A Level Edexcel 2024 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes A-Level Mathematics Exam Board: Edexcel Practice Exam Papers

Edexcel A-Level Mathematics Practice Set 1 Paper 1: Pure Mathematics 1 Time allowed: 2 hours Centre name Surname Centre number Other names Candidate number Candidate signature In addition to this paper you should have: • An Edexcel mathematical formula booklet • A calculator For examiner’s use Q Mark Q Mark 1 9 2 10 3 11 4 12 5 13 6 14 7 15 8 16 Total Instructions to candidates • Use black ink or ball-point pen. • A pencil may be used for diagrams, sketches and graphs. • Write your name and other details in the spaces provided above. • Answer all questions in the spaces provided. • Show clearly how you worked out your answers. • Round answers to 3 significant figures unless otherwise stated. Information for candidates • There are 16 questions in this paper. • There are 100 marks available for this paper. • The marks available are given in brackets at the end of each question. • You may get marks for method, even if your answer is incorrect. Advice to candidates • Work steadily through the paper and try to answer every question. • Don’t spend too long on one question. • If you have time at the end, go back and check your answers. Exam Set MEP71 © CGP 2018 — copying more than 5% of this paper is not permitted Answer ALL the questions. Write your answers in the spaces provided. Leave blank 1 a) Given that f(x) = x3 – 4x2 – 3x + 7, find f '(x). (2) b) Hence find the values of x for which f(x) is a decreasing function, giving your answer in the form {x : x > a}  {x : x < b} where a and b are real numbers to be found. (3) 2 A helicopter flies between 3 locations, A, B and C, which are positioned such that AB = 9 km, AC = 5 km and angle ABC = 24°. Find the possible values of angle ACB to 1 decimal place. (3) A-level Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 2 5% of this paper is not permitted 3 a) Express 2 x 31 i 6 in the form axb where a and b are integers to be found. (2) b) Hence find the x-coordinates of the points where the line y = 1 – 3x intersects the curve with equation 2 x 31 i 6 + 6x – 18 = y 2 – y. (4) 4 For each of the following, prove that the statement is false. a) The exterior angles of a regular n-sided polygon are always acute. (1) For n  ℝ, n ≠ –1, nn + 1 ≥ 0. (1) For n < 50, if n is an odd prime then one or both of n + 2 and n + 4 are prime. (1) b) c) Leave blank Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 3 5% of this paper is not permitted 5 The elastic energy stored in a large industrial spring, E, in joules, is directly proportional to the square of how far it is extended, x, in metres. When the spring is extended by ( 2 – 1) m, it has (7 2 – 9) joules of elastic energy. Find the exact amount of elastic energy in the spring when it is extended by 2 m, giving your answer in joules in the form a + b 2, where a and b are integers. (5) 6 The circle C has equation x2 – 8x + y2 + 4y – 29 = 0. The centre of C is at the point X. a) Find: i) the coordinates of the point X. (2) ii) the radius of the circle C. (1) Leave blank A-level Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 4 5% of this paper is not permitted A tangent from the point P(–16, 13) touches the circle at the point Y. b) Find the distance PY. (3) 7 a) Express 4x2 + 8x + 3 as a single fraction in its simplest form. Hence, or otherwise, solve the equation: log3(4x2 + 8x + 3) – log3(4x2 – 9) = 2, x > 1.5 (4) 4x 2 - 9 (2) b) Leave blank Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 5 5% of this paper is not permitted 8 Figure 1 Figure 1 shows the graph of y = f(x) where x ℝ. The graph has stationary points A and B. a) State the nature of the stationary point A, justifying your answer with reference to the shape of the graph. (2) b) Explain why f(x) does not have an inverse function. (1) B is a minimum turning point with coordinates (p, q), where p and q are constants. c) Write down, in terms of p and q, the coordinates of the point B under these transformations: i) y = f(x – 1) (1) ii) y = 3f(2x) (1) Leave blank A-level Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 6 5% of this paper is not permitted d) Given that f(x) = 3x4 – 2x3 – 2: i) find the exact values of p and q. (4) ii) justify that B is a minimum turning point. (2) e) Sketch the graph of y = |f(–x)|. (2) Leave blank Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 7 5% of this paper is not permitted 9 Figure 2 shows how a manufacturer cuts pieces of cheese to sell. Wax D Figure 2 A cylinder of cheese has a layer of wax with negligible thickness applied to its curved surface. The cylinder is sliced horizontally, h cm below the top face. The slice is then cut vertically along two radii, AB and CB, as shown above. Each piece, ABCDEF, has a triangular label, ABC, applied to the top face. For a particular piece, h = 4 cm, angle ABC = 1.02 radians and the area of the label is 85.9 cm2. Find the area of wax on this piece of cheese. (4) Leave blank A-level Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 8 5% of this paper is not permitted 10 a) A student is attempting to find a root, a, of the equation f(x) = 0. The student finds two values, a and b, such that f(a) < 0 and f(b) > 0 and deduces that a < a < b. Explain why the student’s deduction is not necessarily true. (1) b) The student then attempts to use the Newton-Raphson method to find the root a of the equation 2x3 + 9x = 125, using 5 as a first approximation for a. Their working is shown below. f(x) = 2x3 + 9x – 125 f '(x) = 6x2 + 9 x0 = 5 f(x ) 2 (5)3 + 9^5h - 125 x 1 = x0 – f'(x00) = 5 – 6 (5)2 + 9 = 3.9308176... f(x 1 ) 2 (3.9)3 + 9^3.9h - 125 x 2 = x1 – f'(x1) = 3.9 – 6 (3.9)2 + 9 = 3.6133652... so a = 3.613 (3 d.p.) i) Identify two errors that the student has made in their working. (2) ii) Find the correct final value of a, giving your answer to 3 decimal places. (2) Leave blank Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 9 5% of this paper is not permitted 11 A company expects to make a profit of £250 000 in the year 2021. The company’s yearly profit is then expected to increase by 5% per year. a) By forming an inequality and solving it algebraically, work out which year will be the first to have an expected profit of more than £500 000. (4) b) Find the total expected profit for the company from 2021 to 2030 (inclusive), giving your answer to the nearest hundred pounds. (2) Leave blank A-level Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 10 5% of this paper is not permitted 12 a) Express -18 x - 7+3 x) (1in partial fractions. (3) (2 x) The first three terms in the binomial series expansion of 1 +1 x in ascending powers of x are: 1 – x + x2, x < 1 b) Hence find the first three terms, in ascending powers of x, of the binomial series expansion of . (5) x) Leave blank Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 11 5% of this paper is not permitted 13 Prove that: sin 2q + sin