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] equation of line

4 4 3 2663 4 4 25 25 12 2 6 xx x x − + = − = = [M1] substitution [A1] coordinates of Q Q (6,-2) (iii) (3, 2)PQM = Radius = 22(0 3) (6 2) 5− + − = units [B1] centre [M1, A1] (iv) Distance between point and 22 1 (2 4) (6 5) 6.082C = + + − = Since < 9 units therefore lie in 1C Distance between point and 22 2 (2 3) (6 2) 4.123C = − + − = Since < 5 units therefore lie in 2C [B1] [B1] 10 (a) log2 2x – log4(x + 2) = 1 log2 2x − 𝑙𝑜𝑔2( 𝑥+2) 𝑙𝑜𝑔24 = 1 [M1] for changing base 2log2 2x −log2 (x + 2) = 2 [M1] for simplifying 𝑙𝑜𝑔2 ( 4𝑥2 𝑥+2 ) = 2 [M1] / No FT if student write 𝑙𝑜𝑔2 ( 2𝑥2 𝑥+2 ) = 2 4x2 = 4(x + 2) x2 – x – 2 = 0 [M1] ( x – 2) ( x +1) = 0 x = 2 or x = - 1(rejected) [A1] / No A1 if x = -1 is not rejected (b) log 1000 lgz z= lg1000 lglg zz = 2(lg ) 3 lg 3 z z = = 3 lg 3 10 54.0 z z = == 3 lg 3 10 0.0185 z z − =− == [M1] use of change of base [M1] for simplifying [M1] solve for lgz [A1,A1] for answers (c) (i) P = 15000 [ B1] (ii) 28000 = 15000ebt [M1]

5 [Turn Over 28 15 = e3(b) [M1] 3b = ln 28 15 (= 0.624154) b= 1 3 ln 28 15 = 0.208051 P = 15000ebt P = 15000e(7.75 x × 1 3 ln 28 15 ) [M1] for substituting b and t = 7.75/ FT ≈ 75200 (3sf) [A1]/ FT

6 11 x2 + 10 (2x2 + 1)(3 - x) = Ax + B (2x2 + 1) + C (3 + x) x2 + 10 = (Ax + B) ( 3 + x) + C( 2x2 + 1) [M1] when x = - 3, 9 + 10 = C(18 + 1 ) C = 1 [M1]/FT when x = 0, 8 = (B)(3) + 1( -1) 9 = 3B B = 3 [M1]//FT When x = 1, 11 = (A + 3)(4) + 1(3) 8= 4A + 12 A = - 1 [M1]/FT x2 + 8 (2x2+1)(3 + x) = 3- x (2x2 + 1) + 1 (3 + x) [A1] 12 (i) Scale [B1], Points [B1], Best fit line[B1] For (ii) to (iv), answers obtained by calculation are not accepted. (ii) 3.9 ln 3.9 49.4 49 N Ne = = = = Accept 44 60N [M1] vertical intercept value [A1] N value rounded to whole number 3.8 ln 3.8 44.7 45 N Ne = = = = 4.1 ln 4.1 60.34 60 N Ne = = = = (iii) 0ln lnN N kt=+ [M1] form equation 16 8 0.55222 7.5k −== − Accept 0.500 0.600k [A1] (iv) Accepted a range of answers for ln N: ln((44.7)(3)) 4.898= ln((49.4)(3)) 4.998= ln((54.5.)(3)) 5.096= 2th= [M1] find ln N values [A1] find time from the graph

7 [Turn Over 13 (a) grad of AC = − 1 [M1] mid point = ( −2+4 2 , 10+4 2 ) = (1, 7) [M1] grad of P.B = -1 [M1]/ eqn y = x + 6 [A1] no FT if last step is wrong such as student forgetting to find gradient of perp bisector /r mid point (b) y = 0, B = ( -6, 0) [B1] Let D = (x,y) mid point M = ( −6+𝑥 2 , 0+𝑦 2 ) [M1]/ FT from answer for B D = ( 8, 14) [A1]/ FT (c) Area of rhombus = 1 2 | −2 − 6 4 8 − 2 10 0 4 14 10 | [M1]/ FT = 84 units2 [A1]/ FT