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