Cambridge A Level Mathematics - Further 9231 — 2012 May/June Paper 1 · Variant 3

*0973024008* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 May/June 2012 3 hours Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of a calculator is expected, where appropriate. Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 4 printed pages. JC12 06_9231_13/RP © UCLES 2012 [Turn over

2 1 Find the sum of the first n terms of the series 1 1 × 3 + 1 2 × 4 + 1 3 × 5 + . . . and deduce the sum to infinity. [5] 2 For the sequence u1, u2, u3, . . . , it is given that u1 = 1 and ur+1 = 3ur −2 4 for all r. Prove by mathematical induction that un = 43 4n −2, for all positive integers n. [5] 3 The curve C has equation xy + (x + y)3 = 1. Show that dy dx = −3 4 at the point A (1, 0) on C. [3] Find the value of d2y dx2 at A. [5] 4 Let In = ã e 1 x2(ln x)n dx, for n ≥0. Show that, for all n ≥1, In = 1 3e3 −1 3nIn−1. [4] Find the exact value of I3. [4] 5 The matrix A has an eigenvalue λ with corresponding eigenvector e. Prove that the matrix (A + kI), where k is a real constant and I is the identity matrix, has an eigenvalue (λ + k) with corresponding eigenvector e. [2] The matrix B is given by B =  2 2 −3 2 2 3 −3 3 3 . Two of the eigenvalues of B are −3 and 4. Find corresponding eigenvectors. [3] Given that 1 −1 −2 ! is an eigenvector of B, find the corresponding eigenvalue. [1] Hence find the eigenvalues of C, where C =  −1 2 −3 2 −1 3 −3 3 0 , and state corresponding eigenvectors. [3] © UCLES 2012 9231/13/M/J/12

3 6 The curve C has equation y = x2 x −2. Find the equations of the asymptotes of C. [3] Find the coordinates of the turning points on C. [3] Draw a sketch of C. [3] 7 Expand ß + 1 ß 4 ß −1 ß 2 and, by substituting ß = cos θ + i sin θ, find integers p, q, r, s such that 64 sin2θ cos4θ = p + q cos 2θ + r cos 4θ + s cos 6θ. [6] Using the substitution x = 2 cos θ, show that ã 2 1 x4 √(4 −x2) dx = 4 3π + √3. [4] 8 The cubic equation x3 −x2 −3x −10 = 0 has roots α, β, γ. (i) Let u = −α + β + γ . Show that u + 2α = 1, and hence find a cubic equation having roots −α + β + γ, α −β + γ, α + β −γ. [5] (ii) State the value of αβγ and hence find a cubic equation having roots 1 βγ , 1 γ α , 1 αβ . [5] 9 The plane Π1 has parametric equation r = 2i −3j + k + λ(i −2j −k) + µ(i + 2j −2k). Find a cartesian equation of Π1. [4] The plane Π2 has cartesian equation 3x −2y −3ß = 4. Find the acute angle between Π1 and Π2. [3] Find a vector equation of the line of intersection of Π1 and Π2. [4] 10 Find the set of values of a for which the system of equations x −2y −2ß = −7, 2x + (a −9)y −10ß = −11, 3x −6y + 2aß = −29, has a unique solution. [4] Show that the system has no solution in the case a = −3. [2] Given that a = 5, (i) show that the number of solutions is infinite, [2] (ii) find the solution for which x + y + ß = 2. [3] © UCLES 2012 9231/13/M/J/12 [Turn over

4 11 Answer only one of the following two alternatives. EITHER The curve C has cartesian equation (x2 + y2)2 = a2(x2 −y2), where a is a positive constant. Show that C has polar equation r2 = a2 cos 2θ. [2] Sketch C for −π < θ ≤π. [2] Find the area of the sector between θ = −1 4π and θ = 1 4π. [3] Find the polar coordinates of all points of C where the tangent is parallel to the initial line. [7] OR Show that the substitution y = xß reduces the differential equation 1 x d2y dx2 + 6 x −2 x2 dy dx + 9 x −6 x2 + 2 x3 y = 169 sin 2x to the differential equation d2ß dx2 + 6dß dx + 9ß = 169 sin 2x. [4] Find the particular solution for y in terms of x, given that when x = 0, ß = −10 and dß dx = 5. [10] Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2012 9231/13/M/J/12

