pure maths with mechanics and statis

1 a) Give n tha t f(x ) = x 3 – 4x 2 – 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-lev el Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copy ing more than 5% of this paper is not permitted 3 A Level Edexcel 2022 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes i n + 1 1 6 2 3 a) Express 1 6 2 x 3 in the form axb where a and b are integers to be found. (2) Leave blank b) Hence find the x-coordinates of the points where the line y = 1 – 3x intersects the curve with equation 2 x 3 i + 6x – 18 = y – 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) b) For n ℝ, n ≠ –1, n ≥ 0. (1) c) For n 50, if n is an odd prime then one or both of n + 2 and n + 4 are prime. (1) A-lev el Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copy ing more than 5% of this paper is not permitted 4 A Level Edexcel 2022 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes 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) Leave blank 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) A-lev el Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copy ing more than 5% of this paper is not permitted 5 A Level Edexcel 2022 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes 2 - A tangent from the point P(–16, 13) touches the circle at the point Y. b) Find the distance PY. (3) Leave blank 7 a) Express 4x2 + 8x + 3 as a single fraction in its simplest form. 4x 9 (2) b) Hence, or otherwise, solve the equation: log3(4x2 + 8x + 3) – log3(4x2 – 9) = 2, x 1.5 (4) A-lev el Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copy ing more than 5% of this paper is not permitted 6 A Level Edexcel 2022 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes 8 Leave y blank x A B 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) A-lev el Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copy ing more than 5% of this paper is not permitted 7 A Level Edexcel 2022 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes 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 A-lev el Edexcel Maths / Set 1 / Paper 1 © CGP 2018 — copy ing more than 5% of this paper is not permitted 8 A Level Edexcel 2022 Pure Maths Paper 1 and 2 with Mechanics and statistics Papers all included with Mark Schemes 9 Figure 2 shows how a manufacturer cuts pieces of cheese to sell. h cm Leave blank A B C Figure 2 h cm B Label A F C E Wax D 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.1 a) f '(x) = 3 x2 – 8 x – 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  3 x2 – 8 x – 3 = 0  (3 x + 1 )(x – 3 ) = 0 [1 ma r k] x = - 1 and 3 [1 mark] f '(x) is u-shaped so is negative 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 = = - 20 2 + 152 = 625 = 25 [1 mark] Length of XY = radius = 7 PY 2 = P X 2 – X Y 2  PY 2 = 6 25 – 7 2 [1 m a rk ] PY 2 = 576  PY = 576 = 24 [1 mark] [3 marks available in total — as above] when - 1 x 3. So in the required set notation, this is 4x2 + 8x + 3 (2x + 3)(2x + 1) 2x + 1 3 1 7 a) 4x2 - 9 = (2x + 3)(2x - 3) = 2x - 3 {x : x - 3 } {x : x 3} [1 mark]. [3 marks available in total — as above] 2 C [2 marks available — 1 mark for factorising the numerator or denominator correctly, 1 mark for the correct answer] b) log (4x2 + 8x + 3) – log (4x2 – 9) = 23 3 lo g a 4 x2 + 8 x + 3 k = 2 [1 ma r k] 5 km C 3 4x2 - 9 5 km log b 2x + 1 l = 2  2x + 1 = 32 [1 mark] A 9 km B 2x + 1 = 9(2x – 3) 2x + 1 = 18x – 27 [1 mark] sin ACB = sin 24° sin ACB = 9 sin 24° = 0.7321... 16x = 28 x = 7 or 1.75 [1 mark] 9 5 5 4 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 3 = x 3 = 8x [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] 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] Alternati vely, you could have said that f – 1(x) cannot be one-to-one, or that f –1(x) would be one-to-many. c) i) f(x – 1 ) i s a t ran sl ati o n of d1 n B has coordinates (p + 1, q) [1 mark] ii) 3f(2x) is a stretch parallel to the x-axis of scale factor 1 then a stretch parallel to the y-axis of scale factor 3 [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  B h as coo rdi n at es c p , 3q d) i) f ' (x) = 12 x3 – 6 x2 [1 m ar k ] m [1 mar k] [1 mark]. You could also have chosen a regular 4-sided s hape (square) The turning points are found at f '(x) = 0 6x2(2x – 1) = 0 [1 mark] x = 0 and 1 because the exterior angles are 360° ÷ 4 = 90° — not acute. 1 2 b) Choose any value of n such that –1 n 0, e.g. n = –0.2 From the graph, p is positive, so p = 2 [1 mark] and q = 3a 1 k4 – 2a 1 k3 – 2 = - 33 or –2.0625 [1 mark] n + 1 - 0.2 + 1 0.8 [1 mark] c) E.g. n = 23 n + 2 = 25, n + 4 = 27 and 25 and 27 are not [4 marks available in total — as above] ii) f ' '(x) = 3 6 x2 – 1 2 x [1 m a rk ] When x = 1 , f '' (x) = 36 a 1 k2 – 12a 1 k = 3, w hich is 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] greater than 0 so B is a minimum [1 mark]. [2 marks available in total — as above] ( – 1)2 = 2 – 2 2 + 1 = 3 – 2 [1 mark] e) y = |f(–x)| is a reflection of y = f(x) in the y-axis, and a k = 7 2 - 9 3 - 2 2 × 3 + 2 2 3 + 2 2 [1 mark] reflection in the x-axis of the part of the graph where y 0. y = 21 2 - 27 + 28 - 18 2 = 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 y = |f(–x)| y = f(–x) y = f(x) x (x – 4)2 – 1 6 + ( y + 2 )2 – 4 – 2 9 = 0 [1 m ar k ] (x – 4)2 + ( y + 2)2 = 4 9  cen tre X i s (4 , –2 ) [1 m ar k ][2 marks available in total — as above] ii) Radius = 49 = 7 [1 mark] [2 marks available — 1 mark for each correct transformation]