Cambridge A Level Mathematics - Further 9231 — 2018 Oct/Nov Paper 1 · Variant 2

*7959781442* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/12 Paper 1 October/November 2018 3 hours Candidates answer on the Question Paper. Additional Materials: List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions in the space provided. If additional space is required, you should use the lined page at the end of this booklet. The question number(s) must be clearly shown. 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 26 printed pages and 2 blank pages. JC18 11_9231_12/RP © UCLES 2018 [Turn over

2 1 The roots of the cubic equation x3 −5x2 + 13x −4 = 0 are !, ", '. (i) Find the value of !2 + "2 + '2. [3] … … … … … … … … … … … (ii) Find the value of !3 + "3 + '3. [2] … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

3 2 It is given that A = ` 2 3 1 0 −2 1 0 0 1 a . (i) Find the eigenvalue of A corresponding to the eigenvector ` 1 0 0 a . [1] … … … … (ii) Write down the negative eigenvalue of A and find a corresponding eigenvector. [3] … … … … … … … … … … (iii) Find an eigenvalue and a corresponding eigenvector of the matrix A + A6. [2] … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

4 3 The curve C has polar equation r = a cos 31, for −1 60 ≤1 ≤1 60, where a is a positive constant. (i) Sketch C. [2] (ii) Find the area of the region enclosed by C, showing full working. [3] … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

5 … … … … … … … … … … (iii) Using the identity cos 31  4 cos31 −3 cos 1, find a cartesian equation of C. [3] … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

6 4 (i) Find the general solution of the differential equation d2x dt2 + 2dx dt + x = 4 sin t. [7] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

7 … … … … … … … … … … … … … … … … (ii) State an approximate solution for large positive values of t. [1] … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

8 5 The linear transformation T : >4 →>4 is represented by the matrix M, where M = € 3 2 0 1 6 5 −1 3 9 8 −2 5 −3 −2 0 −1  . (i) Find the rank of M. [3] … … … … … … … … … … … … … Let K be the null space of T. (ii) Find a basis for K. [3] … … … … … … © UCLES 2018 9231/12/O/N/18

9 … … … … … … … (iii) Find the general solution of Mx = € 2 5 8 −2  . [3] … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

10 6 It is given that y = exu, where u is a function of x. The rth derivatives dry dxr and dru dxr are denoted by yr and ur respectively. Prove by mathematical induction that, for all positive integers n, yn = ex P@ n 0 A u + @ n 1 A u1 + @n 2 A u2 + à + @n r A ur + à + @n n A un Q . [8] [You may use without proof the result @ k r A + @ k r −1 A = @ k + 1 r A .] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

11 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

12 7 Let SN = N Ð r=1 3r + 13r + 4 and TN = N Ð r=1 1 3r + 13r + 4. (i) Use standard results from the List of Formulae (MF10) to show that SN = N3N2 + 12N + 13. [3] … … … … … … … … … … … … (ii) Use the method of differences to show that TN = 1 12 − 1 33N + 4. [3] … … … … … … © UCLES 2018 9231/12/O/N/18

13 … … … … … … … … … … (iii) Deduce that SN TN is an integer. [2] … … … … … … (iv) Find lim N→∞ SN N3TN . [2] … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

14 8 (i) By considering the binomial expansion of @ z + 1 z A6 , where z = cos 1 + i sin 1, express cos61 in the form 1 32p + q cos 21 + r cos 41 + s cos 61, where p, q, r and s are integers to be determined. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

15 (ii) Hence find the exact value of Ó 1 20 −1 20 cos61 2x dx. [4] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

16 9 The curve C has equation y = 5x2 + 5x + 1 x2 + x + 1 . (i) Find the equation of the asymptote of C. [2] … … … … … (ii) Show that, for all real values of x, −1 3 ≤y < 5. [4] … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

17 (iii) Find the coordinates of any stationary points of C. [2] … … … … … … … … … … … … (iv) Sketch C, stating the coordinates of any intersections with the y-axis. [2] © UCLES 2018 9231/12/O/N/18 [Turn over

