Answers - 3E NV AM EOY 2024

1 North Vista Secondary Sec 3EXP End-of-Year Assessment 2024 Additional Mathematics 4049 Marking Scheme Qn Solutions Total Mark AO 1 2 3 0 2 3 0 2 2 36 2 3 6 2 3 3 3 33 Comparing Indices 2 3 0.........(1) Comparing Indices 2 3 6 2 2 3 4.........(2) (1) (2) 44 1 2 3 xy xy x y xy xy a a a aa xy xy x x y + − + − − − = = += = = − − =− −= + = = =− [4] M1 - Obtains first equation M1 - Obtains second equation A1 - x value A1 - y value (only accept exact value) AO2

2 2(a) y = mx 2 – 5x + 2 y is always positive  b 2 – 4ac < 0 ( ) ( ) 0245 2 −− m 258 m 8 13m [6] M1 – compute D A1 AO1 (b) x 2 – 4x > 0 --------- (1) x(x – 4) > 0 x < 0 or x > 4 x 2 – 4x  2x + 16 -------- (2) x 2 – 6x – 16  0 ( x + 2)( x – 8 )  0 – 2  x  8 The solutions are – 2  x < 0 or 4 < x  8. M1 M1 A2 [only if inequality sign is correct] AO2 0 4 x -2 8 x

3 3(a) When t = 0, 63006000300 =+=R [6] B1 AO1 (b) 63000 =R Given R < 6300 2 ( ) 0.02300 6000 10 3150 t−+ ( ) 0.02 285010 6000 t−  0.0210 0.475t−  Take lg on both sides 0.02 lg 0.475t− lg 0.475 16.165 17 days0.02t = − M1 – form eqn M1 - Take lg on both sides A1 – in days AO2 (c) When ,t→ 0.0210 0 t− → ( ) 0.02300 6000 10 300 tR −= + →  Number of rabbits will not become zero. Need to explain mathematically B1 B1 - Conclusion AO3

4 4(a)(i) [7] B1 – correct shape and close to asymptote B1 - x-intercept: (2,0) B1 - asymptote: x =1 AO2 (ii) 3 1 1 x ex − =− ( ) 3 11ln xx −=− ( ) xx −=− 11ln3 xy −=1 M1 – Take ln both sides A1 AO2 (b) 0100 ln =− pxx e pxx e ln100 = pxx e lnln100ln = pxx ln100ln = 100=p M1 – Take ln both sides A1 AO1

5 5(a) 22 5 3 3 4 2 3 2 2 2 3 27 10 10 3 6 6 3 2 3 6 4 (2 3) 2 3 3 22 2 3 4 8 33 11 3 3 3 3 3 11 3 3 9 3 & 11hk  −+ −+ + + − − += − += += += == [7] M1 - expansion M1 Rationalize A2 AO2 (b) 2 3 2 3 2 3 2 3 3 3 3 3 3 3 2 2 33 3 5 27 125 2 53 3 3 5 25 3 3 5 2 5 3 5 3 5 3 5 2 2730 0.216125 x x x x x x x x x x x x x xx x x x x or +− −  =    =   =   =   = M1 M1 A1 A02

6 6 In the expansion of 11 2 3 2 ,x x  + ( ) 112 1 3 22 5 11 2 11 2 rr r rr Tx r x xr − + −  =  = To get 7, 22 5 7 3x r r − =  = Coefficient of 7x term is = 311 2 1320r  = (Optional step) To get 3 19 4, 22 5 3 3 55x r r − =  = = Since r is not a positive integer, therefore the coefficient of x3 term is 0. Hence the ratio of the coefficient of x7 term to that of the x3 term is 1320 : 0 which is undefined. [4] M1 – gen term M1 - correct expression M1 – value of r B1 - explanation AO3

7 7(a) ( ) ( )( ) 33 33 22 3 24 32 3 2 2 4 xy xy x y x xy y − =− = − + + [8] M1 – diff of cubes A1 AO1 (b) ( )( ) ( ) ( ) ( ) ( )( ) 2 5 1 33121 310 3 2 of tscoefficien Comparing 3 21 0Let 5 525 2Let 1)2)((192 21)2)(1( 192 )2)(1( 192121 103 22 3 2 22 22 2 2 2 2 3 −−+ −+−=−+ −+− = +=− −= +−= = −= =− = ++−+=+−− −++ +=−+ +−− −+ +−−+−=−+ −− xx x xx xx A CA x B CB x C C x xCxBAxxx x C x BAx xx xx xx xx xx xx M1 - Long division M1 - correct form B3 - A, B and C values A1 – conclusion with correct sign AO2

8 8(a) 2g = -8 & 2f = 6 & c = -4 g = -4 f = 3 Centre C = (4, –3) Radius = 16 9 ( 4)+ − − = 29 units CT = 161+ = 17 units < 29 units Hence T lies inside the circle. [9] B1 – centre B1 – radius B1 f.t. for CT, with conclusion B0 for merely substituting (5, 1) into equation AO3 (b)(i) Gradient = tan 45 1− =− (L is slanting downwards) Substitute (11, 0)  y = –x + 11 M1 A1 AO1 (ii) Midpoint y = 5 Substitute y = 5  x = 6  C = (6, 5) M1 A1 AO1 (iii) Radius = 36 4 40+= (x – 6)2 + (y – 5)2 = 40 √M1 (ecf) A1 AO1 (iv) Because CR = 22 55 + = 50  40 Hence, it is impossible for a circle with centre C to pass through all three points, P, Q and R. B1 - explanation AO3

9 9(a) ( ) 32 32 f 3 11 35 By Remainder Theorem, 1Remainder = f 2 1 1 1 3 11 352 2 2 325 85 40 or 40.6258 x x x x= + + − −      = − + − + − −           =− =− − [11] M1 - substitution A1 AO1 (b) 325 5 5 5f 3 11 35 03 3 3 3        = + + − =               By the Factor Theorem , ( )35x− is a factor of ( )f. x [Shown] B1 – cite FT AO1 (c) ( ) ( )( ) 32 2 f 3 11 35 3 5 2 7 x x x x x x x = + + − = − + + ( ) ( )( ) ( )( ) 2 2 2 When f 0, 3 5 2 7 0 5 or 2 7 03 2 4 1 7 24 0 x x x x x x x D = − + + = = + + = =− =−   The equation has only one real root. [Shown] M1 - By long division or by Synthetic Division M1 M1 – discriminant A1 – conclusion AO3

10 (d) ( ) ( ) ( ) ( ) ( ) ( ) 3 1 2 32 32 3 3 35 11 3 3 3 3 11 3 35 0 3 3 3 11 3 35 0 Let = 3 y y y y y y y y y yu + + = − + + − = + + − = From part (ii) and (iii) 53 3 Take lg on both sides, 5lg 3 lg 3 5lg 3 lg 3 0.46497 0.465 (to 3 sf) y y y = =  = =  M1 – attempt to make the same form √M1 (ecf) M1 A1 AO3