(3) (b) Given that f′ = – x + p, show that p = –4. (4) (c) Find f(x). (7) (Total 14 marks) 6. The acceleration, a m s–2, of a particle at time t seconds is given by a = + 3sin 2t, for t ≥ 1. The particle is at rest when t = 1. Find the velocity of the particle when t = 5. (Total 7 marks) 7. The function f is such that f  (x) = 2x – 2. When the graph of f is drawn, it has a minimum point at (3, –7). (a) Show that f (x) = x2 – 2x – 3 and hence find f (x). (6) (b) Find f (0), f (–1) and f (–1). (3)
(c) Hence sketch the graph of f, labelling it with the information obtained in part (b). (4) (Note: It is not necessary to find the coordinates of the points where the graph cuts the x-axis.) (Total 13 marks) 8. Let f(x) = . Line L is the normal to the graph of f at the point (4, 2). (a) Show that the equation of L is y = –4x + 18. (4) (b) A is the x-intercept of L. Find the x-coordinate of A. (2) In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L. (c) Find an expression for the area of R. (3) (d) The region R is rotated 360° the x-axis. Find the volume of the solid formed, giving your answer in terms of π. (8) (Total 17 marks)
5*. 10 units2 (example 5 page 480) 6*. -cos4x + c (example 19a page 467) 1. (a) f (1) = 2 (A1) f ′(x) = 4x A1 evidence of finding the gradient of f at x =1 M1 e.g. substituting x =1 into f ′(x) finding gradient of f at x =1 A1 e.g. f ′(1) = 4 evidence of finding equation of the line M1 e.g. y – 2 = 4(x –1), 2 = 4(1) + b y = 4x – 2 AG N0 5 (b) appropriate approach (M1) e.g. 4x – 2 = 0 x= A1 N2 2 (c) (i) bottom limit x = 0 (seen anywhere) (A1) approach involving subtraction of integrals/areas (M1) e.g. ∫ f (x) – area of triangle, ∫ f – ∫l correct expression A2 N4 e.g. (ii) METHOD 1 (using only integrals) correct integration (A1)(A1)(A1) substitution of limits (M1) e.g. area = A1 N4 METHOD 2 integral and triangle) area of triangle= (A1) correct integration (A1) substitution of limits (M1) e.g. correct simplification (A1) e.g. area = A1 N4 9 [16]
2. f (1) = 12 – 3b + c + 2 = 0 (M1) f (x) = 2x – 3b, f (3) = 6 – 3b = 0 (M1) 3b = 6, b = 2 (A1) 1 – 3(2) + c + 2 = 0, c = 3 (A1) Note: In the event of no working shown, award (C2) for 1 correct answer. [4] 3. (a) y = ex/2 at x = 0 y = e0 = 1 P(0, 1) (A1)(A1) 2 (b) V =  (A4) 4 Notes: Award (A1) for  (A1) for each limit (A1) for (ex/2)2. (c) V = (A1) =  (A1) = [eln2 – e0] (A1) = [2 – 1] =  (A1)(A1) =  (AG) 5 [11] 4. (a) (seen anywhere) A1A1 attempt to substitute into the quotient rule (do not accept product rule) M1 e.g. correct manipulation that clearly leads to result A1 e.g. AG N0 4 (b) evidence of setting the derivative equal to zero (M1) e.g. g′(x) = 0, 1– 2ln x = 0 A1 A1 N2 3 [7] 5. (a) substituting into the second derivative M1 e.g. 3 × f″ = –5 A1 since the second derivative is negative, B is a maximum R1 N0 (b) setting f′(x) equal to zero (M1) evidence of substituting x = 2 (M1) e.g. f′(2) correct substitution A1 e.g. correct simplification e.g. 6 – 2 + p = 0, + p = 0, 4 + p = 0 A1
p = –4 AG N0 (c) evidence of integration (M1) f(x) = A1A1A1 substituting (2, 4) or into their expression (M1) correct equation A1 e.g. A1 N4 [14] 6. evidence of integrating the acceleration function (M1) e.g. correct expression ln t – cos 2t + c A1A1 evidence of substituting (1, 0) (M1) e.g. 0 = ln 1 – cos 2 + c c = –0.624 (A1) v = ln t – (A1) v(5) = 2.24 (accept the exact answer ln 5 – 1.5 cos 10 + 1.5 cos 2) A1 N3 [7] 7. (a) f (x) = 2x – 2 (x) = x2 – 2x + c (M1)(M1) = 0 when x = 3  0 = 9 – 6 + c c = –3 (A1) f (x) = x2 – 2x – 3 (AG) f (x) = – x2 – 3x + d (M1) When x = 3, f (x) = –7  –7 = 9 – 9 – 9 + d (M1)  d = 2 (A1) 6  f (x) = – x2 – 3x + 2 (b) f (0) = 2 (A1) f (–1) = – – 1 + 3 + 2 = 3 (A1) f (–1) = 1 + 2 – 3 = 0 (A1) 3 (c) f (–1) = 0  is a stationary point (A4) 4 Note: Award (A1) for maximum, (A1) for (0, 2) (A1) for (3, –7), (A1) for cubic. [13] 8. (a) finding derivative (A1) e.g. f′(x) = correct of derivative or its negative reciprocal (seen anywhere) A1 e.g. gradient of normal = (seen anywhere) A1 e.g. substituting into equation of line (for normal) M1
e.g. y – 2 = –4(x – 4) y = –4x + 18 AG N0 (b) recognition that y = 0 at A (M1) e.g. –4x + 18 = 0 x = A1 N2 (c) splitting into two appropriate parts (areas and/or integrals) (M1) correct expression for area of R A2 N3 e.g. area of R = (triangle) Note: A1 if dx is missing. (d) correct expression for the volume from x = 0 to x = 4 (A1) e.g. V = V = A1 V = (A1) V = 8π A1 finding the volume from x = 4 to x = 4.5 EITHER recognizing a cone (M1) e.g. V = πr2h V = (A1) = A1 total volume is 8π + A1 N4 OR V = (M1) = = A1 = A1 total volume is 8π + A1 N4 [17] 9. METHOD 1 correct expression for second side, area = 525 (A1) e.g. let AB = x, AD = attempt to set up cost function using $3 for three sides and $11 for one side (M1) e.g. 3(AD + BC + CD) + 11AB correct expression for cost A2 e.g. EITHER sketch of cost function (M1) identifying minimum (A1) e.g. marking point on graph, x = 15 minimum cost is 420 (dollars) A1 N4 OR