2024 FMSS AMATH P2 PRELIM MS

1 Sec 4 Add Math Preliminary Exam 2024 P2 Marking Scheme Qn. Solution Marks AO 1 xx ee 2 1 )3(2 =− x ey 2 1 Let = , 2 2 11 22 2( 3) 2 6 0 (2 3)( 2) 0 3 or 22 3 (rej. as -ve) 22 1 ln 22 xx yy yy yy yy ee x −= − − = + − = =− = =− = = 2ln 2 1. 39 (3 s.f) Therefore, the equation has only one solution where =1.39 x x = = M1 (substitution) M1 (factorization) A1 (reject 3 2− ) (Did not award marks for students who squared both sides and could not justify why they rejected one answer when they ended up with 2 answers) AG1 3

2 2 3 2 3 2 3 5 2 12 5 2 12 2 12 55 2Grad of normal = 5 5Grad of tangent = 2 5 2 At 1, 2 12(1)55 2 At 1, 2 2 (1) 2 ------ (1) 5At 1, 2 5 3 (1) ------ (2)2 5 6 Substituting y ax b dy axdx yx yx yx dy dx x y y xy ab ab dyx dx a a a =+ = += =− + =− + − = = =− + = == =+ += == = = 5 into (1)6 5 26 7 6 b b = += = B1 (find dy dx ) B1 (grad of tangent) M1 (find y = 2) M1 (for either (1) or (2)) A1 (for a) A1 (for b) 2

3 3a 2 3ln(2 )xy x= ( ) 2 22 4 3 3(2) 6 ln 22 3 6 ln 2 3 (1 2ln 2 ) x x xdy x dx x x x x x xx  −= −= =− M1 (Apply quotient rule) M1 (Able to diff 21ln 2 2x xx== .) A1 1 3b 3 3 2 13 3 2 23 2 2 33 2 2 3 6 3ln 2ln 2 d 1 2 ln 2ln 2 d 2ln 2 ln 2 1d 2 ln 2 1 ln 2 1d 22 xx x Cx x x xx x Cx x x xx xCx x x xx xCx x x  − = +  − = + =− − + =− + +     M1 (reverse differentiation – must include + C) M1 (integrate 3 1 x ) A1(must include + C) (Whole question will only deduct once if they did not put + C) 2

4 4a f ( ) 1xx− 2 31x ax x− +  − 2 ( 1) 4 0x a x− + +  Since it is always positive for all real values of x, the graph of the curve 2 ( 1) 4y x a x= − + + lies entirely above the x-axis  the equation 2 ( 1) 4 0x a x− + + = has no real roots Discriminant, 0D 2( 1) 4(1)(4) 0a+ −  22( 1) 4 0a+ −  ( 5)( 3) 0aa+ −  53 a−   M1 (form inequality) (students who equated both eqns together will not get M1 unless they explain that there are no real roots and lead to D<0) M1 (Discriminant less than zero) M1 (Factorization) A1 2 4b f ( ) 4 yx ya    = =+ 2 34x ax a− + = + 2 10x ax a− − − = Since the line 4ya=+ is a tangent to the curve, this equation has equal real roots Discriminant, 0D= 2( ) 4(1)( 1) 0aa− − − − = 2 4 4 0aa+ + = ( 2)( 2) 0aa+ + = 2 a =− M1 (equate equations together) M1 (D = 0) A1 1 5a sin( ) sin cos cos sin cos sin sin cos sin( ) 53 = 88 1 = 4 A B A B A B A B A B A B − = − = − − − M1 (make cos sinAB the subject) A1 1 –5 3

5 5b 2 2sin 2 (sec tan ) 1 sin2(2sin cos ) cos cos 1 sin2(2sin cos ) cos 4sin 4sin         − =−  −=  =− M1 (double angle formula) M1 (bring cos under same denominator) A1 2 5c ( )( ) 2 2 2sin 2 (sec tan ) 3 0 4sin 4sin 3 0 4sin 4sin 3 0 2sin 1 2sin 3 0 13sin or sin (rejected)22 6 ,266 7 11,66             − + = − + = − − = + − = =− = = = + − = M1 (factorize) M1 (show both and must reject one) A1 (must be in terms of  as qn asked for exact solns) 1 6(i) 2 8 2 8 2 At 1, 5 85 2(1)1 15 8Equation of curve: 2 15 y dx x y x c x xy c c yx x =− =− − + == =− − + = =− − +  M1 (with + c) M1 (Sub in values) A1 1 6(ii) 2 2 2 Let 0 8 20 8 2 4 2 or 2 dy dx x x x xx = −= = = = =− M1 A1 (both answers) 1

