Cambridge A Level Mathematics - Further 9231 — 2011 Oct/Nov Paper 1 · Variant 2

*9171987026* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 October/November 2011 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. JC11 11_9231_12/FP © UCLES 2011 [Turn over

2 1 The equation x3 + px + q = 0 has a repeated root. Prove that 4p3 + 27q2 = 0. [5] 2 The position vectors of points A, B, C, relative to the origin O, are a, b, c, where a = 3i + 2j −k, b = 4i −3j + 2k, c = 3i −j −k. Find a × b and deduce the area of the triangle OAB. [3] Hence find the volume of the tetrahedron OABC, given that the volume of a tetrahedron is 1 3 × area of base × perpendicular height. [2] 3 Prove by mathematical induction that, for all positive integers n, dn dxn (ex sin x) = 2 1 2nex sinx + 1 4nπ. [7] 4 The linear transformation T : >4 →>4 is represented by the matrix M, where M =  3 4 2 5 6 7 5 8 9 9 9 9 15 16 14 17 . Find (i) the rank of M and a basis for the range space of T, [4] (ii) a basis for the null space of T. [3] 5 The point P (2, 1) lies on the curve with equation x3 −2y3 = 3xy. Find (i) the value of dy dx at P, [3] (ii) the value of d2y dx2 at P. [4] 6 Let In = ã 1 0 xn(1 −x) 1 2 dx, for n ≥0. Show that, for n ≥1, (3 + 2n)In = 2nIn−1. [5] Hence find the exact value of I3. [3] © UCLES 2011 9231/12/O/N/11

3 7 The curve C has equation y = x2 + px + 1 x −2 , where p is a constant. Given that C has two asymptotes, find the equation of each asymptote. [3] Find the set of values of p for which C has two distinct turning points. [5] Sketch C in the case p = −1. Your sketch should indicate the coordinates of any intersections with the axes, but need not show the coordinates of any turning points. [3] 8 The vector e is an eigenvector of the matrix A, with corresponding eigenvalue λ, and is also an eigenvector of the matrix B, with corresponding eigenvalue µ. Show that e is an eigenvector of the matrix AB with corresponding eigenvalue λµ. [2] State the eigenvalues of the matrix C, where C =  −1 −1 3 0 1 2 0 0 2 , and find corresponding eigenvectors. [4] Show that 1 6 3 ! is an eigenvector of the matrix D, where D =  1 −1 1 −6 −3 4 −9 −3 7 , and state the corresponding eigenvalue. [3] Hence state an eigenvector of the matrix CD and give the corresponding eigenvalue. [2] 9 The curve C has equation y = 1 2ex + e−x for 0 ≤x ≤ln 5. Find (i) the mean value of y with respect to x over the interval 0 ≤x ≤ln 5, [4] (ii) the arc length of C, [4] (iii) the surface area generated when C is rotated through 2π radians about the x-axis. [4] 10 The curve C has polar equation r = 3 + 2 cos θ, for −π < θ ≤π. The straight line l has polar equation r cos θ = 2. Sketch both C and l on a single diagram. [3] Find the polar coordinates of the points of intersection of C and l. [4] The region R is enclosed by C and l, and contains the pole. Find the area of R. [6] [Question 11 is printed on the next page.] © UCLES 2011 9231/12/O/N/11 [Turn over

4 11 Answer only one of the following two alternatives. EITHER Let ω = cos 1 5π + i sin 1 5π. Show that ω5 + 1 = 0 and deduce that ω4 −ω3 + ω2 −ω = −1. [2] Show further that ω −ω4 = 2 cos 1 5π and ω3 −ω2 = 2 cos 3 5π. [4] Hence find the values of cos 1 5π + cos 3 5π and cos 1 5π cos 3 5π. [4] Find a quadratic equation having roots cos 1 5π and cos 3 5π and deduce the exact value of cos 1 5π. [4] OR Given that x2 d2y dx2 + 4x(1 + x)dy dx + 2(1 + 4x + 2x2)y = 8x2 and that x2y = ß, show that d2ß dx2 + 4dß dx + 4ß = 8x2. [4] Find the general solution for y in terms of x. [8] Describe the behaviour of y as x →∞. [2] 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 2011 9231/12/O/N/11

