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1.Let f(x) = ex sin 2x + 10, for 0 ≤ x ≤ 4. Part of the graph of f is given below.There is an x-intercept at the point A, a local maximum point at M, where x = p and a local minimum point at N, where x = q. (a)Write down the x-coordinate of A.(1) (b)Find the value of(i)p;(ii)q.(2) (c)Find . Explain why this is not the area of the shaded region.(3)(Total 6 marks)2.The graph of y = sin 2x from 0? x ? ? is shown below.The area of the shaded region is 0.85. Find the value of k.(Total 6 marks) 3.The following diagram shows part of the graph of the function f(x) = 2x2.diagram not to scaleThe line T is the tangent to the graph of f at x = 1. (a)Show that the equation of T is y = 4x – 2.(5)(b)Find the x-intercept of T.(2) (c)The shaded region R is enclosed by the graph of f, the line T, and the x-axis.(i)Write down an expression for the area of R.(ii)Find the area of R.(9)(Total 16 marks) 4.Let f(x) = e2x cos x, –1 ≤ x ≤ 2.(a)Show that f′(x) = e2x (2 cos x – sin x).(3) Let the line L be the normal to the curve of f at x = 0.(b)Find the equation of L.(5) The graph of f and the line L intersect at the point (0, 1) and at a second point P.(c)(i)Find the x-coordinate of P.(ii)Find the area of the region enclosed by the graph of f and the line L.(6)(Total 14 marks)5.Let f (x) = , where p, q? +.Part of the graph of f, including the asymptotes, is shown below.(a)The equations of the asymptotes are x =1, x = ?1, y = 2. Write down the value of(i)p;(ii)q.(2) (b)Let R be the region bounded by the graph of f, the x-axis, and the y-axis.(i)Find the negative x-intercept of f.(ii)Hence find the volume obtained when R is revolved through 360? about the x-axis.(7) (c)(i)Show that f ′ (x) = .(ii)Hence, show that there are no maximum or minimum points on the graph of f.(8) (d)Let g (x) = f ′ (x). Let A be the area of the region enclosed by the graph of g and the x-axis, between x = 0 and x = a, where a ? 0. Given that A = 2, find the value of a.(7)(Total 24 marks) 6.The diagram shows part of the graph of y = (a)Find the coordinates of the point P, where the graph meets the y-axis. (2)The shaded region between the graph and the x-axis, bounded by x = 0 and x = ln 2, is rotated through 360° about the x-axis.(b)Write down an integral which represents the volume of the solid obtained.(4)(c)Show that this volume is ?.(5)(Total 11 marks) 7.Trains approaching a station start to slow down when they pass a point P. As a train slows down, its velocity is given by v = 40 – at, where t = 0 at P. The station is 500 m from P.(b)A train M slows down so that it comes to a stop at the station.(i)Find the time it takes train M to come to a stop, giving your answer in terms of a.(ii)Hence show that a = .(6) (c)For a different train N, the value of a is 4.Show that this train will stop before it reaches the station.(5)(Total 17 marks) 8.The velocity v m s–1 of a moving body at time t seconds is given by v = 50 – 10t.(a)Find its acceleration in m s–2.(2) (b)The initial displacement s is 40 metres. Find an expression for s in terms of t.(4)(Total 6 marks)Answer key1.(a)2.31A1N1 (b)(i)1.02A1N1 (ii)2.59A1N1(c) = 9.96A1N1split into two regions, make the area below the x-axis positiveR1R1N2[6] 2.Note:There are many approaches possible.However, there must be some evidenceof their method.Area = (must be seen somewhere)(A1)Using area = 0.85(must be seen somewhere)(M1)EITHERIntegrating (A1)Simplifying(A1)Equation = 0.85(cos 2k = ? 0.7)OREvidence of using trial and error on a GDC(M1)(A1)Eg = 0.5 , too small etcORUsing GDC and solver, starting with ? 0.85 = 0(M1)(A1)THENk = 1.17(A2)(N3)[6]3.