AMKSS Prelim 2024 AMath P1 MS

2024 AMKSS 4E5N Prelim AM Paper 1 Solutions Comments: - Minus 1 mark overall for expressing coordinates in improper fraction (Q1 & Q12i) or missing unit such as degree (Q9i, Q10ii) Qn Solutions Marks 1 [5] 22 22 2 2 2 2 22 22 2 2 5 12 7 5 12 7 5 12 7 (1) 57 (2)2 Substitude (2)into (1) 575 12 7 2 25 70 495 12 7 4 5 3(25 70 49) 7 5 75 210 147 7 70 210 140 0 3 2 0 ( 2)( 1) xy y x xy xy xy xy xy xy xy xx xxx x x x x x x xx xx xx −−= −−= − =− − +=− +− =−  ++− =− − + + =− − − − =− + + = + + = ++ 0 2 or 1 1.5 or 1 ( 2, 1.5), ( 1,1) xx yy AB = =− =− =− = − − − M1 (substitution) M1 (expand & simplify to quadratic eqn) M1 (factorise/formula) A2 (coordinates should not be in improper fraction) 2(a) [2] 3 3 2 3 2 3 2 36 32 36 + −+ += − =+ M1 (conjugate) A1

2(b) [4] 22 2 22 2 2 2 2 2 2 30 3 6 2 2 4 30 (3 6) 4 4 16 30 (3 6) 4 30 9 6 6 6 4 45 6 6 4 4(45 6 6) 180 24 6 P P P P P P P     + +=            ++= + + = + + + = += =+ =+ M1 (Pythagoras theorem) M1 (attempt to form an eqn in P) A1, A1 3(i) [1] There is a fixed price of $100 million incurred, even when no submarines were assembled. B1 3(ii) [2] ( ) 2 2 22 2 2 2 5 20 1002 5 8 402 5 8 8 8 402 2 2 5 [( 4) 24]2 5 ( 4) 602 y x x y x x y x x yx yx = − + = − +     = − + − +         = − + = − + M1 A1 3(iii) [1] x = 4 occurs at the minimum point of the curve. Hence, the cost per submarine will be the lowest when we assemble 4 submarines B1

4(a) [4] ( )4 44 44 2 2 log 3log 2 3log 2 2 2log log 8 3 2 8 3 3 4 3 16 16 48 0 ( 4)( 12) 0 4 or 12 yy y y y y yy yy yy yy yy −−=  = − = − −= −= − + = − − = == M1 (quotient law of log) A1 (correct expression or equivalent) M1 (factorise/formula) A1 4(b) [3] 27 9 33 32 33 1 2 33 1 2 33 13 22 33 3 4 33 3 4 log log loglog log 3 log 3 log log 32 3log log 2 log log log log zy yz zy zy zy zy zy  = = = = = = = NOTE: If working end up with 3 2 2 33 44 or (Rej) zy z y z y = = =− M1 (change of base formula) M1 (power law) A1 Minus 1m if never show reject negative answer

5 [5] 32 2 32 2 32 32 2 22 2 2 6 1 ( 1)( 2) 2 6 1 ( 1)( 4 4) 2 6 1 34 Using long division, 92 ( 1)( 2) 9 ( 1)( 2) 1 2 ( 2) 9 ( 2) ( 1)( 2) ( 1) When 1; 9 9 1 When 2; 9 3 3 When 0; 9 4 xx xx xx x x x xx xx xx A B C x x x x x A x B x x C x xA A xC C x ++ −+ ++= − + + ++= +− + −+ = + +− + − + + = + + − + + − == = =− =− =− == 2 23 1 1 1 32 1 2 ( 2) B B x x x −+ =−  + − −− + + B1 2 () R Dx  + M1 (correct case) A2 (minus 1m for each incorrect ans) A1 6(a) [4] 17 22 Gradient of 2 Equation of : 2 6 2 6 (1) 5 6 (2) 5( 2 6) 6 10 30 6 9 36 4 2 (4, 2) yx AC AC y x yx xy xx xx x x y C =+ =− =− + =− + − + =− − + − + =− − + =− − =− = =− − M1 ( 1 2 1line: m m⊥ =− ) M1 (solve simultaneous eqns) M1 A1

6b(i) [3]     At ; 0 6 ( 6,0) Let ( ,6) Area of 1.5 Area of 0 6 4 0 0 4 011 1.5 6 0 2 6 6 2 6 622 1336 ( 36) 24 6 ( 2 24)24 336 (8 )4 6 (6,6) x axis y x D Bx ACD ABC x xx x x B −= =− − = − =− − − = + − − + = =  A1 (find pt D) M1 (must be anticlockwise) A1 6b(ii) [3] 22(0 4) (6 2) 80 1 80 362 72 or 8.05 unit 80 AC d d = − + + =   = = M1 (length formula) M1 (area of ACD) A1 7(i) [4] 2 2 2 40 40 2 2 2 402 2 2 802 2 2 802 xy x xy x P x x y xP x x x P x x x x Px x      += −= = + +  −= + +  = + + − =+ M1 (make y the subj) M1 (correct P exp) M1 (subt in their y) M1 (show expansion)

