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Dubai

Integration IB Mathematics

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Published in: Mathematics
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Practice problems with Solutions

Mukul D / Dubai

6 years of teaching experience

Qualification: Bachelor's of Engineering

Teaches: Chemistry, Mathematics, SAT, ACT, Maths

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  1. Integration by Substitution and Parts 2008-2014 with MS Ia. [5 marks] Let Show that and deduce that fis an increasing function. 1b. [6 marks] Show that the curveY = f (X) has one point of inflexion, and find its coordinates. Ic. [11 marks] — arcsin x Use the substitution C Sin to show that 2a. [1 mark] The function fis defined on the domain T] by f (X) = COS State the two zeros off. 2b. [1 mark] Sketch the graph off. 2c. [7 marks] The region bounded by the graph, the x-axis and they-axis is denoted by A and the region bounded by the graph and the x-axis is denoted by B . Show that the ratio of the area of A to the area of B is 3.[7 marks] By using the substitution X = Sint , find 4.[6 marks] By using an appropriate substitution find tan(ln y) Y 5.[6 marks] x sin 2:rdx Show that 0 8 6.[5 marks] 24. re In Cda Calculate the exact value of 1 7a. [9 marks] (i) Sketch the graphs of Y and Y sin , on the same set of axes, for O (ii) Find the x-coordinates of the points of intersection of the graphs in the domain (iii) Find the area enclosed by the graphs. 7b. [8 marks] Find the value of 0 using the substitution = 4sin20.
  2. 7c. [8 marks] The increasing function f satisfies f (O) = O and f (a) = b wherea > Oandb > (). r = ab — fob (i) By reference to a sketch, show that 0 12 arcsin (f) dc (ii) Hence find the value of O 8a. [8 marks] Prove by mathematical induction that, forn e Z + 1 1 1 1 1+2 +3 2 2 2 2 8b. [1 7 marks] f e2a sinædc = e2z (2 sinc — cos x) + C (a) Using integration by parts, show that dy _ I — y2e2r sin x (b) Solve the differential equation dc , given that y = 0 when x = 0, writing your answer in the form Y = f(x) (c) (i) Sketch the graph of Y = f(x) , found in part (b), for O 'c 1.5 Determine the coordinates of the point P, the first positive intercept on the x-axis, and mark it on your sketch. (ii) The region bounded by the graph of Y = f (a) and the x-axis, between the origin and P, is rotated 3600 about the x-axis to form a solid of revolution. Calculate the volume of this solid. 9a. [6 marks] The integralln is defined by n Show that O 9b. [4 marks] By letting Y = — MT , show that n — 9c. [5 marks] Hence determine the exact value of 0 10a. [6 marks] dc 10b. [3 marks] Find tan3ædx 1 Ia. [7 marks] sin cldx, for n € N Find the value of the integral 11b. [5 marks] Find the value of the integral arcsin Ilc. [7 marks] 2
  3. Using the substitution t tan 9 , find the value of the integral 0 3cos29+ sin2 9 12a. [2 marks] Express 4x2 — + 5 in the form — + k where a, h, k Q. 12b. [3 marks] The graph of Y = is transformed onto the graph of Y = — + 5. Describe a sequence of transformations that does this, making the order of transformations clear. — 16. 12c. [2 marks] —4:r+5 . The functionfis defined by Sketch the graph of Y = f(x). 12d. [2 marks] Find the range off. 12e. [3 marks] By using a suitable substitution show that 12f. [7 marks] r3.5 Prove that Jl 4x2 —4t+5 13.[7 marks] I —L— du (a) Given thata > 1, use the substitution 1 dc = 1 dc. (b) Hence show that arctan a + arctan — 14.[6 marks] fl tln(t + l)dt Find the value of 0 15a. [4 marks] Find æsec2 15b. [2 marks] Determine the value of m if 16.[5 marks] (a) Integrate I—cos O — to show that 0.5 , where m > 0. (b) Given that 17.[8 marks] Using the substitution C = 2 sine , show that 4 — + B arcsm I —COS O 2 and , find the value ofa. • — + constant , 3
