[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 = − + + = 1 3 44C  = 1 3C = cos 4 sin 2 3 1f ( ) 4 2 4 xxxx  −= − + + M1(award marks even without C1) M1(award marks even without C2) M1(for substitution) A1(for either C1 or C2 correct) A1 AO2

5 Qn. No. Solution Marks AO 5(b) 2 sincos 1133f ( )6 4 2 2 4   − = − + + = 13 322 4 2 4−+ = 1 3 3 8 4 4−+ = 73 84− = 7 2 3 8 − (shown) M1(for basic angles) M1 AG1(depends on 73 84− ) AO3 6(a) (a) 2g = 4 and 2f = 6− g = 2 and f = 3− Centre (2, 3− ) Sub 2 and = 3xy=− , ( ) ( )3 3 4 2 . k− − = 17k =− Radius = 4 9 12++ = 5 units B1(for centre) M1(substitution) A1 B1 AO2 6(b) Length between centre of 1C to centre (14, 2) ( ) ( ) 22 14 2 2 3− + + = 13 units Radius of 2C = 8 units Eqn. of 2C ----- ( ) ( ) 22 14 2 64xy− + − = Or 22 28 4 136 0x y x y+ − − + = M1 A1 A1 AO2

6 Qn. No. Solution Marks AO 7(a) LHS = 2 2 2tan 1 tan 1 tan xx x ++ − = ( ) ( )( ) 2 1 tan 1 tan 1 tan x xx + −+ = ( ) ( ) 1 tan 1 tan x x + − = sin1 cos sin1 cos x x x x + − = cos sin cos cos sin cos xx x xx x + − = cos sin cos sin xx xx + − = RHS (proved) M1(change sec2x) M1(either factorization of numerator or denominator) M1(change tan x) AG1 AO3 Or 7(a) LHS = 2 2 2 sin 12 cos cos sin1 cos x xx x x + − = 2 22 2 2sin cos 1 cos cos sin cos xx x xx x + − = 22 2sin cos 1 cos sin xx xx + − = ( )( ) 222sin cos sin cos cos sin cos sin x x x x x x x x ++ +− = ( )( ) ( )( ) cos sin cos sin cos sin cos sin x x x x x x x x ++ +− = cos sin cos sin xx xx + − = RHS(proved) M1(change tan x & sec2 x) M1(simplify fractions) M1(factorization of either numerator or denominator

7 Qn. No. Solution Marks AO 7(b) 2cosec 5cot 5xx− =− ( ) 21 cot 5cot 5 0xx+ − + = 2cot 5cot 6 0xx− + = ( )( )cot 2 cot 3 0xx− − = cot 2 or cot 3xx== 11tan or tan 23xx== 26.56 or 18.43 =   18.4 , 26.6 , 198.4 , 206.6x=     (1 dp) M1(sub. 21 cot x+ ) M1(factorization or use quadratic formula) A2(all values of x) Or A1(for 2 angles) AO1 8(a) Area of triangle OAB = 1 50 50 sin(90 )2     − = 1250cos Area of triangle ODC = 1 80 80 sin2  = 3200sin Total area S = 3200 sin  + 1250 cos  (shown) M1 AG1 AO3

8 8(b) (a) Let ( )3200sin 1250cos sin R   + = + = sin cos cos sinRR    + By comparing, cos 3200 (1), sin 1250 (2)RR == (2)/(1) sin 1250 cos 3200 R R   = 25tan 64 = 21.3 = (3 sf) (1)2 + (2)2, 2 10240000 1562500R =+ 11802500 50 4721R== or 3440 (3sf) ( ) ( ) ( ) 11802500 sin 21.3 or 50 4721sin 21.3 or 3435.5sin 21.3 S     =+ ++ or ( )3440sin 21.3  + M1(for two eqns) B1(for  ) B1(for R) A1 AO1 8(c) Max. value of S When ( )sin 21.34 1  += 21.34 90 += 68.66 68.7or = M1 A1 AO1 9(a) 25xy+ =− ----- (1) From (1), 52yx=− − Sub, into ( )5 2 3 0xx− − + = 22 5 3 0xx− − + = 22 5 3 0xx+ − = ( )( )3 2 1 0xx+ − = 13 (NA) or 2xx=− = When 1 , 62xy= =− B( 1 , 62 − ) M1(substitution) M1(factorization or using quad. formula) A1 AO1

9 9(b) Area of triangle = 1 5 1 692 2 2  +  = 1 1 2 3 dxx− =   1 1 2 3 ln x− = 13ln 2 = 3ln 2− or 1ln 8 = ln8− Area above curve = ln 8 Total area = 9 + ln8 M1(allow FT using their values) M1 M1(correct application of limits) M1(apply law of log. get ln8− ) A1(total area) AO2 9(c) 3y x −= ( ) 23dy xdx −=− − = 23x− At x = 1, 3dy dx = Gradient of normal = 1 3− When x =1 , y = 3− Equation of normal ( )131 3yx+ =− − or 18 33yx=− − B1 B1 B1(find y coordinate of pt. C) A1 AO2

10 10(a) 2 23xx+− = ( )( )13xx−+ f( 1 ) = 1 6 2 3 0a b a+ + + − = 7ba− =− -------- (1) f( 3− ) = ( ) ( ) ( ) ( ) 4 3 2 3 6 3 2 3 3 3 0 a b a− + − + − + − − = 15 3 81 0ab− − = 5 27ab−= ----------- (2) (1) + (2), 4 20a= 5a= Sub. 5a= into (1), 2b=− M1(factorization to obtain 2 factors) M1(sub x =1 and obtain eqn) M1(sub x =-3 and obtain eqn) M1(solve simultaneous eqns) A2 AO1 10(b) f(x) = 4 3 26 10 2 15x x x x+ + − − = ( )( ) 22 2 3 5x x x kx+ − + + By comparing coeff. of 3x , 62 k=+ 4k = The other quadratic factor is 2 45xx++ . M1(comparison or long division method) A1 AO1 10(c) ( )( ) 22 2 3 4 5 0x x x x+ − + + = ( )( )( ) 21 3 4 5 0x x x x− + + + = 1 or 3xx= =− 2 4b ac− = 24 4(1)(5)− = 40− No real roots There is only 2 real distinct roots. M1(two real roots) M1(discriminant) AG1 AO3 11(a) See attached graph. x 15 20 25 30 35 40 lg y 0.82− 0.42−