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2012 question paper for the guidance of teachers 9231 FURTHER MATHEMATICS 9231/13 Paper 1, maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Mark for a correct result or statement independent of method marks. • When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through √" marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 1 Finds partial fractions. Use method of differences. Obtains results.   + − = + 2 1 1 2 1 ) 2 ( 1 r r r r = + ∑ = n r r r 1 ) 2 ( 1          − +     − + +     + − − +     + − 3 1 1 4 1 2 1 ... 1 1 1 1 2 1 1 2 1 n n n n   + − + − = 2 1 1 1 2 3 2 1 n n (acf) 4 3 = ⇒ ∞ S M1A1 M1 A1A1√ 5 [5] 2 (States proposition.) Proves base case. States Inductive hypothesis. Proves inductive step. States conclusion. ) 2 4 3 4 : P ( −       = n n n u Let n = 1 4 × 4 3 – 2 = 3 –2 = 1⇒P1 true. Assume Pk is true for some k. 4 2 6 4 3 . 4 3 .4 4 2 2 4 3 4 3 1 + −       = −       −       = + k k k u 1 1 P P 2 4 3 .4 + + ⇒ ∴ −       = k k k ∴ By PMI Pn is true ∀ positive integers. B1 B1 M1 A1 A1 5 [5] 3 0 d d 1 ) ( 3 d d 2 =      + + + + x y y x x y x y 0 3 3 0 = ′ + + ′ + y y 4 3 − = ′ ⇒y (AG) 0 d d ) ( 3 d d 1 ) ( 6 d d d d d d 2 2 2 2 2 2 = + +      + + + + + x y y x x y y x x y x x y x y 0 3 16 1 6 4 3 4 3 = ′′ + × + ′′ + − − y y 32 9 = ′′ ⇒y N.B. Mark similarly if expression expanded before differentiating. B1B1 B1 B1 B1B1 M1 A1 3 5 [8]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 4 Integrates by parts. Obtains reduction formula. Finds I0 ( or I1) and uses reduction formula. (M1A1 for I1 if found immediately.) Obtains I2 . Obtains I3 . ∫ = e 1 2 d ln x x) ( x I n n x x x x n x x n e n d 3 1 ) (ln 3 ) (ln 3 e 1 1 1 3 ∫ ⋅ ⋅ −       = − 1 3 3 3 e − − = nI n (AG) ∫ − =       = = e 1 3 e 1 3 2 0 3 1 e 3 d x x x I 9 1 e 2 3 1 e 3 1 3 e 3 3 3 1 + =       − − = ⇒I 27 2 e 5 9 1 e 2 3 2 3 e 3 3 3 2 − =       + − = ⇒I 27 2 e 4 27 2 e 5 3 e 3 3 3 3 + =       − − = ⇒I M1A1 A1 A1 B1 M1 A1 A1 4 4 [8] 5 Proves initial result. Finds eigenvectors corresponding to given eigenvalues. Finds corresponding eigenvalue. Recognises the result proved initially. Gives eigenvalues and matches eigenvectors. (A + kI)e = Ae + kIe = λe + ke = (λ + k)e ∴ (A + kI) has eigenvalue (λ + k) with corresponding eigenvector e. Eigenvalues are –3 and 4 (given). Corresponding eigenvectors are           − 1 1 1 ,           0 1 1 Third eigenvalue is 6. C = B – 3I (Stated or implied.) Eigenvalues: –6, 1, 3 . Corresponding eigenvectors are:           − 1 1 1 ,           0 1 1 ,           − − 2 1 1 . (OE) (For ‘non hence’ method, using characteristic equation, award B1 rather than M1A1 for eigenvalues, followed by B1 for eigenvectors.) M1A1 M1A1 A1 B1 M1 A1 A1√ 2 3 1 3 [2] [3] [1] [3]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 6 States vertical asymptote. Finds oblique asymptote. Differentiates and equates to zero. Finds x coordinates. States coordinates of turning points. Deduct at most 1 mark for poor forms at infinity. Vertical asymptote is x = 2. 2 4 2 − + + = x x y Oblique asymptote is y = x + 2. 4 ) 2 ( 0 ) 2 ( 4 1 2 2 = − ⇒ = − − = ′ x x y x = 0, 4. Turning points are (0,0) and (4,8) Axes and both asymptotes correct. Upper branch correct. Lower branch correct. B1 M1 A1 M1 A1 A1 B1 B1 B1 3 3 3 [9] 7 Complete strategy and getting halfway. Fully correct. Grouping. Correct LHS and RHS. Sets up substitution. Uses result obtained above. Obtains result.       + −       + + =      −      + 4 4 2 2 2 4 1 2 1 2 1 1 z z z z z z z z 6 4 2 2 4 6 1 2 1 4 2 z z z z z z + + − − − + = 4 1 1 2 1 2 2 4 4 6 6 −       + −       + +       + = z z z z z z 4 2 cos 2 4 cos 4 6 cos 2 sin 4 cos 16 2 4 − − + = − ⋅ θ θ θ θ θ θ θ θ θ θ 6 cos 2 4 cos 4 2 cos 2 4 sin cos 64 2 4 − − + = θ θ θ sin 2 d d cos 2 − = = x x 0 2 3 1 = ⇒ = = ⇒ = θ π θ x x ∫ ∫ = − 0 3 3 0 2 4 2 4 d sin cos 64 d sin 4. cos 16 π π θ θ θ θ θ θ (LR – allow use of 2π, if seen.). ∫ − − + = 3 0 d ) 6 cos 2 4 cos 4 2 cos 2 4 ( π θ θ θ θ 3 0 6 sin 3 1 4 sin 2 sin 4 π θ θ θ θ     − − + = 3 3 4 2 3 2 3 3 4 + =       + + = π π (AG) M1A1 A1 M1 A1(L) A1(R) M1 M1 A1 A1 6 4 [10]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 8 (i) (ii) Deduces initial result. Substitutes into cubic equation. Deduces new cubic equation. Deduces initial result. Substitutes into cubic equation. Deduces new cubic equation. 