q = tan q cos q ≠ 0, - 1 cos 2q + 1 + cos q 2 (4) 14 a) Use a suitable substitution to show that: 75e -x -x 2 dx = 1 + k3 e -x + C f ^ 1 + 3e h where k is an integer to be determined. (5) Leave blank A-level Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 12 5% of this paper is not permitted A population of kangaroos is being studied. The population is modelled by the differential equation: ddPt = 2P^ 175 +e 3-t e-t h2 t ≥ 0 where P is the population of kangaroos in thousands, and t is the time measured in years since the study began. There were 5500 kangaroos when the study began. b) Solve the differential equation, giving your answer in the form P2 = f(t). (4) c) Find the limit of the size of the population as t  , showing clear algebraic working. (2) Leave blank Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copying more than 13 5% of this paper is not permitted 15 A factory, which makes metal cubes, contains a machine that heats the cubes to strengthen them. When heated, the volume of a cube increases at a constant rate of 1.5 cm3s–1. If the rate of increase of the total surface area of a cube is higher than 1 cm2s–1, the cube will be damaged. Showing your working clearly, find the minimum side length that a cube must have so that it doesn’t become damaged when heated. (5) Leave blank include more jokes, honestly. Write your marks in below Paper 1 Paper 2 Paper 3 Total Grade Published by CGP Editors: Sammy El-Bahrawy and Shaun Harrogate. M EP71 ~ /18713 Contributors: Paul Garrett and Charlotte Young. Text, design, layout and original illustrations With thanks to Allan Graham and Dawn Wright © Coordination Group Publications Ltd. (CGP) 2018 for the proofreading. All rights reserved. Set 1 Paper 3, Pages 2-5 and Set 2 Paper 3, Pages 2, 3 & 6 © Crown Copyright, the Met Office. Contains public sector information licensed under the Open Government Licence v3.0 - • 4 Answers Set 1 Paper 1 — Pure Mathematics 1 1 a) f '(x) = 3x2 – 8x – 3 [2 marks available — 1 mark for differentiating to get a quadratic, 1 mark for the correct answer] b) f(x) is decreasing when f '(x) < 0: f '(x) = 0  3x2 – 8x – 3 = 0  (3x + 1)(x – 3) = 0 [1 mark]  x = - 1 3and 3 [1 mark] f '(x) is u-shaped so is negative when - 1 < x < 3. So in the required set notation, this is {x : x > -3 31 }  {x : x < 3} [1 mark]. [3 marks available in total — as above] sin ACB9 A = sin5 24 ° 9 km sin ACB = 9 Bsin 524 ° = 0.7321...  Angle ACB = sin–1(0.7321...) = 47.06...° or 180° – 47.06...°  Angle ACB = 47.1° (1 d.p.) or Angle ACB = 132.9° (1 d.p.) [3 marks available — 1 mark for using the sine rule correctly, 1 mark for 47.1°, 1 mark for 132.9°] 3 a) _ 2 x 31 i6 = 2 6 x × 6 = 8x 2 [2 marks available — 1 mark for an answer in the form kx2 or 8xk, 1 mark for the correct answer] b) Substitute y = 1 – 3x into 8x2 + 6x – 18 = y2 – y to give 8x2 + 6x – 18 = (1 – 3x)2 – (1 – 3x) [1 mark]  8x2 + 6x – 18 = 1 – 6x + 9x2 – 1 + 3x [1 mark]  x2 – 9x + 18 = 0  (x – 3)(x – 6) = 0 [1 mark]  x = 3 and x = 6 [1 mark] [4 marks available in total — as above] 4 a) E.g. in a regular 3-sided shape (equilateral triangle) the exterior angles are 360° ÷ 3 = 120° which is not acute [1 mark]. You could also have chosen a regular 4-sided shape (square) because the exterior angles are 360° ÷ 4 = 90° — not acute. b) Choose any value of n such that –1 < n < 0, e.g. n = –0.2  n +n 1 = - 0.2- 0.2 + 1 = - 0.80.2 = –0.25 which is less than 0. [1 mark] c) E.g. n = 23  n + 2 = 25, n + 4 = 27 and 25 and 27 are not prime [1 mark]. You could also have chosen n = 31 and n = 47. 