  4. where and are constants whose values you are required to find. 18a. [5 marks] Let f(æ) = -F 2)6. Solve the inequality f (X) > 18b. [5 marks] Find f 19.[7 marks] dc Use the substitution C = a seco to show that 20a. [2 marks] (3v'fi -F — 6) v(t) = 12 Particle A moves such that its velocityvm S— , at time t seconds, is given by Sketch the graph of Y = V Indicate clearly the local maximum and write down its coordinates. 20b. [4 marks] f —I— dt Use the substitution u t2 to find 12+t4 20c. [3 marks] Find the exact distance travelled by particle A between t O and t 6 seconds. karctan(b), k, b € R Give your answer in the form 20d. [3 marks] Particle B moves such that its velocityvms— is related to its displacement , by the equation v(s) = arcsin (v"S). Find the acceleration of particle B when S 0.1m 21.[7 marks] By using the substitution a: = 2 tan u, show that 4
  5. Integration by Substitution and Parts 2008-2014 MS Ia. [5 marks] Markscheme EITHER MIAI derivative of is (I-x)2 MIAI f' (x) > 0 (for all O < < 1) so the function is increasing RI OR f(c) = MIAI Al f' (X) > O (for all O < < 1) so the function is increasing RI [5 marks] 1b. [6 marks] Markscheme MIAI MIAI (X) changes syn at — hence there is a point of inflexion RI the coordinates are 4 vr [6 marks] Ic. [11 marks] Markscheme % = 2sin0cos9 = sin20 MIAI f v/Zdx = f 2 sin 9 cos Od0 I—sin 20 MIAI = 2sin29dO Al = f 1 — cos 29dO MIAI — Sin 29 c Al 9 = arcsin Al MIAI f = arcsin — + c hence [11 marks] Examiners report Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part 5
  6. (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates. 2a. [1 mark] Markscheme e-c COS T Al [1 mark] Examiners report Many candidates stated the two zeros off correctly but the graph off was often incorrectly drawn. In (c), many candidates failed to realise that integration by parts had to be used twice here and even those who did that often made algebraic errors, usually due to the frequent changes of sign. 2b. [1 mark] Markscheme Al Note: Accept any form of concavity for Note: Do not penalize unmarked zeros if given in part (a). Note: Zeros written on diagram can be used to allow the mark in part (a) to be awarded retrospectively. [1 mark] Examiners report Many candidates stated the two zeros off correctly but the graph off was often incorrectly drawn. In (c), many candidates failed to realise that integration by parts had to be used twice here and even those who did that often made algebraic errors, usually due to the frequent changes of sign. 2c. [7 marks] Markscheme attempt at integration by parts Ml EITHER I = re— x cosæda,• = —e—x cosædæ — fe —z sin Al I — —e-x cos:rdæ — [—e—x sin x + fe-z cosxd;r] Al •I — (sinc — cost) + C Al Note: Do not penalize absence of C. OR I f e-c cos;rdc = sin c + f sin cdx Al — f coscdx Al I = e-z sinc — e-t cosx I — (sinc — COST) + C Al Note: Do not penalize absence of C. THEN sm — cos 6
  7. — (sin a — cos ratio ofA:B is [7 marks] Examiners report Al Many candidates stated the two zeros off correctly but the graph off was often incorrectly drawn. In (c), many candidates failed to realise that integration by parts had to be used twice here and even those who did that often made algebraic errors, usually due to the frequent changes of sign. 3.[7 marks] Markscheme = sin t, d a: = cost dt sing t cos t dt I —siu2 t Ml = f sin3tdt (Al) = f sin2tsintdt = (I — cos2t) sin t dt MIAI = f sintdt— f cos2tsintdt — cost -+- -h C AMI [7 marks] Examiners report Just a few candidates got full marks in this question. Substitution was usually incorrectly done and lead to wrong results. A cosine term in the denominator was a popular error. Candidates often chose unhelpful trigonometric identities and attempted integration by parts. Results such as f sin3tdt= + C were often seen along with other misconceptions concerning the manipulation/simplification of integrals were also noticed. Some candidates unsatisfactorily attempted to use arcsmx . However, there were some good solutions involving an expression for the cube of Sin t in terms of Sin t and Sin 3t . Very few candidates re-expressed their final result in terms of x. 4.