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2011 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 October/November 2011 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 – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 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 – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 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 – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total (N.B. Not α, β, γ) Let roots be α, α, and β. Writes down sum of roots, (1) 0 2 = + β α M1 sum of products in pairs (2) p = + 2 2 α αβ A1 and product of roots. (3) q − = β α 2 A1 From (1) α β 2 − = Eliminates β ( or α ). (2) 2 2 2 3 4 α α α − = ⇒ = + − ⇒ p p (3) 3 3 2 2 α α = ⇒ − = − ⇒ q q M1 1 Equates power of α ( or β ) 0 27 4 2 3 2 3 2 3 6 = + ⇒       =      − = q p q p α (AG) A1 5 [5] Finds a × b = − − 2 3 4 1 2 3 k j i k j i 17 10 − − M1 A1 Finds area of base. 390 2 1 ) 17 ( ) 10 ( 1 2 1 2 2 2 = − + − + ( = 9.87) A1 3 Attempts to find height Height = 390 30 ) 17 ( ) 10 ( 1 ) 17 10 ).( 3 ( 2 2 2 = − + − + − − − − k j i k j i ( = 1.519) M1 2 Finds volume 5 390 30 390 2 1 3 1 = × × A1 2 [5]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 3 Proves base case. Hn :       + = 4 sin 2 ) sin ( 2 π n x e x e x d d x n x n n x e e x x e x d d x x x cos sin ) sin ( + = M1       + =       + = 4 sin 2 2 cos 2 sin 2 2 1 π x e x x e x x ⇒ H1 is true. A1 States inductive hypothesis. Assume Hk is true : B1 Proves inductive step.             + +       + = + + 4 cos 4 sin 2 ) sin ( 2 1 1 π π k x e k x e x e x d d x x k x k k M1             + +       + = + 4 cos 2 1 4 sin 2 1 2 2 1 π π k x k x e x k A1             + + = + 4 4 sin 2 2 1 π π k x e x k =       + + + 4 )1 ( sin 2 2 1 π k x e x k A1 States conclusion. x Hk ⇒ Hk+1 Hence true for all positive integers by PMI A1 7 [7]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 4 (i) Reduces matrix to echelon form.             17 14 16 15 9 9 9 9 8 5 7 6 5 2 4 3 ~             − − − − − − 8 4 4 0 6 3 3 0 2 1 1 0 5 2 4 3 ~             − − 0 0 0 0 0 0 0 0 2 1 1 0 5 2 4 3 M1A1 States rank, ⇒ R(M) = 2. A1 Basis for range space is: and basis for range space.                                     16 9 7 4 15 9 6 3 , (OE) A1 4 Alternatively: c1 =             15 9 6 3 c2 =             16 9 7 4 c3 =             14 9 5 2 c4 =             17 9 8 5 Shows linear dependence. Finds a lin. indep. set. States rank and basis for range space. 2c1 = c2 + c3 and c4 = c1 + c2 – c3 ⇒ lin. dep. but any two e.g. c1, and c2 are lin. indep. Hence R(M) = 2 Basis of range space is {c1, c2} M1 A1 A1 A1 (ii) Forms equations. 3x + 4y + 2z + 5t = 0 –y + z – 2t = 0 M1 (Gives two parameter solution.) (t = λ , z = µ , y = µ – 2λ , x = λ – 2µ) States basis of null space. Basis of null space is                        −             − 0 1 1 2 1 0 2 1 , or                         −            − 2 1 3 0 1 2 0 3 , A1A1 3 (Or by reducing transpose to echelon form, or by any other valid method.). or any two of the above four vectors. [7]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 5 (i) One mark for each side. y x y y y x ′ + = ′ − 3 3 6 3 2 2 B1B1 Substitutes (2,1) 4 3 6 3 6 12 = ′ ⇒ ′ + = ′ − y y y B1√ 3 (ii) One mark for differentiating both 1st and 3rd terms. One mark for each of 2nd and 4th terms. { } y x y y y y y y x ′′ + ′ + ′ = ′ + ′′ − 3 3 3 ) ( 12 6 6 2 2 B1B1 B1 Substitute (2,1) and 4 3 ) 2 ( = ′y . 16 1 4 3 12 6 4 9 4 9 ) 4 27 6 ( 12 = ′′ ⇒ = ′′ ⇒ ′′ + + = + ′′ − y y y y B1 4 [7] 6 In ∫ − = 1 0 2 1 ) 1( x d x xn Integrates by parts. ∫ − − +         − − = − 1 0 2 1 1 1 0 2 3 ) 1 )( 1( 3 2 ) 1( 3 2 x d x x nx x x n n M1A1 Rearranges. x d x x n x d x x n n n ∫ ∫ − − − + = − 1 0 2 1 1 0 2 1 1 ) 1( 3 2 ) 1( 3 2 0 M1A1 n n I n I n 3 2 3 2 1 − = − Obtains printed result. 1 2 ) 3 2 ( − = + ⇒ n n nI I n (AG) A1 5 Evaluates I0. 3 2 ) 1( 3 2 ) 1( 1 0 2 3 1 0 2 1 0 =         − − = − =∫ x x d x I B1 Uses reduction formula. 315 32 3 2 5 2 7 4 9 6 3 = × × × = I M1A1 3 [8]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 7 Vertical asymptote. x = 2 B1 3 Divides by (x – 2) 2 5 2 2 − + + + + = x p p x y M1 Oblique asymptote. y = x + p + 2 A1 3 Differentiates. 2 2 ) 2 ( 5 2 4 4 − − − + − = x p x x x d y d M1A1 y' = 0 ⇒ x2 – 4x – (2p + 1) = 0 M1 B2 – 4AC > 0 ⇒ 16 + 4(2p + 1) > 0 M1 ⇒ p > 2 5 − A1 5 Sketches graph. Working to show either x2 –x + 1 = 0 has no real roots, or maximum value. Axes and (0,–0.5) marked.. Upper Branch with minimum. Lower with maximum below x-axis. (Deduct at most 1 for poor forms at infinity.) B1 B1 B1 3 [11]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 8 Shows required result, using Ae = λe. ABe = A µe = µAe = µλe = λµe M1A1 2 States eigenvalues from leading diagonal. Eigenvalues of C are –1, 1 and 2 B1 Finds eigenvectors using cross-product or equations. M1A1 for first correct and A1 for the other two. 1 − = λ : e1 = 2 2 0 3 1 0 − k j i =          − 0 0 8 ~           0 0 1 M1A1 1 = λ : e2 =          − = − − 0 4 2 2 0 0 3 1 2 k j i ~           − 0 2 1 2 = λ : e3 =           = − − 3 6 1 2 1 0 3 1 3 k j i A1 4 Uses De = µe. D           3 6 1 =           − =           − − − 3 6 1 2 6 12 2 M1A1 States eigenvalue. Eigenvalue is –2. A1 3 Recognises that CD has an eigenvector common to C and D and CD has an eigenvector           3 6 1 B1 states the corresponding eigenvalue. and the corresponding eigenvalue is 4 2 2 − = × − . B1√ 2 [11]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 9 (i) Uses mean value formula and integrates. M.V. = ( ) ( ) 5 ln 2 1 0 5 ln 2 1 5 ln 0 5 ln 0     − = − + − − ∫ x x x x e e x d e e M1A1 Substitutes limits and evaluates. 5 ln 5 12 5 ln 5 1 5 2 1 =      − = (= 1.49) M1A1 4 (ii) Differentiates and finds 2) ( 1 y′ + . ( ) ( ) 2 2 2 1 ) ( 1 2 1   + = ′ + ⇒ − = ′ − − x x x x e e y e e y M1A1 Integrates and obtains result. ( ) [ ] 5 ln 0 5 ln 0 2 1 2 1 x x x x e e x d e e s − − − = + = ∫ 5 12 5 1 5 2 1 =    − = M1A1 (iii) Uses surface area formula. S = ( ) ( ) ∫ − − + ⋅ + 5 ln 0 2 1 2 1 2 x d e e e e x x x x π M1 4 Integrates. ∫ − + + = 5 ln 0 2 2 ) 2 ( 2 x d e e x x π 5 ln 0 2 2 2 2 2 2       − + = −x x e x e π M1 Substitutes limits and           − + −     − + = 2 1 0 2 1 50 1 5 ln 2 2 25 2 π A1 evaluates. =       + 5 ln 25 156 π (=24.7) A1 4 [12]