1 2 = + + = + ⇒ + + − = γ β α α γ β α u u      − = ⇒ 2 1 u α 0 10 2 1 3 2 1 2 1 2 3 = −      − −      − −      − ⇒ u u u ⇒… 0 93 13 2 3 = + − − ⇒ u u u 10 = αβγ v v v 10 10 1 1 1 = ⇒ = = ⇒ = ⇒ α αβγ α βγ ( ) ( ) ( ) 0 10 10 3 10 10 2 3 = − − − v v v 0 1 3 10 100 2 3 = − − − ⇒ v v v M1A1 M1 A1 A1 B1 M1A1 M1 A1 5 5 [10] 8 (i) (ii) Alternatively: For final 3 marks in (i): Award M1 for an attempt at formulae for all three coefficients. A1 for any two correct. A1 for completion For final 4 marks in (ii): Award M1 for an attempt at formulae for all three coefficients. A1 for any one correct. A1 for a second one correct. A1 for completion. Let equation be u3 + bu2 + cu + d = 0. –b = ∑α = 1⇒b = –1 c = 4∑αβ – (∑α)2 = 4 × (–3) – 12 = –13 ( ) αβγ α αβ α 8 4 3 − − = − ∑ ∑ ∑ d = 4 × 1 × (–3) – 13 – 8 × 10 = –93 So u3 – u2 – 13u + 93 = 0 Let equation be v3 + bv2 + cv + d = 0. 10 1 10 1 − = ⇒ = Σ = − b b αβγ α 100 3 10 3 ) ( 2 2 − = − = Σ = αβγ αβ c 100 1 10 1 ) ( 1 2 2 − = ⇒ = = − d d αβγ So 0 100 1 100 3 10 1 2 3 = − − − v v v or 100v3 – 10v2 – 3v – 1 = 0. (5) (5) [10]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 9 Finds normal vector to plane. Uses known point to find constant term. Angle between normals is equal to angle between planes. Solve plane equation simultaneously. (Note: may find direction from vector product and use with one point.)           = − − − 4 1 6 2 2 1 1 2 1 k j i or µ λ µ λ µ λ 2 1 2 2 3 2 − − = + − − = + + = z y x Plane equation is 6x + y + 4z = constant. Substitute (2, –3,1)⇒12 – 3 + 4 = 13. Or eliminate λ and µ. ⇒Π1 : 6x + y + 4z = 13 22 53 4 3 2 3 4 1 6 ) k 3 j 2 i3 ).( k 4 j i6 ( cos 2 2 2 2 2 2 = + + + + − − + + = θ = ⇒θ 83.3º or 1.45 rad. 6x + y + 4z = 13 and 3x – 2y – 3z = 4 Obtains e.g. y + 2z = 1 and 3x + z = 6 Or two of (0,–11,6) , (11/6,0,1/2) , (2,1,0)           − − +           =           3 6 1 0 1 2 t z y x (OE) Alternatively: Direction of line from vector product. Finds a point on line. States equation of line. M1A1 M1 A1 M1A1 A1 M1 A1A1 A1 (M1A1) (A1) (A1) 4 3 4 [11]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 10 (i) (ii) Reduces augmented matrix Obtains set of values for a, giving unique solutions. Case of no solutions. Case of infinite solutions. Obtains particular solution.           − − + − − − −           − − − − − − − − 8 3 7 6 2 6 2 0 5 2 0 0 1 ~ 29 11 7 2 10 2 6 9 2 3 2 1 a a a a Unique solution for all real a except a = – 3 or 5 Alternatively for first two marks: 0 ) 3 )( 5 ( 0 2 6 3 10 ) 9 ( 2 2 2 1 ≠ + − ⇒ ≠ − − − − − a a a a a = –3 : 2a + 6 = 0 ⇒ − = ⇒ 8 0z no solutions (AG) 2 1 : 5 − = ⇒ = z a and x – 2y = – 8 (*) ∴infinite number of solutions (AG) 2 1 − = z and x + y + z = 2⇒x + y = 2 5 Solving simultaneously with (*) gives x = –1 y = 2 7 z = – 2 1 M1A1 A1A1 M1A1√ M1A1√ B1 M1A1 4 2 2 3 [11]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 11 EITHER Uses x2 + y2 = r2, x = r cos θ and y = r sin θ. One mark for each loop, or half of whole curve. Uses sector area formula. S.C. Omission of ½ factor, but correct integration gets B1. Differentiates. Puts 0 = ′y . Obtains coordinates. (x2 + y2)2 = a2(x2 – y2)       − = ⇒ 2 2 2 2 2 2 r y r x a r = a2 (cos2 θ – sin2 θ) = a2 cos 2θ (AG) Sketches C. Area = ∫ ∫ −       = 4 4 4 0 2 2 d 2 cos d 2 cos 2 1 π π π θ θ θ θ a a 4 0 2 2 2 sin π θ     = a 2 2 a = ) 2 2 ( ) 2 2 )( ( 2 2 2 2 y y x a y y x y x ′ − = ′ + + x a y x x y 2 2 2 ) ( 2 0 = + ⇒ = ′ ) 0 ( 2 2 2 2 ≥ = ⇒ = ⇒ r a r a r 2 1 2 cos = ⇒ θ 6 π θ ± = ⇒ , 6 5π ± i.e.       ± 6 , 2 π a and       ± 6 5 , 2 π a M1 A1 B2,1,0 M1 A1A1 B1B1 M1 A1 M1 A1A1 2 2 3 7 [14]