[6 marks] Markscheme u = Iny du Y dYA1(A1) Let tan(ln V) dy f tan udu Al — = —Inlcosul + CAI EITHER tan(ln V) —In + c AMI OR tan(ln y) dy In Isec(ln + c AIAI 7
  8. [6 marks] Examiners report Many candidates obtained the first three marks, but then attempted various methods unsuccessfully. Quite a few candidates attempted integration by parts rather than substitution. The candidates who successfully integrated the expression often failed to put the absolute value sign in the final answer. 5.[6 marks] Markscheme Using integration by parts (Ml) — COS 2x (Al) — cos 2m) dc Al Note: Award the AIAI above if the limits are not included. T = — 274 Al sin 2:r) — T Al sin 2ædx — AGNO 8 Note: Allow FT on the last two Al marks if the expressions are the negative of the correct ones. [6 marks] Examiners report This uestion was reasonably well done, with few candidates making the inappropriate choice of u and dz . The main source of a loss of marks was in finding v by integration. A few candidates used the double angle formula for sine, with poor results. 6.[5 marks] Markscheme Recognition of integration by parts MI AMI — Inx] — Inx — x21nxdx = _ 9 [5 marks] Examiners report Al Most candidates recognised that a method of integration by parts was appropriate for this question. However, although a good number of correct answers were seen, a number of candidates made algebraic errors in the process. A number of students were also unable to correctly substitute the limits. 7a. [9 marks] Markscheme 8
  9. A2 Note: Award Al for correct sin a , Al for correct Sin . Note: Award AIAO for two correct shapes with and/or 1 missing. Note: Condone graph outside the domain. (ii) sin2z• = sinc 2 sin C cost — sin c = OMI sinc(2cosæ — 1) = 0 AMI NINI — (sin & — sin Ml (iii) area Note: Award Ml for an integral that contains limits, not necessarily correct, with Sin C and sin 2æ subtracted in either order. — cos 2æ + cos — O Al — COS O + COS O) — COS + COS — (Ml) 2 [9 marks] Examiners report A significant number of candidates did not seem to have the time required to attempt this question satisfactorily. Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes. Part (ii) was done well by most the most common error being to divide the equation by sin a and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution. The final part was done well by many candidates. Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully. Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well. 7b. [8 marks] Markscheme 4sin2 x 8 sin 9cos0d9 4— 4sin2e MIAMI Note: Award MI for substitution and reasonable attempt at finding expression for dx in terms ofd9 , first Al for correct limits, second Al for correct substitution for dx . 8sin20d9 Al 4 - 4cos29d9M1 9
  10. — [40 Al -(2+-2sinZ) O [8 marks] Examiners report A significant number of candidates did not seem to have the time required to attempt this question satisfactorily. Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes. Part (ii) was done well by most the most common error being to divide the equation by Sin a: and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution. The final part was done well by many candidates. Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully. Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well. 7c. [8 marks] Markscheme Ml from the diagram above = = ab — fob f -1 (Y)dy RI the shaded area = 0b — fob f -1 (c)dCAG f(æ) = arcsin f f -l (x) = 4sinx Al (ii) arcsin dc — — 4 sin MIAIAI Note: Award Al for the limit 7 seen anywhere, Al for all else correct. — — Al Note: Award no marks for methods using integration by parts. [8 marks] Examiners report 10
