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 o