18 10 The position vectors of the points A, B, C, D are i + j + 3k, 3i + 4j + 5k, −i + 3k, mj + 4k, respectively, where m is a constant. (i) Show that the lines AB and CD are parallel when m = 3 2. [1] … … … … (ii) Given that m ≠3 2, find the shortest distance between the lines AB and CD. [5] … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

19 (iii) When m = 2, find the acute angle between the planes ABC and ABD, giving your answer in degrees. [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

20 11 Answer only one of the following two alternatives. EITHER The curve C is defined parametrically by x = 18t −t2 and y = 8t 3 2, where 0 < t ≤4. (i) Show that at all points of C, d2y dx2 = 39 + t 2t 1 29 −t3 . [4] … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

21 (ii) Show that the mean value of d2y dx2 with respect to x over the interval 0 < x ≤56 is 3 70. [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

22 (iii) Find the area of the surface generated when C is rotated through 20 radians about the x-axis, showing full working. [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

23 OR Let In = Ó ï2 1 x2 −1n dx. (i) Show that, for n ≥1, 2n + 1In = ï2 −2nIn−1. [5] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

24 (ii) Using the substitution x = sec 1, show that In = Ó 1 40 0 tan2n+11 sec 1 d1. [4] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

25 (iii) Deduce the exact value of Ô 1 40 0 sin71 cos81 d1. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18 [Turn over

26 Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/12/O/N/18

27 BLANK PAGE © UCLES 2018 9231/12/O/N/18

28 BLANK PAGE 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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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 2018 9231/12/O/N/18

This document consists of 20 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 October/November 2018 MARK SCHEME Maximum Mark: 100 Published 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 2 of 20 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 3 of 20 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 4 of 20 Abbreviations AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) 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 SOI Seen or implied 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) Question Answer Marks Guidance 1(i) 5 α β γ + + = , 13 αβ αγ βγ + + = B1 Sum of roots and αβ αγ βγ + + . SOI ( ) 2 2 2 2 5 2 13 α β γ + + = − M1 Uses ( ) ( ) 2 2 2 α α αβ ∑ = ∑ − ∑ = ‒1 A1 www 3 1(ii) ( ) ( ) 3 3 3 2 2 2 5 13 12 α β γ α β γ α β γ + + = + + − + + + M1 Uses 3 2 5 13 4. α α α = − + ( ) ( ) 5 1 13 5 12 58 = − − + = − A1 Alt method: Use formula e.g. 3 α ∑ = ( )( ) 2 3 α α αβ αβγ ∑ ∑ −∑ + Or ( ) α ∑ 3 – 3( ) α ∑ ( αβ ∑ ) +3 αβγ 2

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 5 of 20 Question Answer Marks Guidance 2(i) 2 B1 Stated 1 2(ii) Negative eigenvalue 2 = − B1 Stated 4 3 1 2 0 0 1 4 3 1 0 0 3 0 0 1 i j k     + =       A I M1 Uses vector product (or equations) to find corresponding eigenvector. 3 4 0     −       A1 Accept any non-zero scalar multiple of 3 4 0     −       . 3 2(iii) An eigenvalue of 6 A A + is 6 2 2 66 + = , 62 or 2 B1 Corresponding eigenvector is 1 0 0      3 4 0     −       or 6 1 3 −           oe B1 2

923 © U Qu 31/12 CLES 2018 uestion 3(i) C A Cambridge Inter Answer a rnational A Leve PUBLISHED Page 6 of 20 l – Mark Scheme Marks B1 Jus B1 Co or i 2 Octob Gui st one loop, correc orrect position incl in table. ber/November 2 idance ct shape at extrem luding (a, 0) label 018 mities lled