Page 11 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 10 C: Straight line Closed loop through (5,0) and (1,π) Correct shape near (1,π) Perpendicular to initial line , through (2,0) B1 B1 B1 3 ⇒ (3 + 2cosθ)cosθ = 2 Forms quadratic equation in usual form. ⇒ 2cos2θ + 3cosθ – 2 = 0 (aef) M1 ⇒ (2cosθ – 1)(cosθ + 2) = 0 Solves quadratic equation. ⇒ cosθ = 0.5 (since cosθ > 0) A1 Writes down points of intersection. Intersections at       3 ,4 π and       −3 ,4 π . A1A1 4 Calling points of intersection A and B and the pole O. Required area is two congruent sectors between l and C plus triangle OAB. Finds required area. Two sectors ∫ + + × = π π θ θ θ 3 2 ) cos 4 cos 12 9 ( 2 1 2 d M1 ∫ + + = π π θ θ θ 3 ) 2 cos 2 cos 12 11 ( d A1 [ ] π π θ θ θ 3 2 sin sin 12 11 + + = M1 ) 78 . 11 ( 2 3 13 3 22 = − = π A1 Triangle = ) 928 .6 ( 3 4 2 3 2 = = × B1 Total Area = ) 7. 18 708 . 18 ( 2 3 5 3 22 (3sf) = = − π A1 6 [13]