  11. A significant number of candidates did not seem to have the time required to attempt this question satisfactorily. Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes. Part (ii) was done well by most the most common error being to divide the equation by sin C and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution. The final part was done well by many candidates. Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully. Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well. 8a. [8 marks] Markscheme + _en(ä) prove that for n = 1 LHS = 1, 3 1 so true for n 1 RI assume true for n = k Ml 1+2 2k-l now for n = k +1 LHS 1 +2 (+) MIAI 2(k+2) 2k 2k (or equivalent) Al (k+1)+2 2k ) Al (accept Therefore if it is true for n = k it is true for n = k + 1. It has been shown to be true for n = 1 so it is true for all n (€ 7-'+). RI Note: To obtain the final R mark, a reasonable attempt at induction must have been made. [8 marks] Examiners report Part A: Given that this question is at the easier end ofthe 'proof by induction' spectrum, it was disappointing that so many candidates failed to score full marks. The n = 1 case was generally well done. The whole point of the method is that it involves logic, so 'let n = k' or 'put n = k', instead of 'assume to be true for n = k', gains no marks. The algebraic steps need to be more convincing than some candidates were able to show. It is astonishing that the RI mark for the final statement was so often not awarded. 8b. [1 7 marks] Markscheme METHOD 1 sincdæ = — cosæe2æ + f2e2æ — cosæe2Z + 2e2x sin c — sin adz AIAI 5 sinxdæ = —cos xe2a + 2e2a sintM1 Je2z sincdæ = L e2r (2 sin x — cost) + C METHOD 2 2 MIAIAI 11
  12. sin re & cost e: — f sinx—dæ 4 AIAI sin xdc = e2z sin cos Ml J e2a sin = L e2x (2 sinc — cos a:) + C [6 marks] sin xdx MIAI I e2X(2 sinc — arcsin y = Al when y = sin (L — cost) 4- -k) Al [5 marks] Al P is (1.16, 0) Al Note: Award Al for 1.16 seen anywhere, Al for complete sketch. Note: Allow FT on their answer from (b) v = f01162 ry2dXMIA1 (ii) = 1.05M Note: Allow FT on their answers from (b) and (c)(i). [6 marks] Examiners report Part B: Part (a) was often well done, although some faltered after the first integration. Part (b) was also generally well done, although there were some errors with the constant of integration. In (c) the graph was often attempted, but errors in (b) usually led to manifestly incorrect plots. Many attempted the volume of integration and some obtained the correct value. 9a. [6 marks] Markscheme 10 = for sin :rdæ Ml 10 sin Note: Award MI for Attempt at integration by parts, even if inappropriate modulus signs are present. MI — —[e-x cost]å — for e-c cos:rda or —[e—x sin — cosxda: Al —[e-x — [e-t sin Clor — sinxdx —[e—x sinc + — for Al — —[e-x cosa]å — [e —x sinæ]å — 10 or —[e-x sinc + e-z cosæ]o — 10 Ml Note: Do not penalise absence of limits at this stage 10 +1 — 10 Al 12
  13. 10 Note: If modulus signs are used around cos x , award no accuracy marks but do not penalise modulus signs around sin x . [6 marks] Examiners report Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in O which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen. 9b. [4 marks] Markscheme (n+l)r sin Attempt to use the substitution Y = — MI (putting Y = — dy = dc and (n + 1)7T] [0, sin(y + nar)ldy Al e-YI sin(y + nn)ldYA1 for e-y sin ydy Al = e-nTIo AG [4 marks] Examiners report Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in O which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen. 9c. [5 marks] Markscheme Ml (Al) the term is an infinite geometric series with common ratio e therefore 10 sin cldc — — I-e-t (Al) 2(er-1)) Al — 2(1-e-,) [5 marks] Examiners report Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in O which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen. 10a. [6 marks] Markscheme EITHER let u = tan x; d u = sec2xdx (Ml) consideration of change of limits (MI) dc = du 4 tan (Al) Note: Do not penalize lack of limits. 13
