 13 2022 Preliminary Examination H2 Further Mathematics Paper 1 (9649/01) Marking Scheme s/n Solution 1(a) Source: 2025 Promo HCI Qn 3 ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 f, 1 22 1 2 x y x ky f x ky xx f x ky ky − − =+  =+  =+ At ( )2, 2 , ( ) ( ) ( ) 1 2 1 2 f , 4 2 2 4 2 422 x y k f kx fk ky − − =+  =+  =+ ( ) ( ) ( )2, 4 2 2 2 4 2 2 4 2 kL x y k x y kk = + + − + − ++ (b) Consider 2k = , Sub 2, 2.1, 1.95k x y= = = , ( ) ( ) ( ) ( ) ( ) ( ) 22, 8 2.1 2 1.95 2 8 2 8 21 = 2 2 0.1 0.05 2 2 2 2 1 = 2 2 0.15 22 3 = 2 2 2 80 163 = 280 L x y = + − + − + + − + + 2(i) 3 1 3 2 1 2 1 1 0 1 1 0 1 2 0 1 2 0 1 1 1 0 0 1 1 1 0 0 3 0 0 0 0 1 R R R R R R aa aa b a b a ab →− + →− +        − ⎯⎯⎯⎯⎯ → −       −     ⎯⎯⎯⎯⎯ → − −   1Sa v=    and 0b (ii) S is not a subspace of 3 a  and 0b Possible sets of S : 1. 1a and 0b implies  S v=  , where 0v  2. 1a= and 0b implies S =

Page 2 of 13 s/n Solution 3(i) From diagram, ML is the directrix where LK KO h== Let e be the eccentricity of P . 1OAe MA = = Since OA r= and AON = , cosON r = 2 cosMA LO ON h r  = + = + Hence 12 cos OA re MA h r = = = + 2 cos 2 1 cos r h r hr   =+ = − A polar equation of P is 2 1 cos hr = − (ii) Let 1OA r= , 2OB r= Since A and B are points on P , using result from (i), 1 2 1 cos hr = − , ( ) 2 2 2 1 cos hr  = −+ Area 12 1 2OAB r r=  ( ) ( )2 1 2 2 2 1 cos 1 cos hh   = − −+ 22 (1 cos )(1 sin ) h = −+ …(1) Let (1 cos )(1 sin )w = − + Since area 0OAB  , w is always positive.  maximum w  minimum area OAB  A y x O r N M K L B A O

Page 3 of 13 s/n Solution d (1 cos )cos (1 sin )sind w     = − + + 22cos cos sin sin cos sin (sin cos )(sin cos ) (cos sin )(1 sin cos )               = − + + = + + + − = + + − When d 0d w  = , cos sin 0+= or 1 sin cos 0+ − = tan 1 =− or ( )42 sin 1  − =− 3 4 = or 0 = (rejected) or 3 2  = (rejected) Hence from (1), Area OAB  ( )( ) 22 33 1144 22 22 (1 cos )(1 sin ) 11 hh == −+ ++ 2 2 4 (1 2) h= + unit2 4(a) Using GC, (i) 32g( ) 4 6x x x x= − − + diverges. Hence, it is not suitable. (ii) 1 26g( ) 4xx x =− is undefined at 1x . Hence it is not suitable. (iii) 1 26g( ) 4x x = + converges. Hence it is suitable. Suppose 1.0861 = is a root. 32 1 26 4 6 04x x x x =  − − = + Let 32f ( ) 4 6x x x= − − g( )x 32 46x x x− − + 1 26 4xx  − 1 26 4 x  + 0x 1.5 1.5 1.5 1x 4.875− undefined 1.0445 2x 21.9199 1.0906 3x 12426− 1.0857 4x 121.9 10 1.0862 5x 367 10− 1.0861 6x 1.0861

Page 4 of 13 s/n Solution f ( )x is continuous in [1.08605,1.08615] and f (1.08605)f (1.08615) ( 0.0009806)(0.000242)0= −  Hence 1.0861 = (4 d.p.) (b) 2( 1)1 21f ( ) e 2   −− −= 2( 0.5)1 21f ( 0.5 ) e 2   −− −= 2(0)1 21f (0) e 2 − = 2(0.5)1 21f (0.5 ) e 2   − = 2(1)1 21f ( ) e 2   − = By Simpson’s Rule, ( ) ( )( ) 22 2 2 2 22 ( 1) (1) ( 0.5) (0.5) (0) ( 1) (1) 0 11 22 1 1 1 2 2 2 111 822 f ( ) d 11 ee 221 32 1 1 14 e e 2 e 2 2 2 11 e e 8 e 2e6 2 xx              − − − − − − − − −− − −  +  + + +  = + + +  0.683 (3 d.p.) 5(i) 3 5 7 9LHS cos cos cos cos cos10 10 10 10 10 9 7 3 5cos cos cos cos cos10 10 10 10 10 42cos cos 2cos co2 10 2                       = + + + +                               = + + + +                               =+             2s0 10 0 RHS   +  == (ii) 10 10 10 10 10 (13 1) 0 (13 1) 13 1 1 ...(1) zz zz z z + − = − =− − =− 10 i( 2 ) 2i 10 2i 10 113 e 113 e , 5,0, 1, 2, 3, 4 1 13 e , 5,0, 1, 2, 3, 4 ...(2) k k k z kz kz    + + +  −= − = =−     = − =−    