6 6(iii) 2 23 2 23 2 23 16 At 2, 16 202 Maximum point at 2 At 2, 16 16 20( 2) 8 Minimum point at 2 dy dx x x dy dx x x dy dx x =− = =− =−  = =− =− =− = −−  =− M1 (2nd derivative) A1 A1 (for students who got part (ii) wrong, maximum mark is 1M if M1 shown) 1 7a ( ) ( ) 9 1 2 872 18 2 2 99 ....12 m m mx x x xx x x         + = + + +                   18 15 2 129 m 36 ....x x m x= + + + B2 (3 terms all correct) B1(2 terms correct) 1 7b(i) For 9 2 mx x  + , general term is (r + 1)th term = ( ) ( ) ( ) 9219 rr r x m xr − −  ( ) ( )( ) 18 29 r rrm x xr −−= ( ) ( ) 18 39 r rmxr −= 3318 =− r 5=r 59126 5 m− =  5 5 126 126 1 m m −= =− 1m=− M1 (general term) A1 ( r = 5) AG1 3

7 7b(ii) For 9 2 mx x  + , the term independent of x: 0318 =− r 6=r 669 84( 1) 846 m = − = For 9 2 3 12 axxx   −+     , the term independent of x: ( ) 93 3 12 .... 126 84 .... xx  − − + + = 2 84 126+ = 294 M1 A1 M1 A1 1 8a t (years) 6 9 12 15 18 25 33 P 274 203 151 112 83 41 18 Pln 5.61 5.31 5.02 4.72 4.42 3.71 2.89 P2 – Plot points accurately. L1 – Plot straight line graph (See graph attached.) 1 8b ktP Ae −= ln ln ln ln ln ln ln ln ln kt kt P Ae P A e P A kt P kt A − − = =+ =− =− + 0.1 5.61 4.72Grad = 6 15 = 0.0989 (3 s.f) 0.0989 0.1(1 d.p) ln 6.2 492.75 500 (nearest 100) 500 t k k A A A Pe − − − − − =− = = = = = M1 (product law or if evidence shown in transformation from eqn of graph to ktP Ae −= ) M1 (gradient) A1 B1 (If students did not use the gradient of line to solve for k and A, maximum mark is 1M as question mentioned hence.) 2

8 8c ln100 4.6 When ln 4.6 16 (nearest year) P t = = = M1 A1 1 9a 4sinEB = 9cosBF = 4sin 9cosd =+ M1 AG 1 3 9b 81 16R=+ 97R= 9cos 4sin (cos cos sin sin )R     + = + 9 cosR = 4 sinR = 4tan 9 = 23.962 (3 . )dp = 24.0 (1 . )dp = 9cos 4sin 97 cos( 24.0 )  + = −  M1 (Find R) M1 (No M1 given if student do not show this) M1 (Find  ) A1 1 9c 97 cos( 23.96 ) 6−  = 6cos( 23.96 ) 97  −  = 52.467 23.96 = +  76.427 76.4 =   M1 A1 (no A1 if 76.5) 1 E F  d m

9 9d Maximum value of d = 97 cos( 24.0 ) 1−  = 24.0 = Maximum value of d = 97 and occurs when 24.0 = B1 B1 1 10a 6cos 4 When 0, 6 Initial velocity of the particle is 6m/s. vt tv = == B1 1 10b 24sin 4 24sin 4 8 1sin 4 3 4 3.4814 0.870 (3 s.f) dvat dt t t t t = =− −= =− = = M1 ( dv dt ) A1 2 10c 6cos 4 6 sin 44 3 sin 42 When = 0, = 0, = 0 3 sin 42 When = 4, 3 sin162 0.432 Displacement = 0.432m (3 sf) s t dt s t c s t c ts c st t s s = =+ =+ = = =− −  M1 (integration with + c) A1 (conclude c=0) (students who used definite integral must indicate when t = 0, s=0 ) B1 1 10d At instantaneous rest, 0v= 6cos 4 0 cos 4 0 34, 22 3,88 t t t t   = = = = M1 (v= 0) M1 (values of t) 2

10 When , 8 3 sin 1.5m22 3When , 8 33sin 1.5m22 t s t s     = == = = =− Total distance travelled = (1.5 2) 1.5 4.5m + = M1 M1 A1 11a Solution 1 1 2 ( 6)Grad E 4 11 4 2 1Grad D 7 ( 1) 4 1Since Grad Grad 4 1 4 90 (Right angle in semi-circle) is the diameter of F F DF EF DF EF DFE DE C −−= =−−− −== −−  =−  =− ⊥  =   Since PQ is the diameter of 1C Centre of 1C = 1 7 6 4,22 + − +  = (4, 1)− M1 (Find grad) B1 (Show – 1) B1(Conclude 90  ) AG 1 (State reason angle in semi- circle) B1 3 11a Solution 2 22 22 22 (7 1) (4 2) 68 (1 1) ( 6 2) 68 (7 1) (4 6) 136 DF EF DE = + + − = = + + − − = = − + + = ( ) ( ) ( ) 2222 22 2 2 2 68 68 136 136 136 DF EF DE DF EF DE + = + = == += By converse of Pythagoras theorem, triangle DFE is a right angled triangle. Therefore 90DFE =  . M1 (Find distance) B1 (Show this statement) B1(Conclude 90  - must state Pythagoras thm) 3