5 E µ x2  E = kx2  7 2 – 9 = k( 2 – 1)2 [1 mark] ( 2 – 1)2 = 2 – 2 2 + 1 = 3 – 2 2 [1 mark] 7 2 - 9 3 + 2 2 k = × [1 mark] 3 - 2 2 3 + 2 2 21 2 - 27 + 28 - 18 2 = 9 - 8 = 1 + 3 2 [1 mark] When x = 2 m, E = (1 + 3 2)( 2) 2 = 2(1 + 3 2) = 2 + 6 2 joules [1 mark] [5 marks available in total — as above] 6 a) i) x2 – 8x + y2 + 4y – 29 = 0  (x – 4)2 – 16 + ( y + 2)2 – 4 – 29 = 0 [1 mark]  (x – 4)2 + ( y + 2)2 = 49  centre X is (4, –2) [1 mark] [2 marks available in total — as above] ii) Radius = 49 = 7 [1 mark] You could also have put the circle equation into the form b) P (–16, 13), X (4, –2) and Y form a right-angled triangle, because a tangent (PY ) meets a radius (XY ) at 90°. Length of PX = ^- 16 - 4h2 + ^13 - (- 2)h2 = ^- 20 2h + 152 = 625 = 25 [1 mark] Length of XY = radius = 7 PY 2 = PX 2 – XY 2  PY 2 = 625 – 72 [1 mark]  PY 2 = 576  PY = 576 = 24 [1 mark] [3 marks available in total — as above] 7 a) 4x2 + 82 x + 3 = (2(2xx ++ 3)(2 3)(2xx + - 1)3) = 22xx +- 31 4x - 9 [2 marks available — 1 mark for factorising the numerator or denominator correctly, 1 mark for the correct answer] b) log (43 x2 + 82 x + 3) – log (43 x2 – 9) = 2  log a3 4x +4 8xx2 -+ 93 k = 2 [1 mark]  log b3 22xx + - 1 3 l = 2  2x 2+x 1 -= 3 32 [1 mark]  2x + 1 = 9(2x – 3)  2x + 1 = 18x – 27 [1 mark]  16x = 28  x = 7 or 1.75 [1 mark] 4 [4 marks available in total — as above] 8 a) Point A is a (stationary) point of inflection [1 mark] as e.g. the graph goes from convex to concave [1 mark]. [2 marks available in total — as above] b) f(x) doesn’t have an inverse (unless the domain is restricted) because f(x) is many-to-one, not one-to-one. [1 mark] Alternatively, you could have said that f – 1(x) cannot be one-to-one, or that f –1(x) would be one-to-many. 1 c) i) f(x – 1) is a translation of d 0n  B has coordinates (p + 1, q) [1 mark] ii) 3then a stretch parallel to the f(2x) is a stretch parallel to the y-axisx-axis of scale factor of scale factor 3 12  B has coordinates c p 2, 3q m [1 mark] d) i) f '(x) = 12x3 – 6x2 [1 mark] The turning points 6x2(2x – 1) = 0 [1 are found at mark]  x = f0 '( andx) = 10 2 1 From the graph, p is positive, so p = 2 [1 mark] and q = 3a 12 k4 – 2a 12 k3 – 2 = - 3316 or –2.0625 [1 mark] [4 marks available in total — as above] ii) f ''(x) = 36x2 – 12x [1 mark] When x = 12 , f ''(x) = 36a 12 k2 – 12a 12 k = 3, which is greater than 0 so B is a minimum [1 mark]. [2 marks available in total — as above] e) y = |f(–x)| is a reflection of y = f(x) in the y-axis, and a reflection in the x-axis of the part of the graph where y < 0. y = |f(–x)| y = f(x) y = f(–x) [2 marks available — 1 mark for each correct transformation] 9 LetThen the radius of the the area of triangle cylinder be ABC = r1. r2 sin 1.02 = 85.9 [1 mark] x 2 + 2gx + y2 + 2fy + c = 0 and then A Level Edexcel 2022 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes the centre is at 2  r = 2sin × 85.91.02 = 14.199... cm [1 mark] (–g, –f) and the radius is g2 + f 2 - c . Arc length = rq = 14.199... × 1.02 = 14.483... cm [1 mark] Area of wax = arc length × h = 14.483... × 4 = 57.932... cm2 = 57.9 cm2 (3 s.f.) [1 mark] [4 marks available in total — as above] 10 a) E.en f(x) amight and b have. [1 mark] an asymptote / might not be continuous 1 E.g. if f(x) = , f(–1) = –1 > 0 and f(1) = 1 < 0 but x there is no root of f(x) = 0 between –1 and 1. b) i) The student prematurely rounded the answer for x1 when finding x2 [1 mark] and they have concluded that the root is at 3.613 (3 d.p.) before checking that this agrees with the next value of xn. [1 mark] [2 marks available in total — as above] ii) x = 3.6176... x = 3.5919... x = 3.5917... 