2024 FMSS AMATH P1 PRELIM MS

1 FMS(S) SEC 4Exp Additional Mathematics Paper 1 Preliminary Exam 2024 Marking Scheme 1i 24 8 5xx+− 24( 2 ) 5xx= + − 24 ( 1) 1 5x= + − − ( ) 2 4 1 9x= + − M1 M1 complete square A1 AO1 ii Turning point = (–1, –9) B1 AO1 2 24 3 6xx =+ ( )24 3 6x −= ( ) 6 24 3 x= − ( ) 6 2 6 3 2 6 32 6 3 x += +− ( ) ( ) 2 6 18 4 6 3 += − 12 3 2 21 += 42 7 += a = 4 and b = 2 M1 factorise x M1 Multiply by conjugate surd M1 simplification A1, A1 AO1 3i (alt segment theorem/tangent chord thm)CAE ABC = (alternate s, // )ACB CAE BC AE =  (base s of isos )ABC ACB =   ( )AB AC shown= B1 B1 AG1 AO3 3ii 180 2 ( sum of )BAC ABC = −    ( s in same segment)BDC BAC =  180 (adj s on a straight line)CDE BDC = −  180 (180 2 ) (adj s on a straight line)CDE ABC = − −   2 (shown)CDE ABC =  M1 B1 AG1 AO3

2 4a i Since the period = 8 8= b  1 8b= B1 AG1 AO3 ii c = 3 7 tan 3 4a =+  a = 4 4 tan 38 xy =+  M1 A1 AO1 4b B1 correct sinusoidal shape with correct turning points B1 two cycles AO1 5i Length of rectangle = 20 3 2 x− Area = 220 3 1 sin 6022 x xx− − 223 1 310 2 2 2x x x= − − = 26310 4xx  +− B1 M1 AG1 AO3 5ii 6310 2 4 dA xdx  +=−  For stationary point, 6310 2 0 4 x +−= ( )6 3 20 x+= ( ) 20 =2.5866 = 2.6 (2 s.f.) 63 x= + ( ) 26310(2.5866) 2.5866 = 12.933 4A  +=−  B1 - diff M1 – equate to 0 A1 A1 AO1 (45º,7) (135º,1 ) (225º,7) (315º,1)

3 13A= 6a ( )( ) ( ) ( ) ( )( ) ( ) 22 2 8 13Let 1 2 21 2 2 2 8 13 2 1 2 2 1 2 x A B C xxx x x x A x B x x C x + = + ++++ + + + = + + + + + + ( ) 2 3 3 1 19 924 4 0 13 4 2 13 4 4 2 1 xC C xA A x A B C B =−  − =− = =−  = = =  = + + = + + 2B=− ( )( ) ( ) ( ) ( ) 22 8 13 4 2 1 1 2 21 2 2 2 x xxx x x + = − + +++ + + B1 M1 – sub or compare coefficient A1 (anyone of A, B or C correct) A1 (all 3 values A, B & C) A1 AO1 6b ( )( ) ( ) ( ) ( ) ( ) ( ) 2 21 2 21 2 1 8 13 d 1 2 2 4 2 1 d1 2 2 2 41ln 1 2 2ln 222 112ln 5 2ln 4 2ln 3 2ln 343 x x xx xxx x xx x + ++ = − +++ + = + − + − +    = − − − − −         5 1 25 12ln or ln or 0.5304 12 16 12= + + (to 3 sig fig) B2 [B1 for 2 correct ln term; B1 for the 3rd term] B1 AO2 7a 2'( ) 18 2 50f x x ax= + − Since '(1) 6f = 18 2 50 6a+ − = a = 19 Given 3 02f  = 32 3 3 36 19 50 02 2 2 b     + − + =           20.25 42.75 75 0 b+ − + = B1 M1 forming equation A1 M1 forming equation A1 AO2

4 12b= 7b ( ) ( )( ) 22 3 3 14 4f x x x x= − + − When ( ) 0fx = ( )( ) 22 3 3 14 4 0x x x− + − = ( )2 3 0x−= 3 2x= ( ) 23 14 4 0xx+ − = ( ) ( )( ) ( ) 2 14 14 4 3 4 23x −  − −= 0.270x= or 4.94x=− (3 sig fig) B1 found using any valid method B1 B1 AO1 8a 3 4 2 1 32 5 16 5+− = x x x x 3 4 2 1 4 32 2 5 5 2 5 −   = x x x x 4 3 3 216 2 2 5 55 x x x x=    16 255 xx= 16 105 x= 16lg lg105 x= 16lg 5x= (shown) M1: simplify base 2 and base 5 terms correctly M1: introduce log on both sides AG1 AO3 8b 222log log ( 4) 3xx− − = 2 22log log ( 4) 3xx− − = 2 2log 3 4 x x =− 2 324 x x =− 2 8( 4)xx=− 2 8 32 0xx− + = 2( 8) ( 8) 4(1)(32) 2(1)x − −  − −= M1: power/quotient law M1: change to index form M1: form quadratic eqn M1 solve for x OR show 2 4 64b ac− =− AO3

