Answers - 3E NV AM EOY 2022

Marking Scheme Q Working Marks 1 (i) Remainder = f(1) = − 4 [B1] (ii) Remainder = f(3) = 0 [B1] Since remainder = 0, (x – 3) is a factor of f(x). (iii) by long division, f(x) = (x – 3)(2x2 + x – 1) [M2] for performing long division/ comparison method = (x – 3)(2x – 1) ( x + 1) [A1] for f(x) = 0, (x – 3)(2x – 1) ( x + 1) = 0 x = 3, x = 0.5, , x = −1 [A1] [ - 1 for not writing f(x) = 0 ] [B1 for giving final answers only] 2 (a) sin A = 55 8 [B1] (b) cos (−B) = cos B = 19 10− [B1] (c) tan Atan B = 55 9 3 1045 3 19 19  − − =   [M1, A1] 3 2 2 42 2 2 4x x x++ + = + 22 × 22x + 24 × 2x = 2x + 4 [M1] Let 2x = u 4u2 + 15u – 4 = 0 [M1] (4u – 1) ( u + 4) = 0 u = 1 4 or u = − 4 [M1] / [M1] 2x = 1 4 [ NO solution] [ - 1 for not writing no solution] 2x = 2-2 x = - 2 [A1] 4(a) B1 (correct shape) B1 (indicate asymptote & y- intercept) 2 1 5 y=3x+5

2 4 (b) 3ln(3 4)xx=+ [M1] ln(3 4)3 x x=+ 3 34 x ex =+ 3 1 3 5 x ex + = + [M1] Therefore the equation of the line is 35yx=+ [A1] (c) Line drawn and labelled on graph showing y-intercept of 5 [M1] No of solutions = 2 [A1] 5 (i) b2 – 4ac = 0 [M1] (-2)2 – 4 ( 3) (1 – p) = 0 p = 2 3 [A1] 5 (ii) x2 + y2 = 10 y = 3x + k x2 + ( 3x + k) 2 = 10 [M1] 10x2 + 6kx + k2 – 10 = 0 [M1] b2 – 4ac < 0 [M1] − 4k2 + 400 < 0 [M1] 4k2 – 400 > 0 4( k+ 10)(k – 10) > 0 k < -10 or k > 10 [A1] 6 (a) Tr+ 1 = (7 𝑟) (𝑥2)7−𝑟 (− 3 𝑥) 𝑟 [B1] – no need to simplify ** = (7 𝑟) (𝑥)14−2𝑟 (−3)𝑟 ( 1 𝑥) 𝑟 power of x = 14 – 3r [B1] (b) Let x0 = x 14 – 3r 14 – 3r = 0 r = 4 2 3 r is not an integer hence the constant term is not valid in the expansion. [M1]/FT [A1]

3 [Turn Over 7 1 𝟑 × (√2 + √3) 2 × ℎ = 22 + 9√6 [M1] (2 + 2√6 + 3) h = 3 (22 + 9√6) [M1] for correct expansion h = 3 (22 + 9√6) 5 + 2√6 = 3 (22 + 9√6) 5 + 2√6 × 5 − 2√6 5 − 2√6 [M1] / for rationalizing = 6 + 3√6------[A1] [A1] 8 √34x + 8 = 202 -2x. (3)2x + 4 = 202 × 20-2x [M1] for changing square root to half / (3)2x 34 = 202 202𝑥 [M1] (3)2x × (20)2x = 202 34 (60)2x = 202 34 [M1] (60)x = 20 9 [A1] method 2 √34x + 8 = 202 -2x 34x + 8 = (202 −2𝑥) 2 [M1] for squaring 202 -2x (3)4x × (20)4x = 204 38 [M1] (60)4x = 204 38 [M1] (60)x = 20 9 [A1] 9 (i) 22( 4) ( 5) 81xy+ + − = [M1] for standard form/formula mtd Centre (-4,5) [A1] Radius = 9 units [A1] (ii) tan 3 4m = 4 3PQm =− [M1] either one of the gradient Equation of PQ: 4 63yx=− + [A1