Page 5 of 13 s/n Solution Hence i 10 1 1 13 ez =− , i 10 * 1 1 13 ez −=− 3i 10 2 1 13 ez =− , 3i 10 * 2 1 13 ez −=− 5i 10 3 1 13 ez =− , 5i 10 * 3 1 13 ez −=− 7i 10 4 1 13 ez =− , 7i 10 * 4 1 13 ez −=− 9i 10 5 1 13 ez =− , 9i 10 * 5 1 13 ez −=− ii 10 10 ** 11 1 1 ii 10 10 1 1 1 . [13 e ][13 e ] 169 13[e e ] 1 170 26cos 10 zz z z        −           −       = = − − = − + + =−  Similarly, * 22 * 33 * 44 * 55 13 170 26cos 10 15 170 26cos 10 17 170 26cos 10 19 170 26cos 10 zz zz zz zz     =−  =−  =−  =−  Hence * * * * * 1 1 2 2 3 3 4 4 5 5 1 1 1 1 1 z z z z z z z z z z+ + + + 3 5 7 95(170) 26 cos cos cos cos cos10 10 10 10 10 850               = − + + + +                     = 6(i) 2 0 12 0 1 2 2 2 00 1 2 2 2 2 0 1 2 2 2 2 00 12 1 2 1 2 33 22 3 1 22 3 11 2 2 2 11 33 11 33 111 3 3 3 1 3 (1 ) d (1 ) d ( )(1 ) ( 1) ( )(1 ) d (1 ) (1 )(1 ) d (1 ) (1 ) d (1 ) d (1 ) t n n t n t tnn tnn ttn n n n n nn I x x x x x x x x x n x x x t t x x x x t t x x x x x x tt − −− −− −− − − −− =+ =+ = + − − + = + − + + = + − + − + =+      2 3 2 11 33 nn nn II− −−−−

Page 6 of 13 s/n Solution 12 2 12 2 3 2 3 2 21 1 3 3 3 (1 ) ( 2) (1 ) ( 1) (shown) n nn n nn nn I t t I n I t t n I − − − − +− = + −  + = + − − (ii) 2 2 22 (2 )11 2d1 11l 1 ( ) 1 nd x xxxx x x x += + ++= + + + 2 22 2 2 22 2 2 22 2 0 0 444 333 44 33 44 33 44 33 4 3 00 00 00 0 0 2 1 1 20 1 9 11 20 19 1 201 1 ) 9 20 4 5ln9 3 3 10 101, 2 l d 1 d 1+ d d 1+ d d 2 d n ( 11d ln 3 where 99 x x x x x x x x x x xx x x x x xx xx x xm n + + ++ + + + + + ++  + + = + − =− =− =+  =  = + = =      (iii) 22 2 1 1 1 1 cos2 1 2 2 2cos 1( )secyx   + == +== Surface area of revolution about x-axis ( ) 2 22 22 22 2 2 14 23 4 13 2 44 1133 22 344 11 233 22 0 0 00 00 41 d 4 d 4 33 2 1 d )(1 ) d (1 d () ( ) d (1 ) d (1 ) ( 1 1 1 ) d y x x x x y xx xx x x x x xx     =+  =+ = + + +   = + + − + +  +     2 4 13 2 0 125 3 81 4 (1 ) dx x  ++=  ( )125 3 10 1 81 4 9 2 385 3 162 8 ln 3 ln 3    ++ =+  = 7(i) 2 2 d 4 5sin 3d r r  += …(1) Characteristic equation is 2 40m += 2im = Hence complementary function is cos 2 sin 2r A B =+ , ,AB   . Let a particular integral be sin 3 cos3r    =+ , ,   .

Page 7 of 13 s/n Solution 2 2 d 3 cos3 3 sin 3 ...(2)d d 9 sin 3 9 cos3 ...(3)d r r          = − =− − Substitute (2), (3) into (1): ( 9 sin 3 9 cos3 ) 4( sin 3 cos3 ) 5sin 3        − − + + = Comparing coefficients: 9 4 5 9 4 0 1 , 0    − + = − + =  =− = Hence a particular integral is sin 3r =− cos 2 sin 2 sin 3 d 2 sin 2 2 cos 2 3cos3d r A B r AB        = + − = + + When 2  = , 1r= and d 2d r  =− . 31 cos sin sin 0 2 32 2 sin 2 cos 3cos 1 2 A B A A B B    = + −  = − = + +  = Hence sin 2 sin 3r =− (ii) (iii) Area 2 3 5 5 1 2 (sin 2 sin 3 ) d     =− ( ) ( ) 22 3 5 5 3 5 5 3 5 5 1 2 1 cos4 1 cos61 2 2 2 1 4 sin 2sin 2 sin 3 sin d (cos5 cos ) d 2 cos 4 2cos5 2cos cos 2 6d 3                     −− = − + = + − + = − + − −    ( ) ( ) 3 5 5 12 18 55 46 55 sin4 2sin5 sin61 54 4 6 sin sin63 55 461 4 sin sin2 55 46 2 2sin 0 2sin 0 2sin          = − + − −  − + − −=  − − + − −

Page 8 of 13 s/n Solution ( ) ( ) 4 2 2 211 5 5 5 5461 4 11 5 5 546 sin 2sin sin sin 2sin sin         − − += − − − + 2 4 25 2 251 5 5 54 12 12 25 2 25 1 5 5 5 548 48 sin sin sin sin