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 7 of 20 Question Answer Marks Guidance 3(ii) 6 2 2 6 1 3 d 2 a cos π π θ θ −∫ M1 For using correct formula ( ) 6 2 6 cos6 1 d 4 a π π θ θ − + ∫ M1 Using double angle formula correctly 2 2 6 6 1 sin6 4 6 12 a a π π π θ θ −   = + =     A1 3 3(iii) ( ) 2 2 cos 4cos 3 4 3 x x r a r a r r θ θ       = − ⇒ = −               B1 Uses cos x r θ = and 2 2 2 x y r + = . ( ) ( ) ( ) ( ) 2 4 2 2 2 2 2 2 2 4 3 4 3 r ax x r x y ax x x y ⇒ = − ⇒ + = − + M1 For eliminating θ ( ) ( ) 2 2 2 2 2 3 x y ax x y ⇒ + = − A1 Any equivalent cartesian form without fractions. 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 8 of 20 Question Answer Marks Guidance 4(i) ( ) 2 2 2 1 0 1 0 1 m m m m + + = ⇒ + = ⇒ = − M1 Forms and solves auxiliary equation. CF: ( ) t A Bt e− + A1 States CF. PI: sin cos x p t q t = + M1 Uses correct form of PI and differentiates twice. ¨ cos sin sin cos x p t q t x p t q t ⇒ = − ⇒ = − −  A1 ( ) sin cos 2 cos sin sin cos 4 p t q t p t q t p t q t sint − − + − + + = M1 Compares coefficients and attempts to solve 2 0 0 p p = ⇒ = . 2 4 2 q q − = ⇒ = −. A1 GS: ( ) 2cos t A Bt e t − = + − x A1FT States general solution. FT on correct form only 7 4(ii) 2cos x t ≈− B1FT 1

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 9 of 20 Question Answer Marks Guidance 5(i) 3 2 0 1 3 2 0 1 6 5 1 3 0 1 1 1 9 8 2 5 0 2 2 2 3 2 0 1 0 0 0 0         − −     →     − −     − − −     M1 Attempt to row reduction. 3 2 0 1 0 1 1 1 0 0 0 0 0 0 0 0     −   →      A1 Two correct rows only ( ) 4 2 2 r = − = M A1 Obtains rank. 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 10 of 20 Question Answer Marks Guidance 5(ii) 3 2 0 x y t + + = 0 y z t − + = M1 Solves homogeneous system of equations. 2 1 , , , 3 3 t z y x µ λ λ µ λ µ ⇒ = = = − = − + M1 Using 2 parameters A basis is 2 1 1 0 3 3 0 1 , or , 3 0 3 1 0 3 3 2     − −                 − −                                             or equivalent A1 AEF 3 5(iii) 0 2 1 5 0 8 0 2          =       −    M so a particular solution is 0 1 0 0       B1 Finds a particular solution. General solution: ( ) 0 2 1 1 3 3 0 3 0 0 0 3 λ µ −           −      = + +                x M1 Using correct format A1FT 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 11 of 20 Question Answer Marks Guidance 6 ( ) ( ) ( ) 1 1 1 1 1 1 0 1 x x x y e u ue e u u H     = + = + ⇒         is true M1A1 Shows base case using product rule Assume that ( ) ( ) ( ) ( ) 1 : 0 1 k r k x k k k k k H y e u u u u r k       = + + + + +             " " B1 States inductive hypothesis. Then ( ) ( ) ( ) ( ) 1 1 0 1 k r k x k k k k y e u u u u r k +       = + + + + + +             " " M1 Differentiates using product rule ( ) ( ) ( ) ( ) 1 2 1 1 0 1 r k x k k k k e u u u u r k + +       + + + + +             " " = ( ) 1 0 1 k x r k k k k e u u u r r k +          +… + +…            −          M1A1 Shows application of 1 1 k k k r r r +      + =      −      . = ( )1 1 1 1 0 1 k x r k k k e u u u r k +   + + +       +… +…         +         B1 Shows reasoning for first and last term correctly So 1 k k H implies H + so, by induction, n H is true for all 1 n. . A1 States conclusion. 8

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 12 of 20 Question Answer Marks Guidance 7(i) ( )( ) 2 1 1 1 3 1 3 4 9 15 4 N N N r r r r r r r N = = = + + = + + ∑ ∑ ∑ M1 Expands ( )( ) ( ) 1 1 9 1 2 1 15 1 4 6 2 N N N N N N     + + + + +         M1 Substitutes formulae for r ∑ and 2 r ∑ . ( ) 2 9 15 15 2 3 1 4 6 2 2 N N N N   = + + + + +     ( ) 2 3 12 13 N N N = + + A1 Shows simplification to the given answer (AG). 3 7(ii) ( )( ) 1 1 1 1 3 1 3 4 3 3 1 3 4 r r r r   = −   + + + +   B1 Finds partial fractions. ( ) 1 1 1 1 1 1 1 3 4 7 7 10 3 1 1 3 4 N T N N   = − + − + + −     − + +   " M1 Expresses terms as differences. ( ) 1 1 1 1 1 3 4 3 4 12 3 3 4 N N   − = −   + +   A1 Cancels to given answer (AG). 3 7(iii) ( ) ( )( ) 2 4 3 4 3 12 13 4 3 4 N N N S N T N N N N T = ⇒ = + + + + M1 Writes N N S T as a polynomial So N N S T is an integer because all terms are integers A1 Justifies expression being integer 2