Page 12 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 11 EITHER Verifies that ω is a root. 0 1 sin cos 1 5 sin 5 cos 5 = + + = +       + π π π π i i B1 Factorises. 0 )1 )( 1 ( )1 ( 2 3 4 5 = + − + − + = + ω ω ω ω ω ω 0 1 1 2 3 4 = + − + − ⇒ − ≠ ω ω ω ω ω 1 2 3 4 − = − + − ⇒ ω ω ω ω B1 2 5 sin 5 cos π π ω i + = Finds ω 4 5 sin 5 cos 5 4 sin 5 4 cos 4 π π π π ω + − = + = ⇒ i M1 and subtracts. 5 cos 2 4 π ω ω = − ⇒ A1 Finds 3 ω 5 3 sin 5 3 cos 3 π π ω i + = and 2 ω 5 3 sin 5 3 cos 5 2 sin 5 2 cos 2 π π π π ω i i − = + = M1 and subtracts 5 3 cos 2 2 3 π ω ω = − A1 4 2 1 5 3 cos 5 cos 1 5 3 cos 2 5 cos 2 = + ⇒ − = − − π π π π M1A1 ) )( ( 4 1 5 3 cos 5 cos 2 3 4 ω ω ω ω π π − − = M1 ) ( 4 1 6 7 3 4 ω ω ω ω + − − = 4 1 ) ( 4 1 2 3 4 − = − + − = ω ω ω ω A1 4 Equation with roots 5 cosπ and 5 3 cos π is: Finds required quadratic equation. 0 4 1 2 1 2 = − − x x or 0 1 2 4 2 = − −x x M1 Solves for x. 8 5 2 2 ± = ⇒x M1A1 States required value. 4 5 1 5 cos + = ⇒ π (since 0 < 5 cosπ < 1) A1 4 [14]

Page 13 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 11 OR Differentiates xy dx dy x dx dz y x z 2 2 2 + = ⇒ = M1 twice. y dx dy x dx y d x dx z d 2 4 2 2 2 2 2 + + = ⇒ A1 Rearranges LHS of DE. ( ) ( ) = + + + + + ∴ y x x dx dy x x dx y d x 2 2 2 2 4 8 2 1 4 M1 y x xy dx dy x y dx dy x dx y d x 2 2 2 2 2 4 2 4 2 4 +       + +       + + 2 2 2 8 4 4 x z dx dz dx z d = + + = (AG) A1 4 Finds and solves AQE. 0 ) 2 ( 0 4 4 2 2 = + ⇒ = + + m m m 2 − = ⇒m M1 States CF CF: x x e Bx e A z 2 2 − − + = A1 States form of PI. PI: c bx ax z + + = 2 Differentiates twice, a z b ax z 2 2 = ′′ ⇒ + = ′ ⇒ M1 substitutes 2 2 8 4 4 4 4 8 2 x c bx ax b ax a = + + + + + A1 and equates coefficients. 0 4 4 2 = + + c b a M1 0 4 8 = + b a 8 4 = a Solves. 3 4 2 = − = = c b a A1 States GS for z–x. 3 4 2 2 2 2 + − + + = − − x x e Bx e A z x x M1 States GS for y–x. 2 2 2 3 4 2 x x e x B e x A y 2x - x + − + + = − A1 8 Considers the effect of ∞ → x . As 0 1 1 , 2 2 → ∞ → − x and x e , x x M1 2 → ∴y A1 2 [14]