2 3 4 So a = 3.592 (3 d.p.) You can stop here because both 3.5919... and 3.5917... round to 3.592 (3 d.p.). [2 marks available — 1 mark for iterating until two values round to the same number to 3 d.p., 1 mark for the correct final answer] 11 a) This is a geometric sequence with r = 1.05 The nth term = arn – 1 = 250 000 × 1.05n – 1 > 500 000  1.05n – 1 > 2  (n – 1)ln 1.05 > ln 2  n – 1 > 14.2...  n > 15.2...  n = 16 n = 1 in 2021, so the expected profit will be over £500 000 in 2036. [4 marks available — 1 mark for the correct values of r and a, 1 mark for setting up an inequality correctly, 1 mark for solving by taking logs of both sides, 1 mark for the correct year] - rn) then b) a = 250 000, r = 1.05, n = 10 so since S = a (1 250 000 (1 - 1.05 10) n 1 - r S 10 = 1 - 1.05 [1 mark] = 3 144 473.13... = £3 144 500 (nearest £100) [1 mark] [2 marks available in total — as above] 12 a) 18x - 7 º A + B (2 - 3x)(1 + x) 2 - 3x 1 + x  18x – 7 º A(1 + x) + B(2 – 3x). Using x = 2  12 – 7 = 5 A  A = 3 3 3 and x = –1  –18 – 7 = 5B  B = –5 So 18x - 7 º 3 (2 - 3x)(1 + x) x [3 marks available — 1 mark for splitting the expression into two fractions with the correct denominators, 1 mark for a correct method to find A or B, 1 mark for the correct answer] 18x - 7 3 5 b) = 2 - 3x – 1 + x - 1 = 3(2 – 3x)–1 – 5(1 + x)–1 = 3 b - 3 l – 5(1 + x)–1 [1 mark] 2 1 2 x -1 2 =b1 1- +32 3x x l += 9 1 x +2 + ...(–1) [1b- mark 32 x l ]+ - 1 ×2! - 2 b- 32 x l + ... [1 mark] So 3 b2 - 3 4 l-1 – 5(1 + x)–1 2 1 2 x == 233 b+1 9+ x 2 3+ x 27 + x492 x–2 5+ +... 5l –x –5(1 5x –2 + ...x + x2 + ...) [1 mark] 2 7 4 29 8 13 = - 2 + 4 x – 8 x2 + ... [1 mark] [5 marks available in total — as above] 13 Using sin 2q  2 sin q cos q and cos 2q  2 cos2 q – 1 sin 2q + sin q 2 sin q cos q + sin q cos 2q + 1 + cos q = (2 cos2 q - 1) + 1 + cos q = sin q (2 cos q + 1) = sin q = tan q cos q (2 cos q + 1) cos q 5 14 a) Let u = 1 + 3e–x  du = –3e–x  du = –3e–x dx -x dx -25 (-3e-x) =So ff ^1 +75 d3eue -=x h2 f d -x 25 = uf- 2 d^u1 += –325(e-xh–2 u–d1)x + C = 25 + C - 25 u2 u so f ^1 75+ 3ee--xx h2 dx = 1 + 25 - + C (i.e. k = 25) [5 marks available —x 31e mark for any suitable substitution, 1 mark for differentiating the substitution with respect to x, 1 mark for forming a new integral in terms of u, 1 mark for a correct method to solve the integral in terms of u, 1 mark b) fordP the correct answer in= 75e-t  2P terms of x] dP = 75 e-t dt 2P^1 + 3e-th2 ^1 + 3e-th2 dt 75e-t  f 2P dP = f ^1 + 3e-t h2 d t [1 mark] When P 2 = t =1 0, +25 3P e = -t +5.5 C [1 5.5mark]2 = 25 + C [1 mark] 1 + 3  C = 30.25 – 6.25  C = 24 so P2 = 1 +25 3e -t + 24 [1 mark] [4 marks available in total — as above] c) As t  , e–t = 1  0  P 2  25 + 24 [1 mark]  P2  49  P e 7t 1 + 3 (0) So the limit of the size of the population is 7000 [1 mark]. [2 marks available in total — as above] 15 Let the side length of the cube be x cm. V = x3  dV = 3x2 [1 mark] dx AUsing = 6x 2  dddAxV =× 12 dxx [1with mark] dV = 1.5 and dV = 3x2 dV = dt dx dt dt dx gives 1.5 = 3x2 × dx  dx = 1 [1 mark] Using dA = dA ×d td x withdt dx =2x 2 1 and dA = 12x dt dx dt dt 2x2 dx dA 1 6 givesThe cube dt will= 12 bex × damaged 2x2 = x if[1 d dAmark]t > 1  6x > 1  x < 6. So 6 cm is the minimum side length that a cube must have so that it is not