Cambridge A Level Mathematics - Further 9231 — 2017 Oct/Nov Paper 2 · Variant 2

*0817630226* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/22 Paper 2 October/November 2017 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. 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. Where a numerical value is necessary, take the acceleration due to gravity to be 10 m s−2. 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 23 printed pages and 1 blank page. JC17 11_9231_22/3R © UCLES 2017 [Turn over

2 1 A particle P is moving in a circle of radius 0.8 m. At time t s its velocity is 8 −pt + t2 m s−1, where p is a constant. The magnitude of the transverse component of the acceleration of P when t = 2 is zero. Find the magnitude of the radial component of the acceleration of P when t = 2. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17

3 2 The piston in a large engine rises and falls in simple harmonic motion. When the piston is 1.6 m below its highest level, the rate of change of its height is 3 50 metres per second. When the piston is 0.2 m below its highest level, the rate of change of its height is 1 40 metres per second. Find the amplitude and period of the motion. [7] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

4 3 Three uniform small smooth spheres A, B and C have equal radii and masses m, km and m respectively, where k is a constant. The spheres are moving in the same direction along a straight line on a smooth horizontal surface, with B between A and C. The speeds of A, B and C are 2u, u and 4 3u respectively. The coefficient of restitution between any pair of the spheres is 1 2. After sphere A has collided with sphere B, sphere B collides with sphere C. (i) Find an inequality satisfied by k. [5] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17

5 (ii) Given that k = 2, show that after B has collided with C there are no further collisions between any of the three spheres. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

6 4 1 A B P Q a 2a A small ring P of weight W is free to slide on a rough horizontal wire, one end of which is attached to a vertical wall at Q. The end A of a thin uniform rod AB of length 2a and weight 5 2W is freely hinged to the wall at the point A which is a distance a vertically below Q. A light elastic string of natural length 2a has one end attached to the ring P and the other end attached to the rod at B. The string is at right angles to the rod and A, B, P and Q lie in a vertical plane. The system is in limiting equilibrium with AB making an angle 1 with the horizontal, where sin 1 = 3 5 (see diagram). (i) Find the tension in the string in terms of W. [2] … … … … … (ii) Find the coefficient of friction between the ring and the wire. [2] … … … … … (iii) Find the magnitude of the resultant force on the rod at the hinge in terms of W. [3] … … … © UCLES 2017 9231/22/O/N/17

7 … … … … … … … (iv) Find the modulus of elasticity of the string in terms of W. [3] … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

8 5 O A B C D E F G H A uniform picture frame of mass m is made by removing a rectangular lamina EFGH in which EF = 4a and FG = 2a from a larger rectangular lamina ABCD in which AB = 6a and BC = 4a. The side EF is parallel to the side AB. The point of intersection of the diagonals AC and BD coincides with the point of intersection of the diagonals EG and FH. One end of a light inextensible string of length 10a is attached to A and the other end is attached to B. The frame is suspended from the mid-point O of the string. A small object of mass 11 12m is fixed to the mid-point of AB (see diagram). (i) Show that the moment of inertia of the system, consisting of frame and small object, about an axis through O perpendicular to the plane of the frame, is 169 3 ma2. [7] … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17

9 … … … … … … … (ii) Show that small oscillations of the system about this axis are approximately simple harmonic and state their period. [5] … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

10 6 A pair of fair dice is thrown repeatedly until a pair of sixes is obtained. The number of throws taken is denoted by the random variable X. (i) Find the mean value of X. [2] … … … … … … … (ii) Find the probability that exactly 12 throws are required to obtain a pair of sixes. [2] … … … … … … … (iii) Find the probability that more than 12 throws are required to obtain a pair of sixes. [2] … … … … … … … © UCLES 2017 9231/22/O/N/17

11 7 The random variable X has probability density function f given by fx = T 0.2e−0.2x x ≥0, 0 otherwise. (i) Find the distribution function of X. [2] … … … … … … (ii) Find PX > 2. [2] … … … … … (iii) Find the median of X. [3] … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

12 8 Members of a Statistics club are voting to elect a new president of the club. Members must choose to vote either by post or by text or by email. The method of voting chosen by a random sample of 60 male members and 40 female members is given in the following table. Method of voting Post Text Email Male 10 12 38 Gender Female 5 21 14 Test, at the 1% significance level, whether there is an association between method of voting and gender. [8] … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17