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 13 of 20 Question Answer Marks Guidance 7(iv) ( )( ) 2 3 3 4 3 4 3 16 9 N N N N N S N T N + + + = M1 Divides expression in (iii) by 3 N and takes limit ( )( ) 4 3 3 36 → = A1 2 Question Answer Marks Guidance 8(i) 1 2cos z z θ − + = B1 Use of 1 2cos z z θ − + = . ( ) ( ) ( ) ( ) 6 1 6 6 4 4 2 2 6 15 20 z z z z z z z z − − − − + = + + + + + + M1A1 Expands and groups. 6 64cos 2cos6 12cos4 30cos2 20 θ θ θ θ = + + + M1A1 Substitutes 2cos n n z z nθ − + = . ( ) 6 1 cos 10 15cos2 6cos4 cos6 32 θ θ θ θ ⇒ = + + + A1 (Allow 10 p = , 15, 6, 1 q r s = = = .) 6

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 14 of 20 Question Answer Marks Guidance 8(ii) 1 1 2 2 6 1 1 2 2 1 cos d 10 15cos 6cos2 cos3 d 2 32 x x x x x x π π π π − − = + + + ∫ ∫ M1 Applies part (i) M1 Integrates correctly (3/4 terms correct). 1 2 1 2 1 1 10 15sin 3sin 2 sin3 32 3 x x x x π π −   + + +     M1 Inserts limits and evaluates. 1 1 1 1 44 5 15 0 5 15 0 5 32 3 3 16 3 π π π         = + + − −− − + + = +                 A1 4 Question Answer Marks Guidance 9(i) 2 4 5 1 y x x = − + + M1 Alt method: Finding limit As x →±∞, 5 5 y y →∴ = CAO A1 2

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 15 of 20 Question Answer Marks Guidance 9(ii) 2 2 5 5 1 yx yx y x x + + = + + ( ) ( ) ( ) 2 5 5 1 0 y x y x y ⇒ − + − + − = B1 Forms quadratic equation in x . For real x , ( ) ( )( ) 2 5 4 5 1 0 y y y − − − − . (condone >) M1 Uses discriminant ( )( ) 5 3 1 0 y y ⇒ − + - M1 Factorising 1 5, 3 y ⇒− < - because y = 5 is an asymptote (www) A1 Explaining strict upper inequality (AG) 4 9(iii) ( )( ) ( )( ) 2 2 0 1 10 5 5 5 1 2 1 0 y x x x x x x ′ = ⇒ + + + − + + + = M1 Differentiates and equates to 0. ( ) 1 1 4 2 1 0 , 2 3 x x y ⇒ + = ⇒ = − = − A1 2

923 © U Qu 9 Qu 1 31/12 CLES 2018 uestion 9(iv) uestion 10(i) AB   =  JJJG 2 1 3 , 1.5 2 1 CD       = =          JJJG C A A 1 2 AB = JJJG Cambridge Inter Answer Answer rnational A Leve PUBLISHED Page 16 of 20 l – Mark Scheme Marks B1FT Pos min B1 Co gra 2 Marks B1 Or 1 Octob Gui sitive y-intercept a nimum point from orrect asymptote a aph. Gui shows if parallel, ber/November 2 idance at (0,1), FT dep o m (iii). and completely cor idance , then m=3/2 018 on rrect