damaged when heated. [1 mark]. [5 marks available in total — as above] 16 Differentiating the implicit equation for C with respect to x: 3x2  6x, 2y2  4y dy and 18  0 [1 mark] –2xy 2 –– 22xxy d dx+ y –2 y22 y= [1 d18x mark]  6x – 2x dy – 2y + 4y dy = 0 So 3x d y dx dx  (4y – 2x) = 2y – 6x  dy = y - 3x = d 3x - y n [1 mark] When x = 2, d3(2)x 2 – 2(2)y + 2yd2 x= 182 y -12 x – 4yx + - 2 2yy2 = 18  y2 – 2y – 3 = 0 [1 mark]  (y + 1)(y – 3) = 0  y = –1 and y = 3. From the graph, the y-coordinate is positive at x = 2, so the coordinates of the point where line L meets curve C are (2, 3) [1 mark]. dy 3 - 3 (2) 3 Gradient of line L = dx at (2, 3) = 2 (3) - 2 = - 4 [1 mark] so L has equation y – 3 = - 34 (x – 2) [1 mark] 3 9  y = - 4 x + 2  3x + 4y = 18 Line L crosses x-axis at y = 0  3x + 4(0) = 18  3x = 18  x = 6 [1 mark] Curve  3x2 =C 18crosses  x2 x=- axis at 6  x y = =± 0  x = – [1 mark] 6 6 [4 marks available — 1 mark for using the identity for sin 2q, 1 mark for A Level Edexcel 2022 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes using the identity for cos 2q, 1 mark for factorising the (using the graph, take the negative solution) expression, 1 mark for the correct identity for tan q] [10 marks available in total — as above] So w = 6 – - 6 = 8.449... = 8.45 (3 s.f.) [1 mark] 6 Set 1 Paper 2 — Pure Mathematics 2 b) 1 a) 25 = 32 [1 mark] b) 24 × 5C × p = 240 [1 mark]  80p = 240  p = 3 [1 mark] [2 1 marks available in total — as above] c) q = 23 × 5C × p2 [1 mark] = 8 × 10 × 32 = 720 [1 mark] [2 marks available in total — as above] 2 2 a) A O To get from X to Y, you can go from X to O to A to Y, i.e. XY = XO + OA + AY [1 mark] Y is 3 times further away from B than A, so AY = AB ((124 i – 3j) – (4i + 9j)) 14 (8i – 12j) = 2i – 3j [1 mark] So XY = XO + OA + AY = (–4j) + (4i + 9j) + (2i – 3j) = (4 + 2)i + (–4 + 9 – 3)j = (6i + 2j) km [1 mark] [3 marks available in total — as above] There are lots of ways to do this question, e.g. you could use XY BA or start by finding OY . Award the marks for the correct answer with stages of working clearly explained. b) Distance XY = 62 + 22 [1 mark] = 36 + 4 = 40 = 4 10 = 2 10 km [1 mark] [2 marks available in total — as above] 3 a) t ≥ 0 since time is measured from today, so it cannot have a negative value [1 mark]. t = 9 is the solution to S = 0, so when t > 9, S is negative, but you can’t have a negative number of subscribers [1 mark]. [2 marks available in total — as above] b) dS = 7 – 2t [1 mark] and after 6 months t = 0.5 dt ddSt = 7 – 2(0.5) = 6 so the rate of change is 6 million subscribers per year. [1 mark] [2 marks available in total — as above] c) S = 18 + 7t – t2 = –(t2 – 7t – 18) = – bat - 72 k2 - 494 - 18l [1 mark] = – bat - 72 k2 - 1241 l = 1241 – at - 72 k2 or 30.25 – (t – 3.5)2 [1 mark] [2 marks available in total — as above] d) When t = 3.5, S = 30.25, i.e. subscribers [1 mark]. 4 a) f(–2) = 0  x + 2 is a factor. Take out a factor of (x + 2) using e.g. algebraic long division: x2 – x – 6 x + 2 x3 + x2 – 8x – 12 x3 + 2x2 –x2 – 8x –x2 – 2x –6x – 12 –6x – 12 0 So f(x) = (x + 2)(x2 – x – 6)  f(x) = (x + 2)(x + 2)(x – 3) or (x + 2)2(x – 3) [4 marks available — 1 mark for using the factor theorem, 1 mark for a correct method to take out a factor of x + 2, 1 mark for correctly factorising out x + 2, 1 mark for the correct answer] [3 marks available — 1 mark for the correct shape of a cubic graph with a positive x3 coefficient, 1 mark for the y-axis intercept of (0, –12), 1 mark for the