MFSS AMath Prelim P2 Solutions for Students

1 2024 4E5N AM Prelim P2 MS Qn Solution 1a 2 3 2 3 2 2 31 1 3 4 2 3 3 2 1 3 x x x x x xx xx x x − + − + + + − + + −− + 1b 3( 2) ( 2) 1 8 2 1 93 3 aa aa a a − + − = + − − = + −= =− 2 19 3 5 2 2 5 19 3 5 2 2 5 2 2 5 2 2 5 38 38 5 6 5 30 4 4(5) 68 44 5 16 17 11 544 − + −−= +− − − += − −= − =− + 3i ( ) ( ) 92 1 18 3 7 8 3 9 2 9 2 0 18 3 6 5376 3 18 3 7 14608 r r r r r Tx r x xr r r T r r T x − + −  =−  =− =− = = − = − = =−  3ii Term indep of 3(5376) ( 1)( 4608) 20736 x= + − − =

2 4i 3tan 4 15tan 8 3 15 48tan( ) 3 151 48 84 6 or 613 13 A B AB = = + += −  =− − 4ii ( ) 2 2 cos 2cos 1 2 1cos cos 122 18 12 17 9 34 3cos 2 34 180 270 90 1352 BB B B B B B =− =+ = − + = =−   5a 2 2 2 2 23 34 (3 4)2 (2 3)3 (3 4) 17 (3 4) 17 17 25 (3 4) (3 4) 25 1 3 7 15 17( , )3 15 xy x dy x x dx x x x x x y P −= + +−−= + = + = + += = =− −

3 5b 2 2 22 2 4 2 3 43 2 1143 44 1 114 44 dy xxdx xx x x = − + = − +    = − − +       = − + 2 2 1 04 1 11 114 4 4 4 0 0 x x dy dx  −  − +    Therefore, y is always increasing for all real values of x 6i 2 2 2 12 2 2 6 4 0 2 4 16 0 2 8 0 ( 2)( 4) 2 or 4 16 or 4 x x x xx xx xx x y − = − − = − − = − − = + − =− = (−2, 16) and (4, 4) 6ii 2 4 16 4midpt ( , ) 22 (1,10) 16 4grad 24 2 − + += = −= −− =− 1grad of perpen bisector 2 eqn of perpen bisector: 110 ( 1)2 1 19 22 yx yx = − = − =+

4 7a ( ) 2 2 5 3 2 5 5 3 10 2 2 5 3 0 (2 1)( 3) 0 x x x x x x xx xx −  − −  − − −  + −  1 32 x−   7b ( ) ( ) 2 2 2 50 10 40 ( 50) 10 3 250 3 50 10250 y a x a a yx = − + = − + = = − + 8i A is a point of inflexion B is a minimum point 0.9x= 1x= 1.1x= 0dy dx  0dy dx = 0dy dx  3.9x= 4x= 4.1x= 0dy dx  0dy dx = 0dy dx 

5 8ii ( ) ( ) 2 2 32 4 3 2 4 3 2 4 3 2 14 ( 2 1)( 4) ( 6 9 4) 19( 2 4 )42 0 (64 128 72 16) 8 195 ( 2 4)42 558 4 20 27 160 27 20 1 7 160( 2 4 )27 4 2 27 5 40 70 80 160 27 27 27 27 29 dy a x xdx a x x x a x x x y a x x x x c ac ac ac aa a c y x x x x x x x x = − − = − + − = − + − = − + − + = − + − + = = − + − + =− + = = = − + − + = − + − + 9i 12 12 5 1212 5 1212 5 hr rh rh − = −= =− 9ii 2 2 23 12(12 ) 5 1212 5 V r h rr rr    = =− =−

6 9iii 23 2 2 2 2 2 2 10 3 1212 5 3624 5 360 24 5 30 12 (2 ) 5 100 (rej) or 3 7224 5 24 0 r V r r dV rrdr rr rr r dV rdr dV dr       = =− =− =− =− = =− =−  10 is max when 3Vr= 400 9V = or 139.62…… 10a 2sin 2 3cos 0 4sin cos 3cos 0 cos (4sin 3) 0 3cos 0 or sin 4 PV of 90 or 48.6 xx x x x xx xx x += += += = =− =− 10bi 4 5 2 2 a c b b  = = = = 10bii mk =+ 10biii 24 kl + = or 2kl += etc 11ai The centre of C1 looks like (−r, r)

7 11aii 2 2 2 2 2 2 2 22 ( 8) ( 1) 16 64 2 1 18 65 0 ( 5)( 13) 0 5 or 13 (rej) ( 5) ( 5) 25 r r r r r r r r rr rr r xy − + + − = − + + − + = − + = − − = = + + − = 11bi P (7, 10) 227 10 113 6 r = + − = 11bii QSR max  SQ and SR are tangents to circle  SQ = SR since they met at an external point 12i 0.3 0.3 0.3 0 0.3 0.3 0.3 0.3 10 3 103 3 1 3 10 1 33 10 10 33 1 10 1ln 0.310 10 1ln3 10 10 ln103 t t t t t t v e dt e c ec ec c ve e e t t − − − − − − =− −=+− =+ =+ =− =− =− = =− =− = 

8 12ii 0.3 0.3 0.3 0 0.3 ln10 10 1 33 10 1 33 100 1 93 100 10 (0) 93 100 9 100 1 100 9 3 9 10 ln103 100 1 10 100 ln109 3 3 9 1010 ln109 7.44 t t t t ve s e dt e t c ec c s e t t se − − − − − =− =− =− − + =− − + = =− − + = =− − +  =− =  12iii When t = 33, 0.111s= When t = 34, 0.223s=− Since displacement changes sign from t = 33 to t =34, the particle is again at O during the 34th second.