5 8 64 2x −=  No real solution (shown) AG1: explain 64− does not exist or 2 4b ac− < 0 no real solution 9a 3 7,=+y k x ( ) 1 2 d3 37d2 − =+yk xx Given dd 3dd yx tt= dy dy dx dt dx dt= 3d d d dd x y x t x dt= ( ) 1 2 33 3 72 − =+ k x Sub x = 3 ( ) 1 21 3 3 72 − =  +k 11 24=k 8=k B1: correct differentiation M1: write the correct ratio M1: form chain rule correctly and sub x = 3 A1 AO2 9b i Sub t = 0 , 24 gm= B1 AO1 ii 0.0224e 12t− = 0.02lne ln 0.5t− = 0.02 ln 0.5t−= ln 0.5 34.7 days0.02t==− M1 A1 AO1 iii 0.02d 0.48ed tm t −=− Sub 0.5t = , 0.02(0.5)d 0.48e 0.475 (3 s.f.)d m t −=− =− Mass is decreasing at 0.475 g/day B1 B1 (must write statement) AO1

6 10 i LHS 1 1 sin11cos cos cos cos sinx x x x x x       =  + + +              sin cos cos 1 cos cos cos 1 cos sin x x x x x x x x    +    =  +        +       sin 1 cos cos 1 sin xx xx +=+ + = 22sin (1 cos ) sin (1 cos ) xx xx ++ + 22sin 1 2cos cos sin (1 cos ) + + += + x x x xx 2 2cos sin (1 cos ) += + x xx 2(1 cos ) sin (1 cos ) += + x xx 2 sin= x (shown) OR LHS 22tan (1 sec ) tan (1 sec ) ++= + xx xx 22tan 1 2sec sec tan (1 sec ) + + += + x x x xx 22sec 2sec tan (1 sec ) += + xx xx 2sec (sec 1) tan (1 sec ) += + xx xx 2 sin cos cos= x xx 2 sin= x (shown) M1: sintan cos xx x= & 1sec cosx x= M1: simplify both [ ] correctly M1: add the fractions and expand correctly M1: factorise numerator AG1 OR M1: add fractions correctly M1: expand and add correctly M1: factorise numerator M1: 1sec cosx x= & sintan cos xx x= AG1 AO2 10 ii tan 1 sec 1 3sin1 sec tan xx xxx ++ = ++ 2 1 3sinsin xx =+ 23sin sin 2 0xx+ − = ( )( )3sin 2 sin 1 0xx− + = M1: form quadratic eqn M1: factorisation or general formula AO1

7 2sin 3x= or sin 1x=− 41.8 ,138.2x= or 270x= (reject (to 1 dp) as tan 270 is undefined) A1, A1 11 i (9 )( 3)y x x= − − Sub (k, k-3) into (9 )( 3)y x x= − − 3 (9 )( 3)− = − −k k k * 23 9 27 3− = − − +k k k k 2 11 24 0− + =kk ( 3)( 8) 0− − =kk 3( . .) or 8==k N A k *OR (9 )( 3) ( 3) 0k k k− − − − = ( 3)(9 1) 0kk− − − = ( 3)(8 ) 0kk− − = 3( . .) or 8==k N A k M1 substitution M1 form quadratic eqn M1 factorisation AG1 must state N.A. for x = 3 AO3

8 11 i Let y = 0 (9 )( 3) 0xx− − = 3 or 9xx== x-coordinate of B = 9 x-coordinate of A = 39 62 + = ** OR use 2 12dy xdx =− + At turning point, 0dy dx = 2 12 0x− + = 6Ax = Area of APCQ 25 2 10 units=  = Area CQB ( ) 9 2 8 12 27 d= − + − x x x 9 32 8 1 6 273 = − + − x x x ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 2119 6 9 27 9 8 6 8 27 833    = − + − − − + −       222 units3= Shaded area 22210 2 12 units33= + = M1 : either or ABxx B1 (area of rectangle) M1: Integrate all terms correctly A1 A1 AO2 12 i Gradient of PQ = gradient of OR= 1 2 Eqn of PQ: ( )134 2yx− = + 1 52yx=+ ----- (1) Gradient of QR = 2− Eqn of QR: 2 2( 4)yx− =− −