13 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

14 9 The land areas x (in suitable units) and populations y (in millions) for a sample of 8 randomly chosen cities are given in the following table. Land area (x) 1.0 4.5 2.4 1.6 3.8 8.6 7.5 6.5 Population (y) 0.8 8.4 4.2 1.6 2.2 10.2 4.2 5.2 [Σ x = 35.9, Σx2 = 216.47, Σy = 36.8, Σ y2 = 244.96, Σ xy = 212.62.] (i) Find, showing all necessary working, the value of the product moment correlation coefficient for this sample. [3] … … … … … … … … … … … … … (ii) Using a 1% significance level, test whether there is positive correlation between land area and population of cities. [4] … … … … … … © UCLES 2017 9231/22/O/N/17

15 … … … … … … … … … … … … The land areas and populations for another randomly chosen sample of cities, this time of size n, give a product moment correlation coefficient of 0.651. Using a test at the 1% significance level, there is evidence of non-zero correlation between the variables. (iii) Find the least possible value of n, justifying your answer. [2] … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

16 10 A factory produces bottles of an energy juice. Two different machines are used to fill empty bottles with the juice. The manager chooses a random sample of 50 bottles filled by machine X and a random sample of 60 bottles filled by machine Y. The volumes of juice, x and y respectively, measured in appropriate units, are summarised by Σx = 45.5, Σx −x2 = 19.56, Σ y = 72.3, Σy −y2 = 30.25, where x and y are the sample means of the volume of juice in the bottles filled by X and Y respectively. (i) Find a 90% confidence interval for the difference between the mean volume of juice in bottles filled by machine X and the mean volume of juice in bottles filled by machine Y. [7] … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17

17 A test at the !% significance level does not provide evidence that there is any difference in the means of the volume of juice in bottles filled by machine X and the volume of juice in bottles filled by machine Y. (ii) Find the set of possible values of !. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

18 11 Answer only one of the following two alternatives. EITHER ! " " O A A′ B B′ a A particle P of mass m is free to move on the smooth inner surface of a fixed hollow sphere of radius a. The centre of the sphere is O. The points A and A′ are on the inner surface of the sphere, on opposite sides of the vertical through O; the radius OA makes an angle ! with the downward vertical and the radius OA′ makes an angle " with the upward vertical. The point B is on the inner surface of the sphere, vertically below O. The point B′ is on the inner surface of the sphere and such that OB′ makes an angle 2" with the upward vertical through O (see diagram). It is given that cos ! = 1 16. (i) P is projected from A with speed u along the surface of the sphere downwards towards B. Subsequently it loses contact with the sphere at A′. Show that u2 = 1 8ag1 + 24 cos ". [5] … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17

19 (ii) P is now projected from B with speed u along the surface of the sphere towards B′. Subsequently it loses contact with the sphere at B′. Find cos ". [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

20 (iii) In part (i), the reaction of the sphere on P when it is initially projected at A is R. Find R in terms of m and g. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17

21 OR A large number of people attended a course to improve the speed of their logical thinking. The times taken to complete a particular type of logic puzzle at the beginning of the course and at the end of the course are recorded for each person. The time taken, in minutes, at the beginning of the course is denoted by x and the time taken, in minutes, at the end of the course is denoted by y. For a random sample of 9 people, the results are summarised as follows. Σ x = 45.3 Σx2 = 245.59 Σy = 40.5 Σy2 = 195.11 Σxy = 218.72 Ken attended the course, but his time to complete the puzzle at the beginning of the course was not recorded. His time to complete the puzzle at the end of the course was 4.2 minutes. (i) By finding, showing all necessary working, the equation of a suitable regression line, find an estimate for the time that Ken would have taken to complete the puzzle at the beginning of the course. [5] … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17 [Turn over

22 The values of x −y for the sample of 9 people are as follows. 0.2 0.8 0.5 1.0 0.2 0.6 0.2 0.5 0.8 The organiser of the course believes that, on average, the time taken to complete the puzzle decreases between the beginning and the end of the course by more than 0.3 minutes. (ii) Stating suitable hypotheses and assuming a normal distribution, test the organiser’s belief at the 21 2% significance level. [9] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/O/N/17

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24 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 2017 9231/22/O/N/17

