2022 AMKSS AMath Prelims P2 Ans

1 AMKSS 4E/5NA PRELIM 4049/02/2022 [Turn Over 2022 AMath Prelim Paper 2 Answer Scheme 1(a) 2 2 2 cossec 1 sin 1 cos cos 1 sin 1 sin cos cos (1 sin ) 1 sin (1 sin ) cos (1 sin ) sin sin cos (1 sin ) sin (1 sin ) cos (1 sin ) sin cos tan                − + =− + +−= + + − −= + += + += + = = M1 M1 M1 M1 1(b) tan 2sin tan 2sin 0 sin 2sin 0cos sin 2sin cos 0 sin (1 2cos ) 0 sin 0 1 2cos 0 10(NA), , 2 (NA) cos 2 Acute 3 5,33 or                = −= −= −= −= = − = == = = M1 M1 A1 A1 2(i) r = 5 22( 5) ( 7) 25xy− + − = B1 B1

2 AMKSS 4E/5NA PRELIM 4049/02/2022 [Turn Over 2(ii) 4 16 33 3Gradient 4 3 4 13At(5,7), 4 3 13 44 yx CQ y x c c yx =− + = =+ = =+ M1 M1 A1 2(iii) 4 16 3 13 3 3 4 4 1 4 (1, 4) xx x y Q − + = + = = M1 A1 2(iv) Let ( , ) 14 , (5,7)22 14 5 or 722 9 or = 10 (9,10) Subst (9,10)into 4 3 4(9) 3(10) 66 R x y xy xy xy R x y k k k ++ = ++ == = += += = M1 M1 A1 3(i) 32 3 3 3 32 3 2 32 2 2 2 0 Since 0, by factor theorem, 2 3is a factor f Rx        = − + −               = =− B1 3(ii) ( ) 2 By Long division, (2 3)( 1)f x x x= − + M1 A1

3 AMKSS 4E/5NA PRELIM 4049/02/2022 [Turn Over 3(iii) 2 3 2 2 22 2 78 2 3 2 3 2 3 1 7 8 ( 1) (2 3) When 1.5; 16.25 3.25 5 When 0; 8 5 3 1 When 1; 14 5(2) ( 1)( 1) 3 5 3 1 2 3 1 x x A Bx C x x x x x x x A x Bx C x xA A xC C xB B x xx − + + + =+− + − − + − + + = + + + − == = = = − =− = = + − − =− +−−+ M1 M1 M1 – substitution of values/comparing coefficients M1 – 2 values correct A1 4(a) ( ) ( ) ( ) ( ) 21 2 2 2 2 2 2 log 3 2 log 1 1 log ( 1)log 3 2 1 1log 2 log 3 2 ( 1) 1 3 4 0 (3 4)( 1) 0 4 or 1(NA)3 xx xx xx xx xx xx + − − = −+ − = + − = − − = − + = = =− M1 M1 M1 M1 A1 4(b) 2 22 12lg ( )3 1lg ( )9 1lg ( 2 )9 1lg (9 )9 lg lg lg ab ab a ab b ab ab ab + =+ = + + = = =+ M1 M1 M1 M1 5(i) 2 2 2 2 9 tan 24sec 25 9(sec 1) 24sec 25 0 9sec 24sec 16 0 (3sec 4) 0 4sec 3 3cos 4 41.4 , 318.6 xx xx xx x x x x =− − − − = − + = −= = = =   M1 M1 M1 A1

4 AMKSS 4E/5NA PRELIM 4049/02/2022 [Turn Over 5(ii) 2 2 2 3cos ec 2 4 4 3 4 sin 2 4 3sin 2 44 3sin 2 42 Acute 3 242 , , 4 3 3 3 5 13,,24 24 24 x x x x x x              += = +  +=  + = = += = M1 M1 M1 A1 6(i) ln ln( ) ln ln ln ln ln ln Plot ln against ln 8 2980(Accept 2700 3000) 7.4 6.5ln 1.5 3.7 1.51(Accept 1.45 1.55) tN Ab tN A b N A t b Nt A A b b −= −=+ =− = =− −−= − =− M1 M1 M1 A1 M1 A1