Page 11 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total Alternatively for last 7 marks: Obtains condition for tangent parallel to initial line. Differentiates equation of C. Forms equation. Solves for tanθ, and θ. Writes coordinates of points. 0 sin d d cos 0 d d 0 d d = + ⇒ = ⇒ = θ θ θ θ r r y x y θ θ θ θ 2 sin d d 2 sin 2 d d 2 2 2 r a r a r r − = ⇒ − = 0 sin 2 sin cos 2 = − ∴ θ θ θ r a r θ θ θ θ sin 2 sin cos 2 cos 2 2 a a = ∴ θ θ tan 2 tan 1 = ∴ 1 3 1 2 2 1 2 2 2 2 = ⇒ − = ⇒ = − ∴ t t t t t t 3 1 ± = ∴t 6 5 , 6 π π θ ± ± = ∴       ± 6 , 2 π a ,       ± 6 5 , 2 π a (Award final A1 if r found but result not written out.) (M1A1) (A1) (M1) (M1) (A1) (A1) (7) [14]

Page 12 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 13 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 11 OR Differentiates once. Differentiates again. Substitutes. Obtains result Finds CF. Finds PI. Finds GS. Evaluates coefficients from initial conditions. Finds particular solution. x z x z x y d d d d + = 2 2 2 2 d d d d 2 d d x z x x z x y + =       + − + − +       − + + 2 2 2 2 2 6 9 2 6 d d 2 6 2 d d x z x z z x z x z x z x x x z x 2 sin 169 = x z x z x z 2 sin 169 9 d d 6 2 d 2 d = + + ⇒ (AG) (Mark similarly if substitution is rearranged to x y z = .) 3 0 ) 3 ( 9 6 2 2 − = ⇒ = + = + + m m m m CF: Ae–3x + Bxe–3x PI: y = p sin 2x + q cos 2x y′ = 2p cos 2x – 2q sin 2x y″ = –4p sin 2x – 4q cos 2x 169 12 5 = − q p 0 5 12 = + q p ⇒ p = 5 and q = – 12 GS: z = Ae–3x + Bxe–3x + 5 sin 2x – 12 cos 2x 2 12 10 = ⇒ − = − A A z′ = –6e–3x + Be–3x – 3Bxe–3x + 10 cos 2x + 24 sin 2x 1 10 6 5 = ⇒ + + − = B B x x x z x x 2 cos 12 2 sin 5 e e 2 3 3 − + + = − − x x x x x x y x x 2 cos 12 2 sin 5 e e 2 3 2 3 − + + = ∴ − − B1 B1 M1 A1 M1 A1 M1 M1 A1 A1 B1 M1 A1 A1 4 10 [14]