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 17 of 20 Question Answer Marks Guidance 10(ii) 2 1 3 , 2 1 AB CD m       = =          JJJG JJJG 2 1 0 AC −     = −       and JJJG or AD = 1 1 1 m −     −       or BC= 4 4 2 −     −     −   B1 n = 3 2 2 3 2 0 1 1 2 3 m m m −     =     −   i j k ( 1 so parallel to 0 1         −   ) M1A1 Finds common perpendicular using cross product. ( ) ( ) ( ) 2 2 2 3 2 0 0 . 3 2 2 3 m AC m m − − + + = − + − n n o.e. M1 Uses formula for shortest distance. 2 2 2 = = A1 5 10(iii) 2 1 2 3 2 4 ~ 2 2 1 0 4 2         = − −         − −     i j k o.e. M1A1 Finds normal to plane ABC (AEF). 1 2 3 2 4 1 1 1 5     = −     −   i j k o.e. A1 Finds normal to ABD (AEF). 2 2 2 2 2 2 1 8 10 19 cos 378 1 2 2 1 4 5 θ + +   = =     + + + + M1A1FT Uses formula for angle between two lines. 12.2 θ ⇒ = ° A1 CAO. 6

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 18 of 20 Question Answer Marks Guidance 11E(i) 1 1 2 2 d d d 12 6 d d d 18 2 9 y y t t t x t x t t = × = = − − B1 AEF. 2 2 d d d d d d d d y y t t x x x   = ×     M1 Uses chain rule again to find second derivative. ( ) ( ) ( ) 1 1 2 2 2 3 9 6 9 18 2 t t t t t − − + = − − dM1 Uses quotient (or product rule) ( ) ( ) ( ) ( ) 1 2 3 1 3 2 3 9 2 3 9 2 9 2 9 t t t t t t t − −+ + = = − − A1 AG. 4 11E(ii) 56 2 2 0 1 d d 56 d y x x ∫ B1 Uses correct formula for mean value 4 1 2 0 1 6 56 9 t t t t = =     =   −     M1 Finding limits correctly M1 Using expression 1 6 4 3 . 56 9 4 70   = =     −   A1 Inserts correct values of t and obtains answer (AG.) 4

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 19 of 20 Question Answer Marks Guidance 11E(iii) ( ) ( ) 2 2 2 2 d d 18 2 144 4 9 d d x y t t t t t     + = − + = +         M1A1 Simplifies 2 2 d d d d x y t t     +         . ( ) ( ) 4 4 3 5 3 2 2 2 0 0 2 8 2 9 d 32 9 d t t t t t t π π   + = +       ∫ ∫ M1A1 Uses d 2 d d s y t t π ∫ . 4 7 5 2 2 0 2 18 32 7 5 t t π   = +       M1 Integrates term by term. 11 2 83 169984 15300 35 35 π π × = = = A1 Accept exact answer or decimal rounding to 15300. 6 11O(i) ( ) ( ) 2 2 1 2 2 2 1 1 1 2 1 d n n nI x x n x x x −   = − − −     ∫ M1A1 Integrates by parts. ( )( ) 2 1 2 2 1 2 2 1 1 1 d n n x x x − = − −+ − ∫ M1 Uses 2 2 1 1 x x = −+ . 1 2 2 2 n n nI nI − = − − A1 ( ) 1 2 1 2 2 n n n I nI − ⇒ + = − A1 AG. 5

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 20 of 20 Question Answer Marks Guidance 11O(ii) tan sec dx d θ θ θ = M1A1 Differentiates secθ . sec 2 4 π θ θ = ⇒ = sec 1 0 θ θ = ⇒ = B1 Changes limits. ( ) 4 4 2 2n 1 0 0 sec 1 tan sec d tan sec d n nI π π θ θ θ θ θ θ θ + = − = ∫ ∫ B1 Uses 2 2 sec 1 tan θ θ −= . (AG.) 4 11O(iii) 4 7 3 8 0 sin d cos I π θ θ θ = ∫ B1 Deduces that integral is 3I . 0 2 1 I = − (or 1I = 2 2 3 − ) B1 Calculates 0I or 1 I 1 0 1 2 2 3 2 2 . 3 I I I − = − ⇒ = M1A1 Uses reduction formula to find 1 2 I or I ……… 2 1 2 4 7 2 8 5 5 15 I I − = − = Finds 2I . 3 2 2 6 16 9 2 7 7 35 I I − = − = A1 Finds 3I . 5