(a)f (1) = 2(A1)f ′(x) = 4xA1evidence of finding the gradient of f at x =1M1e.g. substituting x =1 into f ′(x)finding gradient of f at x =1A1e.g. f ′(1) = 4evidence of finding equation of the lineM1e.g. y – 2 = 4(x –1), 2 = 4(1) + by = 4x – 2AGN05 (b)appropriate approach(M1)e.g. 4x – 2 = 0x=A1N22 (c)(i)bottom limit x = 0 (seen anywhere)(A1)approach involving subtraction of integrals/areas(M1)e.g. ∫ f (x) – area of triangle, ∫ f – ∫lcorrect expressionA2N4e.g. (ii)METHOD 1 (using only integrals)correct integration(A1)(A1)(A1)substitution of limits(M1)e.g. area = A1N4METHOD 2 (using integral and triangle)area of triangle=(A1)correct integration(A1)substitution of limits(M1)e.g. correct simplification(A1)e.g. area = A1N49[16] 4.(a)correctly finding the derivative of e2x, i.e. 2e2xA1correctly finding the derivative of cos x, i.e. –sin xA1evidence of using the product rule, seen anywhereM1e.g. f′(x) = 2e2x cos x – e2x sin xf′(x) = e2x(2 cos x – sin x)AGN0 (b)evidence of finding f(0) = 1, seen anywhereA1attempt to find the gradient of f(M1)e.g. substituting x = 0 into f′(x)value of the gradient of fA1e.g. f′(0) = 2, equation of tangent is y = 2x + 1gradient of normal = (A1)y – 1 = A1N3 (c)(i)evidence of equating correct functionsM1e.g. e2x cos x = , sketch showing intersection of graphsx = 1.56A1N1(ii)evidence of approach involving subtraction of integrals/areas(M1)e.g. – area under trapeziumfully correct integral expressionA2e.g. area = 3.28A1N2[14] 5.(a)(i)p = 2A1N1(ii)q = 1A1N1 (b)(i)f (x) = 0(M1)2 ? = 0(2x2 ? 3x ? 2 = 0)A1x = x = 2A1N2(ii)Using V = ?y2dx (limits not required)(M1)V = ?A2V = 2.52A1N2 (c)(i)Evidence of appropriate methodM1eg Product or quotient ruleCorrect derivatives of 3x and x2 ? 1A1A1Correct substitutionA1eg f ′ (x) = A1f ′ (x) = = AGN0 (ii)METHOD 1Evidence of using f ′(x) = 0 at max/min(M1)3 (x2 + 1) = 0 (3x2 + 3 = 0)A1no (real) solutionR1Therefore, no maximum or minimum.AGN0METHOD 2Evidence of using f ′(x) = 0 at max/min(M1)Sketch of f ′(x) with good asymptotic behaviourA1Never crosses the x-axisR1Therefore, no maximum or minimum.AGN0METHOD 3Evidence of using f ′ (x) = 0 at max/min(M1)Evidence of considering the sign of f ′ (x)A1f ′ (x) is an increasing function (f ′ (x) ? 0, always)R1Therefore, no maximum or minimum.AGN0 (d)For using integral(M1)Area = A1Recognizing that A2Setting up equation (seen anywhere)(M1)Correct equationA1eg dx = 2, ? = 2, 2a2 + 3a ? 2 = 0a = a = ? 2a = A1N2[24] 6.(a)y = ex/2 at x = 0 y = e0 = 1 P(0, 1)(A1)(A1)2(b)V = ?(A4)4Notes: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]7.Note: In this question, do not penalize absence of units.(a)(i)s = (M1)s = 40t (A1)(A1)substituting s = 100 when t = 0 (c = 100)(M1)s = 40t A1N5 (ii)s = 40t A1N1 (b)(i)stops at station, so v = 0(M1)t = (seconds)A1N2 (ii)evidence of choosing formula for s from (a) (ii)(M1)substituting t = (M1)e.g. 40 × setting up equationM1e.g. 500 = s, 500 = 40 × evidence of simplification to an expression which obviouslyleads to a = A1e.g. 500a = 800, 5 = , 1000a = 3200 – 1600a = AGN0 (c)METHOD 1v = 40 – 4t, stops when v = 040 – 4t = 0(A1)t = 10A1substituting into expression for sM1s = 40 × 10 – × 4 × 102s = 200A1since 200 < 500 (allow FT on their s, if s < 500)R1train stops before the stationAGN0 METHOD 2from (b) t = = 10A2substituting into expression for se.g. s = 40 × 10 × 4 × 102M1s = 200A1since 200 < 500,R1train stops before the stationAGN0 METHOD 3a is decelerationA24 > A1so stops in shorter time(A1)so less distance travelledR1so stops before stationAGN0[17] 8.(a)a = (M1)= –10 (m s-2)A1N2 (b)s = ∫v dt(M1)= 50t – 5t2 + cA140 = 50(0) – 5(0) + c c = 40A1s = 50t – 5t2 + 40A1N2Note:Award (M1) and the first A1 in part (b) if c ismissing, but do not award the final 2 marks.[6] ................
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