Cambridge A Level Mathematics - Further 9231 — 2012 May/June Paper 1 · Variant 2

*9088454742* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/12 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_12/FP © UCLES 2012 [Turn over

2 1 The roots of the cubic equation x3 −7x2 + 2x −3 = 0 are α, β, γ. Find the values of (i) α2 + β2 + γ 2, [2] (ii) α3 + β3 + γ 3. [3] 2 Prove, by mathematical induction, that, for integers n ≥2, 4n > 2n + 3n. [5] 3 Given that f(r) = 1 (r + 1)(r + 2), show that f(r −1) −f(r) = 2 r(r + 1)(r + 2). [2] Hence find n ∑ r=1 1 r(r + 1)(r + 2). [3] Deduce the value of ∞ ∑ r=1 1 r(r + 1)(r + 2). [1] 4 The curve C has polar equation r = 2 + 2 cos θ, for 0 ≤θ ≤π. Sketch the graph of C. [2] Find the area of the region R enclosed by C and the initial line. [4] The half-line θ = 1 5π divides R into two parts. Find the area of each part, correct to 3 decimal places. [3] 5 A matrix A has eigenvalues −1, 1 and 2, with corresponding eigenvectors 0 1 −2 !, −1 −1 3 ! and 2 −3 5 !, respectively. Find A. [9] 6 Write down the values of θ, in the interval 0 ≤θ < 2π, for which cos θ + i sin θ is a fifth root of unity. [2] By writing the equation (ß + 1)5 = ß5 in the form ß + 1 ß  5 = 1, show that its roots are −1 2n1 + i cotkπ 5 o, k = 1, 2, 3, 4. [7] © UCLES 2012 9231/12/M/J/12

3 7 The linear transformations T1 : >4 →>4 and T2 : >4 →>4 are represented by the matrices M1 =  1 1 1 4 2 1 4 11 3 4 1 9 4 −3 18 37  and M2 =  1 1 1 −1 2 3 0 1 3 4 1 0 4 5 2 0 , respectively. The null space of T1 is denoted by K1 and the null space of T2 is denoted by K2. Show that the dimension of K1 is 2 and that the dimension of K2 is 1. [5] Find the basis of K1 which has the form    p q 1 0 ,  r s 0 1    and show that K2 is a subspace of K1. [5] 8 Find the particular solution of the differential equation d2y dx2 + 2dy dx + 5y = 10e−2x, given that y = 5 and dy dx = 1 when x = 0. [11] 9 The curve C has equation y = 2x2 + 2x + 3 x2 + 2 . Show that, for all x, 1 ≤y ≤5 2. [4] Find the coordinates of the turning points on C. [3] Find the equation of the asymptote of C. [2] Sketch the graph of C, stating the coordinates of any intersections with the y-axis and the asymptote. [2] 10 The curve C has equation y = 2x 3 3 2, where 0 ≤x ≤3. Show that the arc length of C is 2(2 √2 −1). [4] Find the coordinates of the centroid of the region enclosed by C, the x-axis and the line x = 3. [7] © UCLES 2012 9231/12/M/J/12 [Turn over

4 11 Answer only one of the following two alternatives. EITHER Show that ã π 0 ex sin x dx = 1 + eπ 2 . [4] Given that In = ã π 0 ex sinnx dx, show that, for n ≥2, In = n(n −1) ã π 0 ex cos2x sinn−2x dx −nIn, and deduce that (n2 + 1)In = n(n −1)In−2. [6] A curve has equation y = ex sin5x. Find, in an exact form, the mean value of y over the interval 0 ≤x ≤π. [4] OR The position vectors of the points A, B, C, D are 2i + 4j −3k, −2i + 5j −4k, i + 4j + k, i + 5j + mk, respectively, where m is an integer. It is given that the shortest distance between the line through A and B and the line through C and D is 3. Show that the only possible value of m is 2. [7] Find the shortest distance of D from the line through A and C. [3] Show that the acute angle between the planes ACD and BCD is cos−1 1 √3 . [4] 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 9231/12/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 9231/12 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 12 © 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 12 © 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 12 © University of Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Mark Total 1 States ∑α and ∑αβ Uses formula for correctly. Uses formula for ∑ 3 α to obtain result. ∑ = 7 α ∑ = 2 αβ ∑ = × − = 45 2 2 72 2 α ∑ ∑ ∑ + − = 9 2 7 2 3 α α α = 315-14+9 = 310 B1 B1 M1 A1A1 2 3 [5] 2 (States proposition.) Proves base case. States inductive hypothesis. Proves inductive step. States conclusion. (Pn: 4n > 2n + 3n) Let n = 2, 16 > 4 + 9⇒P2 is true. Assume Pk is true ⇒4k > 2k + 3k k k k k k k 3.4 2.4 ) 3 2 ( 4 4.4 4 1 + = + > = + 1 1 3 2 3.3 2.2 + + + = + > k k k k ∴Pk⇒Pk+1 Hence result true, by PMI, for all integers n ≥ 2. B1 B1 M1 A1 A1 (CWO) 5 [5] 3 Proves initial result. Sets up method of differences. Shows cancellation to get result. States sum to infinity. f(r – 1) – f(r) = ) 2 )( 1 ( 1 )1 ( 1 + + − + r r r r ) 2 )( 1 ( 2 ) 2 )( 1 ( 2 + + = + + − + = r r r r r r r r (AG)   × − × = + + ∑ 3 2 1 2 1 1 2 1 ) 2 )( 1 ( 1 1 n r r r ……       + + − + + ) 2 )( 1 ( 1 )1 ( 1 2 1 n n n n       + + − = ) 2 )( 1 ( 1 2 1 4 1 n n (OE) 4 1 ) 2 )( 1 ( 1 1 = + + ∴∑ ∞ r r r M1 A1 M1A1 A1 A1√ 2 3 1 [6] ‘Non hence’ method for last two parts i.e. penalty of 1 mark. ) 2 ( 2 1 )1 ( 1 2 1 ) 2 )( 1 ( 1 + + + − = − + r r r r r r ⇒ ⋅⋅⋅ ⇒ ) 2 ( 2 1 )1 ( 1 )1 ( 2 1 ... 4 1 2 1 2 1 + + + − + + + − n n n       + + − = ) 2 )( 1 ( 1 2 1 4 1 n n (OE) 4 1 ) 2 )( 1 ( 1 1 = + + ∴∑ ∞ r r r (M1) (A1) (A1) (A1√) (3) (1)