  14. 30 2 1 Al 2 OR 2 [6 marks] 3ü-3 2 AMINO 3(tan x) M2A2 .3ü-3 2 J AMINO Examiners report Quite a variety of methods were successfully employed to solve part (a). 10b. [3 marks] Markscheme J tan3xdæ = f tan a(sec2æ — l)dc Ml = (tanz x sec2æ — tanæ)dæ — —In Isecæl + CAIAI Note: Do not penalize the absence of absolute value or C. [3 marks] Examiners report Many candidates did not attempt part (b). 1 Ia. [7 marks] Markscheme letc = 2sin0M1 dc = 2cos9d9A1 4 cos29d9) 1 = 2cose x 2cos0d9 AIAI Note: Award Al for limits and Al for expression. = (1 + cos20)d9A1 = 2[0+ O Al Al [7 marks] Examiners report 11b. [5 marks] Markscheme I = arcsin — x = arcsin x + f -IAI — 12 [5 marks] MIAMI 0.5 O Al Examiners report Ilc. [7 marks] Markscheme sec29dO, [0, f] [0 1] , Al(A1) 14
  15. MI(AI) 1 dt 0 3+t2 Al — 60 Al [7 marks] Examiners report 12a. [2 marks] Markscheme — 0.5)2 + 4 AIAI Note: Al for two correct parameters, A2 for all three correct. [2 marks] Examiners report This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) - (e). 12b. [3 marks] Markscheme 0.5 O (allow "0.5 to the right") Al translation stretch parallel toy-axis, scale factor 4 (allow vertical stretch or similar) Al 4 (allow "4 up") Al translation Note: All transformations must state magnitude and direction. Note: First two transformations can be in either order. (05) It could be a stretch followed by a single translation of 4 . If the vertical translation is before 1 the stretch it is [3 marks] Examiners report This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) - (e) but the following points caused some difficulties. (b) Exam technique would have helped those candidates who could not get part (a) correct as any solution of the form given in the question could have led to full marks in part (b). Several candidates obtained expressions which were not of this form in (a) and so were unable to receive any marks in (b) Many missed the fact that if a vertical translation is performed before the vertical stretch it has a different magnitude to if it is done afterwards. Though on this occasion the markscheme was fairly flexible in the words it allowed to be used by candidates to describe the transformations it would be less risky to use the correct expressions. 12c. [2 marks] Markscheme 15
  16. general shape including asymptote and single maximum in first quadrant), Al S or maximum 2 shown Al intercept [2 marks] Examiners report This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) - (e) but the following points caused some difficulties. (c) Generally the sketches were poor. The general rule for all sketch questions should be that any asymptotes or intercepts should be clearly labelled. Sketches do not need to be done on graph paper, but a ruler should be used, particularly when asymptotes are involved. 12d. [2 marks] Markscheme O < < a AMI 7, Al for() Note: Al for [2 marks] Examiners report This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) - (e). 12e. [3 marks] Markscheme letu = I (Or du = dc) Al 4x2 —4x+5 +4 Al I —I— du 4 u2+1 AG Note: If following through an incorrect answer to part (a), do not award final Al mark. [3 marks] Examiners report This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) - (e) but the following points caused some difficulties. (e) and (f) were well done up to the final part of (f), in which candidates did not realise they needed to use the compound angle formula. 12f. [7 marks] Markscheme dc 1 4r2 —4x4-5 I —L— du — JO.5 Al Note: Al for correct change of limits. Award also if they do not change limits but go back to x values when substituting the limit (even if there is an error in the integral). 16
