[FFM] [2023] 4E AM 4049 Prelims P2_MS

1 Sec 4 Add Math Preliminary Exam 2023 P2 Marking Scheme Qn. No. Solution Marks AO 1(a) 221x + 323 2 6 3x x x x+ + + + - ( ) 3226xx+ 3x+ - ( )3x+ 0 Or 3 2 22 6 3 3 2 1x x x x x+ + +  + = + B1 B1(either presentation) AO1 1(b) 2 32 9 10 16 2 6 3 xx x x x −− + + + = ( )( ) 2 2 9 10 16 3 2 1 xx xx −− ++ = ( ) ( ) 23 21 A Bx C x x +++ + 29 10 16xx−− = ( ) ( )( ) 22 1 3A x Bx C x+ + + + Let 3x=− , 81 30 16 19 A+ − = 95 19 A= 5A= Let 0x= , 16 5 3 C− = + 21 3 C−= 7C =− Let 1x= , ( )17 15 4 7 B− = + − 17 13 4 B− =− + 1B=− 2 32 9 10 16 2 6 3 xx x x x −− + + + = ( ) ( ) ( ) 2 75 3 21 x x x −−++ + or ( ) ( ) 2 57 3 21 x x x +−+ + M1 M1(substitution or comparison method) A1(1st unknown) Allow FT2 A1(for the remaining unknown) B1 AO1 2(a) ( )( ) 2 54xd exdx − = ( ) 225 4 2 5 xxx e e−+ = ( ) 2 10 8 5xex −+ = ( ) 2 10 3xex − (shown) M2(M1 for each part of using product rule) AG1 AO3

2 Qn. No. Solution Marks AO 2(b) ( ) ( ) 2 210 3 5 4x xe x dx e x C − = − + ( ) 22 210 3 5 4xx xxe e dx e x C− = − + ( ) 22 210 3 5 4xx xxe dx e dx e x C− = − + ( ) 22 210 5 4 3xx xxe dx e x e dx C= − + + ( ) 2 2 210 5 4 3 2 x x x exe dx e x C = − + +  ( ) 2 2 210 4 5 4 342 x x x exe dx e x C = − + +  2 3 5 4 2 xe x C= − + + 2 5 5 2 xe x C= − + 2 2254 5 52 x xxe dx e x C = − +  or = ( ) 2 2 1xe x C −+ 3 0 24 xxe dx = ( ) 3 0 2 2 1xex − = ( ) 6 5 1e −− = 651e + or 2020 (3sf) M1 M1(correct integration) M1(manipulation) M1(simplify and obtain 24 xxe dx ) A1 AO2 3(a) 3 log 27p= 3 27p = 3p= 32 log q−= 23 q− = 1 9q= B1 B1 AO1

3 Qn. No. Solution Marks AO 3(b) C2(for sketch of 2 curves correctly) P1(the x-intercept and y-intercept clearly indicated) Minus 1 mark if axes not labelled AO1 3(c) 1 solution B1 AO2 4(a) ( ) ( )66log 2 1 log 2 4 1yy+ − − = ( ) ( ) 6 21 log 1 24 y y + = − or 6log 6 ( ) ( ) 121 6 24 y y + = − ( )2 1 6 2 4yy+ = − ( )2 1 6 2 24yy+ = − ( )5 2 25y = 25y = ln 2 ln 5y = ln 5 ln 2y = 2.32y= (3 sf) M1(quotient law) M1(simplify) M1(ln on both sides and power law) A1 AO1 x y

4 Qn. No. Solution Marks AO 4(b) ( ) 6 loglog log 8 log x xx x xxy y +=   ( ) 61 log 8 log x x y y +=   6 68logx y += 6 2logx y = 6log 2 x y= 3yx= M1(product law) M1(change of base) M1(change log. to exponential form) A1 AO1 5(a) ( )f '( ) 4cos 4 2sin 2 x x x dx=+ = 1sin 4 cos2x x C−+ ( )1f ( ) sin 4 cos 2 x x x C dx= − + = 12 cos 4 sin 2 42 xx C x C− − + + f(0) = 2 1 04 C− += 2 1 4C = 1 1 1 1 3f( )4 4 2 4 4 4 C