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 4 Draws sketch of C. Uses θ β α d 2 1 2 ∫ r Uses double angle formula. Integrates and obtains area. Finds areas. Shows (4,0) and (0,π) lie on C. Correct shape. (Full cardioid is B1 unless clear evidence of plotting up to 2π or –π to π.) ∫ + + π θ θ θ 0 2 d ) cos 4 cos 8 4 ( 2 1 ∫ + + = π θ θ θ 0 2 d ) cos 2 cos 4 2 ( ∫ + + = π θ θ θ 0 d ) 2 cos cos 4 3 ( π θ θ θ π 3 2 2 sin sin 4 3 0 =     + + = (A1 for correct integral) 712 .4 5 cos 5 sin 5 sin 4 5 3 = + + π π π π 713 .4 712 .4 3 = − π B1 B1 M1 M1 A1A1 CWO M1A1 A1 2 4 3 [9] 5 Identifies matrices P and D. Finds inverse of P. Uses appropriate result to obtain A. (First mark can be implied by correct working.) P =           − − − − 5 3 2 3 1 1 2 1 0 D =          − 2 0 0 0 1 0 0 0 1 Det P =1 P–1 = Adj P =           1 2 1 2 4 1 5 11 4 P–1AP = D ⇒ A = PDP–1 A =                    −           − − − − − 1 2 1 2 4 1 5 11 4 2 0 0 0 1 0 0 0 1 5 3 2 3 1 1 2 1 0 =                     − − − − 1 2 1 2 4 1 5 11 4 10 3 2 6 1 1 4 1 0 =           − − − 26 54 21 13 27 11 2 4 3 B1B1 B1 M1A1 M1 M1A1√ A1 9 [9] 5 Alternative Approach: A =           i h g f e d c b a Use of Ae = λe Obtains 3 sets of 3 linear equations: One set Other two sets Solves one set Solves other sets (M1) (M1A1) (A1A1) (M1A1) (A1A1) (9) [9]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 6 Obtains fifth roots of unity by de M’s Thm. Rewrites and factorises. Isolates z. Obtains purely imaginary denominator and obtains result. 1 = cos(2kπ) + i sin(2kπ) k = 0, 1, 2, 3, 4.       = 5 2 π θ k k = 0, 1, 2, 3, 4. 1 1 1 )1 ( )1 ( 5 5 5 5 5 =       + ⇒ = + ⇒ = + z z z z z z       = + ⇒       = + 5 2 cis 1 5 2 cis 1 π π k z z k z z 1 5 2 cis 1 − =             − ⇒ π k z       −      −            − − =       − − = ⇒ 5 cis 5 cis 5 cis 5 2 cis 1 1 π π π π k k k k z ,3 ,2 ,1 5 cot i 2 1 2 1 5 sin i 2 5 sin i 5 cos =      π + − =      π −      π +      π − = k k k k k (Alternatively for the above three marks – rationalise denominator.)             + − = 5 cot i 1 2 1 π k k = 1, 2, 3, 4. (AG) Observes that original equation is a quartic with real coefficients, so roots occur in conjugate pairs and k = 0 must be rejected. M1 A1 M1 A1 M1A1 A1 A1 B1 2 7 [9]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 7 Reduces M1 to echelon form. Finds. Dim(K1) Reduces M2 to echelon form. Finds Dim(K2) Obtains basis for K1. Obtains basis for K2, and shows 1 2 K K ⊂ .             − 37 18 3 4 9 1 4 3 11 4 1 2 4 1 1 1 →…→             − 0 0 0 0 0 0 0 0 3 2 1 0 4 1 1 1 Dim(K1) = 4 – 2 =2 (AG)             − 0 2 5 4 0 1 4 3 1 0 3 2 1 1 1 1 →…→             − − 0 0 0 0 1 0 0 0 3 2 1 0 1 1 1 1 (aef) Dim(K2) = 4 – 3 = 1 (AG) 0 4 = + + + t z y x 0 3 2 = + + − t z y Legitimately obtains: Basis for K1 is                        −            − 1 0 3 7 , 0 1 2 3 (OE) 0 3 2 0 = + − = − + + t z y t z y x 0 = t Legitimately obtains: Basis for K2 is                        − 0 1 2 3 (OE) 1 2 K K ⊂ ⇒ M1A1 A1 A1 A1 M1 A1 A1 M1 A1 5 5 [10] 8 Forms AQE and solves. Writes CF. Correct form for PI and differentiates twice. Substitutes. Writes PI. Writes GS. Uses y(0) = 5 to find A. Uses y′(0) = 1 to find B. Writes particular solution. m2 + 2m + 5 = 0 ⇒m = –1 ± 2i CF: y = e–x (A cos 2x + B sin 2x) y = ke–2x ⇒y′ = –2ke–2x ⇒y″ = 4e–2x ⇒4k – 4k + 5k = 10 ⇒ k = 2 PI: y = 2e–2x GS: y = e–x (A cos 2x + B sin 2x) +2e–2x y = 5, x = 0⇒5 = A + 2⇒A = 3 y′ = –e–x (A cos 2x + B sin 2x) +e–x (–2A sin 2x + 2B cos 2x) –4e–2x y′ = 1, x = 0⇒1 = –3 + 2B – 4⇒B = 4 ∴ y = e–x (3 cos 2x + 4 sin 2x) +2e–2x M1A1 A1 M1 M1 A1 A1 B1 M1 A1 A1 CAO 11 [11]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 9(i) and (ii) Possible approach for first two parts together. (i) can come after finding turning points: Continuous function (implied by graph) ⇒ (2,2.5) Max and (–1,1) Min 2 5 1 ≤ ≤ ⇒ y (AG) N.B. Award B1 if Max and Min assumed without proof. i.e. 1/4 . Writes 2 )1 ( 1 2 3 2 2 2 2 2 2 + + + = + + + = x x x x x y States 1 0 2 )1 ( 2 2 ≥ ⇒ ≥ + + y x x From this it is clear that (–1, 1) is a turning point. Writes ) 2 ( 2 ) 2 ( 2 5 2 3 2 2 2 2 2 2 + − − = + + + = x x x x x y States 2 5 0 ) 2 ( 2 ) 2 ( 2 2 ≤ ⇒ ≥ + − y x x From this it is clear that (2, 2 2 1 ) is the other turning point. (B1) (B1) (M1A1) (B1) (B1) (A1) (M1) (M1A1) (A1) (7) (4)