7(ii) [6] 2 2 2 2 23 3 2 23 802 8020 40 6.325 6.235(NA) 80Stationary value of 2(6.235) 6.235 25.3 cm 802( 2) 160 When 6.325; 160 0(6.235) is minimum dP dx x x x x or P dP dx x x x dP dx P =− −= = =− =+ = =− −  = = =  M1 (-1m for any incorrect term) M1 ( dP dx = 0) A1 (must show reject negative x value) A1 M1 (accept 1st or 2nd derivative test) A1 (explanation + conclude min) 8(i) [4] ( ) ( 1)( 2)( 5) (3) (4)(1)( 2) 30 8 15 4 15( ) ( 1)( 2)( 5)4 ( 3) 300 300 F x k x x x Fk k k F x x x x F R = + − − =− =− =− =− + − − −= = M1 (do not award if coeff of x3 assume 1) M1 (remainder theorem) M1 (subt 3x=− into their F(x) or correct long division) A1 8(ii) [2] 15 ( 1)( 2)( 5) 04 1 or 2 or 5 (Rej) 4 25 m m m m m m mm − + − − = =− = = == M1 A1 (must reject one ans)

9(i) [2] Amplitude = 3 Period = 1800 B1 B1 (must write degree) 9(ii) [4] For y = 3cos2x + 1 B1- 2 cycles with correct cosine shape through (0o, 4) & (360o, 4) & of amplitude 3 B1- correct turning points & middle values (45o, 135o, 225o, 315o) For y = sin (x/2) B1- half cycle of sine through (0o, 0) & (360o, 0) & of amplitude 1 B1- correct max point (180o, 1) 9(iii) [1] k = 3 A1 (given only if both graphs in (ii) are sketched correctly) 10(i) [1] p – q B1 10(ii) [3] 2 2 80 80 (5) 1.02cm / s dV dt dV hdh dV dV dh dt dh dt dh dt dh dt   = = = = = M1 M1 A1 (3sf) y =3cos2x+ 1 y = sin(x/2)

11(a) [3] Let ( in alternate segment) ( is bisector of ) ( sin same segment) (proven) DBE x BAD DBE x s BAE CAF x EA BAC CBE CAF x CBD DBE x =  = =   = =   = =   = = M1 M1 M1 11(b) (i) [2] 90 (right angle in semi-circle) 90 (common ) is similar to (AA Similarity test) ADB AOF ADB OAF DAB AOF ADB  =   = =   =    M1 M1 (must state AA similarity test) 11(b) (ii) [2] 2 Sin ce AOF is similar to ADB ( is radius and is diameter)2 2( ) ( ) AO AF AD AB AO AF AF FD AB AO AF AO ABAF FD AO AO AF AF FD  = =+ =+ =  + M1 M1 12(i) [5] 1 21 (6 5) (6)2 3 65 Gradient of tangent = 1 3 1 65 6 5 3 6 5 9 12 3 3 12 ,33 dy xdx x x x x x y T − =− = − = − −= −= = =   M1 (chain rule) A1 M1 (dy/dx =1) M1 A1

12(ii) [6] 2 2 15 2 1 5 3 2 1 53 2 1 2 65 65 6 5 0 ( 5)( 1) 0 5 or 1 (6 1) (6 5) 3 62 (6 5) 9 125 1 99 713 9 Area of trapezium 1 (5 1) 42 12 Shaded Area 7= 13 129 71 units9 xx xx xx xx xx x dx x x −= −= − + = − − = == −   −=    −=   =− = = +  = − =  M1 (solve simultaneously) A1 M1 (correct limits) M1 (correct integration) M1 A1 (accepts improper fraction for area)

13(i) [4] 2 22 sin 1LHS 1 cos tan sin 1 sin1 cos cos sin cos 1 cos sin sin cos (1 cos ) sin (1 cos ) sin cos cos sin (1 cos ) 1 cos sin (1 cos ) 1 sin cosec                    =− − =− − =−− −−= − −+= − −= − = = M1 ( sintan cos  = ) M1 (combined fraction) M1( 22sin cos 1+= ) M1 13(ii) [4] 2 2 cosec 2 9sin 2 1 9sin 2sin 2 9sin 2 1 1sin 2 9 1sin 2 3 0.33983 2 0.340, 2.80,3.48,5.94 0.170,1.40,1.74, 2.97 AA AA A A A Acute A A = = = = = = = = M1 M1 (2 values) A1, A1