5 AMKSS 4E/5NA PRELIM 4049/02/2022 [Turn Over 6(ii) 152980.957987(1.05445) 6.48 6 trees(Accept 5 7 trees) N N N −= = − M1 A1

6 AMKSS 4E/5NA PRELIM 4049/02/2022 [Turn Over 7(i) 2 3 3 3 3 3 3 33 33 1 e2 1At 6; e 2 1 e2 1At( 6, ); e ( 6) 2 2 1 e22 1At 0; e 2 0 2 4 ( 4,0) x dy dx dyx dx y x c e e c ce y x e y x e x B − =− =− =− =− + − =− − + =− =− − = − − = =− − M1 M1 M1 A1 7(ii) 3 3 2 2 6 2 2 6 13 3 1 33 1 2 1Area of triangle = 22 2 2 ( 2 ) 22 Area of shaded region 2=2 19.3 units x x e e e dx e ee e e ee e − − − − −  = =− =− − − =− −− =  M1 M1 M1 M1 M1 A1 8(i) By long division M1

7 AMKSS 4E/5NA PRELIM 4049/02/2022 [Turn Over 62 35 x−+ − A1 8(ii) 2( 5) 2ln(3 5 )35 10 2ln(3 5 )35 xx x x xx − +− − −= + −− M1, M1 for each term A1 8(iii) 2 2 2 1 2 2 2 10 2ln(3 5 ) ln(3 5 )35 62 2ln(3 5 ) ln(3 5 )35 62 ln(3 5 ) 2ln(3 5 ) ln(3 5 )5 62ln(3 5 ) ln(3 5 ) 2 ln(3 5 ) 5 ln(3 5 ) 3ln(3 5 ) ln(3 5 ) 25 x x dx x x cx x dx x x cx x x c x dx x x c x dx x x x x c xxx dx x x − + − = − +− − − + + − = − + −  + − + + − = − + − = − − − − + −− = − − −      3 3 3ln(3 5 ) ln(3 5 ) ln(3 5 ) 5 c x dx x x x x c + − = − − − − + M1 M1 M1 M1 A1 9(i) cos 28 28cos sin 4 4sin 4sin 28cos 28sin 4cos 4 28 32 32sin 24cos AP AP CQ CQ L AD DC CB AB          = = = = = + + + = + + − + + = + + M1 M1 M1 9(ii) 22 tan 0.75 36.9 32 24 40 32 40sin( 36.9 ) R R    = = =+ = + +  M1 M1 A1 9(iii) 32 40sin( 36.9 ) 70 sin( 36.9 ) 0.95 36.9 71.8051 ,108.1948 34.9 ,71.3     + +  = +  = + =   =   M1 A1

8 AMKSS 4E/5NA PRELIM 4049/02/2022 [Turn Over 9(iv) Since sin( 36.9 ) 1 40 sin( 36.9 ) 40 32 40sin( 36.9 ) 32 40 72 should not exceed 72. L L    +    +   + +   +   M1 A1 10 ( ) ( ) ( ) 1 1 12 12 2 2 1 1 1 f ' (3cos3 4sin 2 ) 3sin 3 4( cos 2 ) 32 sin 3 2cos 2 f (sin 3 2cos 2 ) cos3 2sin 2 32 cos3 sin 23 f 0 0 100 3 1 3 cos3 1f ( ) sin 2 33 1500 2 3 6 1 cos3() 3 x x x dx xx c x x c x x x c dx xx c x c x x c x c c c xx x c x c c xfx   =− −= − + = + + = + + −= + + + −= + + + = − + = = −= + + + − + + = = −=   1sin 2 3 cos 1 1( ) sin 23 3 3 3 3 31 2 xx f       + + + − = + +  +  =+ A1, A1 A1, A1 M1 M1 M1 M1, M1