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 9 Forms quadratic equation in x. Uses discriminant to obtain condition for real roots. Differentiates and equates to zero. Solves equation. States coordinates of turning points. Expresses y in an appropriate form. (May alternatively divide numerator and denominator by x2. Finds y-intercept and intersection with y = 2. Completes graph. 3 2 2 2 2 2 + + = + x x y yx 0 ) 3 2 ( 2 ) 2 ( 2 = − + − − ⇒ y x x y For real x 4 – 4(y – 2)(2y – 3) ≥ 0 0 )1 )( 5 2 ( ≤ − − ⇒ y y 2 5 1 ≤ ≤ ⇒ y (AG) 0 ) 3 2 2 ( 2 ) 2 4 )( 2 ( 0 2 2 = + + − + + ⇒ = ′ x x x x x y ⇒(x –2)(x + 1) = 0⇒x = –1 or x = 2 (Or substitutes y =1 and 2 5 in equation of C.) Turning points are (–1, 1) and       2 1 2 ,2 2 1 2 2 2 + − + = x x y As x → ±∞ y → 2 ∴ y = 2 Shows       2 1 1 ,0 and       2 , 2 1 Completely correct graph. M1 A1 M1 A1 M1 A1A1 M1 A1 B1 B1 4 3 2 2 [11]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 10 Differentiates and squares. Uses formula for arc length. Integrates and obtains value. Uses formula for x-coordinate of centroid. Integrates both expressions and obtains value. Uses formula for y-coordinate of centroid. Integrates both expressions and obtains value. 3 ) ( 3 1 2 2 1 x y x y = ′ ⇒ = ′ x x s d 3 1 3 0 ∫ + = )1 2 2 ( 2 2 2 4 3 1 2 3 0 2 3 − = − =                + = x (AG) ∫ ∫ = 3 0 2 3 3 0 2 5 d 3 3 2 d 3 3 2 x x x x x 7 15 5 2 7 2 3 0 2 5 3 0 2 7 =                 = x x (= 2.14) ∫ ∫ × = 3 0 2 3 3 0 3 d 3 3 2 d 27 4 2 1 x x x x y 8 5 5 2 3 3 2 4 27 2 3 0 2 5 3 0 4 =               = x x (= 0.625) B1 M1 A1A1 M1 A1A1 A1 M1 A1 A1 4 7 [11]

