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 7