  17. (MI) (arctan(3) arctan G)) Al let the integral = I tan 41 tan (arctan(3) arctan ( ) ) Ml 3-0.5 _ 2.5 _ — (MI)AI 1+3*0.5 41 I = Tä AMG [7 marks] Examiners report This question covered many syllabus areas, completing the square, transformations of graphs, range, integration by substitution and compound angle formulae. There were many good solutions to parts (a) - (e) but the following points caused some difficulties. (e) and (f) were well done up to the final part of (f), in which candidates did not realise they needed to use the compound angle formula. 13.[7 marks] Markscheme — du = du Al r —L— — — JIF Ml du AIMIAI Note: Award Al for correct integrand and MIAI for correct limits. — —L— du (upon interchanging the two limits) AG arctan x? = arctan arctan a 4 arctan Al arctan a --9 arctan [7 marks] Examiners report This question was successfully answered by few candidates. Both parts of the question prescribed the approach which was required - "use the substitution" and "hence". Many candidates ignored these. The majority of the candidates failed to use substitution properly to change the integration variables and in many cases the limits were fudged. The logic of part (b) was missing in many cases. 14.[6 marks] Markscheme EITHER attempt at integration by substitution (Ml) u = t -i- l, du = dt, the integral becomes using (u 1) In udtLA1 then using integration by parts Ml (u — 1) Inudu = — Jl 2 4 = 7 (accept 0.25) Al OR attempt to integrate by parts (Ml) correct choice of variables to integrate and differentiate Ml t In(t + l)dt — Al 22 In(t + Tot—I + Al 17
  18. t + In(t + 1)] O (Al) — 7 (accept 0.25) Al [6 marks] Examiners report Again very few candidates gained full marks on this question. The most common approach was to begin by integrating by parts, which was done correctly, but very few candidates then knew how to integrate t+l . Those who began with a substitution often made more progress. Again a number of candidates were let down by their inability to simplify appropriately. 15a. [4 marks] Markscheme J tsec2cdx = x tan c — f 1 x tan adz MIAI x tan x + In Icosxl (+c) ctanx —In Isecxl (+c)) MIAI [4 marks] Examiners report In part (a), a large number of candidates were able to use integration by parts correctly but were unable to use integration by substitution to then find the indefinite integral of tan x. In part (b), a large number of candidates attempted to solve the equation without direct use of a GDC's numerical solve command. Some candidates stated more than one solution for m and some specified m correct to two significant figures only. 15b. [2 marks] Markscheme m tan m + In(cos m) 0.5 (Ml) attempting to solve an appropriate equation eg m = 0.822 Al Note: Award Al if m = 0.822 is specified with other positive solutions. [2 marks] Examiners report In part (a), a large number of candidates were able to use integration by parts correctly but were unable to use integration by substitution to then find the indefinite integral of tan x. In part (b), a large number of candidates attempted to solve the equation without direct use of a GDC's numerical solve command. Some candidates stated more than one solution for m and some specified m correct to two significant figures only. 16.[5 marks] Markscheme — f (I—COS 9)' dB = In (1 —cos9)+C (MI)AIAI I —cos O Note: Award Al for In (1 — cos 9) and Al for C. — [In (1 — (b) Ml 1 — cosa = eä a = arccos (1 — VIE)) or 2.28AIN2 [5 marks] Examiners report Generally well answered, although many students did not include the constant of integration. 17.[8 marks] Markscheme a: = 2sin0 dc = 2cos9d9A1 = 4 — 4sin20 x 2 cos MIAI = f2cos9 x2cosOd9 = 4 f cos2 9dO now f cos29dO f ( cos 29 + B)dO MIAI 18