correct x-axis intercepts] c) g(e2zz =) =3 f(e[1 2mark]z) and since 2z e=2 zln > 3 0 the z only= 1 lnsolution 3 or ln is3 at[1 mark] 2 [2 marks available in total — as above] 5 a) Discriminant = b2 – 4ac = k2 – 4(2)(–2) = k2 + 16 [1 mark], which is always positive as k2 ≥ 0, so the equation has real solutions for all values of k [1 mark]. [2 marks available in total — as above] b) k tan q – 2 cos q = 0  k sincos q q– 2 cos q = 0 [1 mark]  k sin q – 2 cos2 q = 0  k sin q – 2(1 – sin2 q ) = 0 [1 mark]  k sin q – 2 + 2 sin2 q = 0  2 sin2 q + k sin q – 2 = 0 as required [1 mark] [3 marks available in total — as above] c) For 3 tan q – 2 cos q = 0, k = 3 so 2 sin2 q + 3 sin q – 2 = 0 [1 mark]  (2 sin q – 1)(sin q + 2) = 0  sin q = 1 [12 mark] (sin q cannot equal –2 as –1 ≤ sin q ≤ 1)  q = 30° and 150° [1 mark] 6 a) V[3 µ marks 1  availableV = k in total — as above] P P V = pr2h so: pr2h = k  Pr2 = k [1 mark] As h is fixed, k is constant.P ph ph Let Po and ro be the original pressure and radius. Let P and r be the new pressure and radius. P or o2 =n p kh =n P nr n2 [1 mark] ro = 2 m and Pn = (1 – 0.84)Po = 0.16Po [1 mark] So: P (22) = (0.16P )r 2  4 = 0.16r 2  r 2 =o 25  r = 5o m n [1 mark] n [4 marksn availablen in total — as above] b) Let P ando r beo the original pressure and radius respectively, and Pn and rn be the new pressure and radius. Then rn = 2ro. So: P r 2 = P r 2  P r 2 = P (2r )2 [1 mark] o o n n o o n o  P r 2 = 4P r 2  P = 4P  P = 1 P o o 1 = 25n o% of P o, which is a n 75n % decrease.4 o [1 mark] So P is n 4 o [2 marks available in total — as above] 7 a) u 3= pu4 u2– –3 3so 33 = 45  p – 3  45p = 36  p = 4 5 u = 4 u = 45 + 3 = 48  u = 60 2 5 1 5 1 1 u 4= 45 u 3 – 3 = 45 (33) – 3 = 1175 or 23 25 , so required sum is 60 + 45 + 33 + 117 = 807 or 161 2 5 5 5 [3 marks available — 1 mark for finding the value of p, 1 mark for finding the value of u1 or u4 , 1 mark for the correct answer] b) As n tends to infinity you can assume that un = un+1 = L L = 54 L – 3 [1 mark]  1 5L = –3  L = –15 [1 mark] [2 marks available in total — as above] 8 a) n so take values of x every 0.2: x 0.4 0.6 0.8 1 y 0.1787... 0.4680... 1.0601... 2.4255... 1 So when splitting into 3 equal strips, f tan2 x dx [4 marks available — 1 mark for using the correct x-values, 1 mark for using the correct y-values, 1 mark for using the trapezium rule correctly, 1 mark for the correct answer] b) sec2 x = 1 + tan2 x so f [1 mark] 0.4 = f0.4 tan2 x dx + 5 f 1 dx = f0.4 1 tan2 x dx + 5 x 0.41 [1 mark] 0.4) = 0.5660... + 3 = 3.57 (3 s.f.) [1 mark] [3 marks available in total — as above] 9 When x is small, sin x ≈ x and cos x ≈ 1 – x2 . cosec 2x (1 - cos 4x) 1 - cos 4x 2 6x = 6x sin 2x [1 mark] ≈ 1 - c61 x -× 2^4x2x h 2 m [1 mark] = 128x x22 = 23 as required [1 mark] [3 marks available in total — as above] 10 a) Let u = ln x and ddvx = 1. Then ddux = 1x and v = x [1 mark] Using integration by parts: f1 4 ln x dx = x4 ln x 41 – f1 4 x × 1x dx [1 mark] = x ln x 14 f– 1 dx = x ln x - x 4 1 [1 mark] 1 = (4 ln 4 – 4) – (1 ln 1 – 1) = 4 ln 4 – 3 [1 mark] = ln 44 – 3 = ln 256 – 3 [1 mark] [5 marks available in total — as above] b) A = f1 4 x dx – f 1 4 ln x dx [1 mark] f 4 x dx = 6x 23 @ 4 1 [1 mark] 1 = 2 ^4 - 13 h2 = 2 (8 - 1) = 14 [1 mark] 33 3 So A = 143 – (ln 256 – 3) = 233 – ln 256 [1 mark] [4 marks available in total — as above] 11 f ' b 3p l = limh " 0 ` p3 + hhj - f` p3 j = limh " 0 cos`3 p + hh j - cos p3 [1 mark] f cosb p + hl = cos p cos h – sin p sin h [1 mark] 3 1 3 3 3