4E5N AM MYE P2 2021 Marking Scheme

Additional Mathematics Sec 4 Express MYE Examination 2021 (Paper 2) Marking Scheme (Setter: Teo Hock Siong) Question Solution Marks Total 1 By long division, 8 = A(x2 + 2) + (Bx + C)(x) Using substitution, x = 0, - 8 = 2A A = - 4 Comparing x-coefficient, Comparing -coefficient Therefore, M1 M1 M1 M1 M1 M1 A1 7m

2a 0.83 0.42 0.28 0.21 0.17 0.62 0.50 0.41 0.38 0.36 Correct axes and plotted points. Appropriate straight line drawn. B1 B1 B1 3m

2bi (3s.f.) B1 1m 2bii From the graph, (3 s.f.) Using 2 points on the graph, (0.83, 0.62) & (0, 0.29) M1 A1 M1 A1 4m 2c By plotting against , a and b can be found using M1

gradient = and - intercept = A1 A1 3m 3 ai x-intercept of curve A1 1m y x 0 1 y = 1

aii M1 A1 2m

3b M1 M1 M1 M1

Required shaded area = 6 + 5.480 = 11.5 units2 (3 s.f.) A1 5m 4a Line and Curve intersect at 2 distinct points, Largest integer k is 5. M1 M1 M1 M1 A1 5m 4b Since y > 0, M1

But y > 0 means the coefficient must be positive, that is Since m must be and for , ∴there are no values of m for which y > 0 (is always positive). M1 M1 A1 4m 5a M1 M1 M1

(3 s.f.) A1 4m 5bi = (shown) M1 M1 A1 3m 5bii = 1.77 (3 s.f.) M1 M1 M1 M1 A1 5m