Page 11 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 11 EITHER Note: (1) the parts can be either way round. (2) Insertion of limits in [ex sin x] causes the term to vanish. ∫ ∫ + − = = x x x x x I x x x d cos e cos e d sin e ∫ − + − = x x x x x x x d sin e sin e cos e ∴2I = ex(sin x – cos x) π π 0 0 ) cos (sin e 2 1 d sin e ∫     − = ∴ x x x x x x 2 e 1 2 1 2 e π π + =      − − = (AG) ∫ = π 0 d sin e x x I n x n [ ] ∫ − − = π π 0 1 0 d ) cos sin ( e e. sin x x x n x n x x n [ ]       − − + − = ∫ − − − π π 0 1 2 2 0 1 d ) sin sin sin )1 ( (cos e e. cos sin 0 x x x x n x n x x n n n x x n ∫ − − + = − π 0 2 2 d sin cos e )1 ( 0 n n x nI x x x n n (AG) ∫ − − − = − π 0 2 2 d sin ) sin 1( e )1 ( n n x nI x x x n n n n n I n n I n n I n )1 ( )1 ( )1 ( 2 − − − = + ∴ − ( ) 2 )1 ( 1 )1 ( − − = + + − ∴ n n I n n I n n n 2 2 )1 ( )1 ( − − = + ∴ n n I n n I n (AG) 1 3 5 10 6 26 20 26 20 I I I × = = ( ) π π e 1 13 3 2 e 1 13 6 5 + =      + × = ⇒I Mean value = ) e 1( 13 3 0 d sin e 0 5 π π π π + = − ∫ x x x M1 A1 M1 A1 M1 A1 A1 M1A1 A1 M1 A1 M1A1 4 6 4 [14]

Page 12 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 11 OR Obtains direction of common perpendicular. Uses result for shortest distance between lines. Solves equation. Finds relevant vectors. Use of cross-product. Obtains shortest distance. Finds 2nd vector in BCD (CD may already have been found.) Finds normal vector to BCD. (Normal to ACD already found.) Finds angle between planes = angle between normal vectors. n = 1 1 0 1 1 4 − − m k j i = –mi + 4(1– m)j + 4k 3 16 ) 2 1( 16 4 4 4 4 0 1 2 2 = + + − +           − −           − m m m m m ⇒… 0 4 40 19 2 = + − ⇒ m m 0 ) 2 )( 2 19 ( = − − ⇒ m m ⇒m = 2, since m is an integer. (AG) CA =           −4 0 1 and CD =           1 1 0 or AD =          − 5 1 1           − − = − 1 1 4 17 1 4 0 1 1 1 0 17 1 k j i 17 18 1 1 4 17 1 2 2 2 = + + (= 1.03) BC = 3i – j +5k n = 1 1 0 5 1 3 − k j i = –6i – 3j +3k ~ 2i + j – k 3 1 6 18 6 1 1 4 1 1 16 ) k j i2 ( ) k j i4 ( cos = = + + + + − + ⋅ + − = θ ∴Angle between planes = cos–1       3 1 (AG) M1A1 M1A1 A1 M1 A1 B1 M1 A1 B1 M1 M1 A1 7 3 4 [14]

Page 13 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 12 © University of Cambridge International Examinations 2012 11 OR Alternatives for middle part: Or (a) Vector from D to any point on AC Uses orthogonality to obtain t. Finds magnitude of perpendicular. Or (b) Finds length of AD (or CD) Finds projection of AD (or CD) onto AC. Finds perpendicular by Pythagoras.           − − − + t t 4 5 1 1 17 21 0 4 0 1 . 4 5 1 1 − = ⇒ =           −           − − − + t t t 17 18 1 1 4 17 1 2 2 2 = + + (= 1.03) 27 = AD 17 21 1 4 4 0 1 . 5 1 1 2 2 = +           −          − 17 18 17 441 27 = − (= 1.03) (B1) (M1) (A1) (B1) (M1) (A1) (3) (3)