  19. 2 Al so original integral = + 29 2 sin 9 cos0 20 4- + 2 arcsin — + 2 arcsin (f) + C AMI Note: Do not penalise omission Ofc. [8 marks] Examiners report For many candidates this was an all or nothing question. Examiners were surprised at the number of candidates who were unable to change the variable in the integral using the given substitution. Another stumbling block, for some candidates, was a lack of care with the application of the trigonometric version of Pythagoras' Theorem to reduce the integrand to a multiple of COS2 9 However, candidates who obtained the latter were generally successful in completing the question. 18a. [5 marks] Markscheme METHOD 1 sketch showing where the lines cross or zeros of Y = •T(æ + 2) 6 — (MI) c —l and x — the solution is —3 < < —l or X > O AIAI Note: Do not award either final Al mark if strict inequalities are not given. METHOD 2 separating into two cases C > O and X < O (Ml) if C > O then (X + 2) 6 > I always true (Ml) if X < Othen + 2) 6 < I —3 —l (Ml) so the solution is —3 < C < —l or X > OAIAI Note: Do not award either final Al mark if strict inequalities are not given. METHOD 3 f(æ) = -F 12x6 + 60æ5 + 160c4 + 240æ3 + 192æ2 + 64t (Al) solutions tox7 + 12æ6 + 60x5 + 160x4 + 240x3 + 192c2 + 63x = Oare (Ml) O, — —l and c = —3 (Al) so the solution is —3 < —l or X > OAIAI Note: Do not award either final Al mark if strict inequalities are not given. METHOD 4 f (X) = when -+- = either X O or (C 4- 2) if(c = I then X ±lsoc the solution is —3 < —l or X > O AIAI Note: Do not award either final Al mark if strict inequalities are not given. [5 marks] Examiners report 18b. [5 marks] Markscheme METHOD 1 (by substitution) substituting U = 4- 2 (Ml) du = dr J (u — 2)u6du MIAI 81 us — (A-c) (Al) Al 19
  20. METHOD 2 (by parts) du _ 1, tv = (x +2)6 } (c +2)7 f + 2)6dæ = + 2) 7 — f (r + 2)7dTM1 AIM METHOD 3 (by expansion) = (x 7 + 12x6 + -F 160x4 + 240æ3 + 192æ2 + 64æ) dc MIAI = + + 10æ6 + 32c5 + + 64c3 + 32æ2(+c) MIA2 Note: Award MIAI if at least four terms are correct. [5 marks] Examiners report 19.[7 marks] Markscheme a sec 0 d9 = asec9tan0 (Al) new limits: = avo 0 = and c = 2a a Sec 9 tan de a3 seca 9 Ml Al = i (Al) cos20 using (cos 29 I) Ml 7 or equivalent Al 27—1— 4a3 2 2 or equivalent Al [7 marks] Examiners report 20a. [2 marks] Markscheme Al for correct sha e and correct domain (1.41, 0.0884) v/ä, 0) 16 Al [2 marks] Examiners report 20b. [4 marks] Markscheme EITHER Al dt Al 20
  21. OR du THEN 12+t4 Al 1 r du dt— 2 J Ml 1 arctan ( (A-c) Ml _ 1 arctan (e) ( + c) or equivalent Al [4 marks] Examiners report 20c. [3 marks] Markscheme J 6 dt 0 12+t4 (Ml) 6 arctan (60) Note: Accept 12 or equivalent. [3 marks] Examiners report 20d. [3 marks] Markscheme a — arcsin ( VG) X a = arcsin (VOÄ) x a = 0.536 (ms-2) Al [3 marks] Examiners report 21.[7 marks] Markscheme EITHER 2sec u Al 2 seca udu 2 sec2 udu ( f 4tan2ux2 sec u u tan u •V I OR u — arctan du 2 4tan2 u -+- 4du 2 Sec J 2x4tan2u Al THEN _ I sec udu — J tan2u 2 sec2 udu 4tan2 u A I 21
  22. = f cosec u cot udu L du cosecu(+C) — u or an appropriate tri onometric identity Ml use of either sinu — or cosec U either [7 marks] Examiners report +4 (or equivalent) Al Most candidates found this a challenging question. A large majority of candidates were able to change variable from x to u but were not able to make any further progress. 22

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