® IGCSE is a registered trademark. This document consists of 15 printed pages. © UCLES 2017 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/22 Paper 2 October/November 2017 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 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 2 of 15 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 FT 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.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 3 of 15 The following abbreviations may be used in a mark scheme or used on the scripts: 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) 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.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 4 of 15 Question Answer Marks Guidance 1 – p + 2t = 0, p = 4 M1 A1 Find p by equating transverse acceln. to 0 at t = 2 aR = (8 – pt + t 2) 2 / 0⋅8 M1 Find radial acceleration aR at t = 2 in terms of p from v2/r = 42 / 0⋅8 = 20 [m s-2] A1 Evaluate with p = 4 4 Question Answer Marks Guidance 2 (3π/5)2 = ω2 (a2 – (a – 1⋅6)2 ) (AEF) M1 A1 Use v2 = ω2 (a2 – x2) in each posn. (M1 for either) (π/4)2 = ω2 (a2 – (a – 0⋅2)2 ) (3π/5)2 (0⋅4a – 0⋅04) = (π/4)2 (3⋅2a – 2⋅56) A1 Combine to find amplitude a and ω2 (or ω) a = 2⋅6 [m] M1 A1 (M1 for either) ω2 = (π/4)2 or ω = π/4 (AEF) A1 Find other unknown T = 2π / (π/4) = 8 [s] B1FT Find period T from T = 2π/ω (√ on ω2 or ω) 7

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 5 of 15 Question Answer Marks Guidance 3(i) mvA + kmvB = 2mu + kmu (AEF) [vA + kvB = 2u + ku] M1 Use conservation of momentum for A & B (allow omission of m in all momentum eqns) vB – vA = ½ (2u – u) [= ½ u] M1 Use Newton’s restitution law with consistent LHS signs vB = u (2k + 5) / 2(k + 1) or u (k + 5/2) / (k + 1) (AEF) A1 Combine to find vB [vA = u (k + 4) / 2(k + 1)] vB > 4u/3 if k < 7/2 M1 A1 Find inequality for k from speeds of B and C after 1st collision 5 3(ii) kmwB + mvC = kmvB + m(4u/3) (AEF) [2wB + vC = 2(3u/2) + 4u/3 = 13u/3 when k = 2] M1 Use conservation of momentum for B & C vC – wB = ½ (vB – 4u/3) [= u/12] (k + 1) wB = (k – ½) vB + 2u M1 Use Newton’s restitution law with consistent LHS signs Combine to find wB 3 wB = (3/2) vB + 2u with vB = 3u/2, so wB = 17u/12 *A1 when k = 2 vA = u, vA < wB DB1 Verify no further collisions between A and B EITHER: (k + 1) vC = (3k/2) vB + (2 – k) (2u/3) 3 vC = 3 vB with vB = 3u/2 so vC = 3u/2 > wB (DB1) EITHER: Find vC and verify no further collisions between B and C OR: B and C cannot meet again since they move apart after colliding (AEF) (DB1) OR: State explicitly that no further collisions between B and C 5

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 6 of 15 Question Answer Marks Guidance 4(i) T × 2a = (5W/2) × a cos θ M1 Take moments for rod about A T = ½ (5W/2) × (4/5) = W A1 to find tension T 2 4(ii) T sin θ = µ (W + T cos θ ) M1 Use FP = µRP at P (3/5) W = µ (1 + 4/5)W = µ (9/5)W, µ = 1/3 A1 to find µ 2 4(iii) EITHER: [±] X = T sin θ = 3W/5 or 0⋅6W [±] Y = 5W/2 – T cos θ or 5W/2 + W – RP (B1 EITHER: Find horizontal component X of force at A …………… Find vertical component Y of force at A = 17W/10 or 1⋅7W B1) OR: [±] X = (5W/2) sin θ = 3W/2 or 1⋅5W (B1 OR: Find component X of force at A along BA [±] Y = (5W/2) cos θ – T = W B1) Find component Y of force at A perp. to BA RA 2 = X2 + Y2 = 13W2/4, RA = ½W√13 or 1⋅80 W B1FT Find magnitude of resultant force RA at A (FT on X, Y) 3 4(iv) PB = (a + 2a sin θ ) / cos θ = 5a/4 + 3a/2 = 11a/4 or 2⋅75a or x = 3a/4 or 0⋅75a B1 Find length PB or extension x of string T = λ (PB – 2a)/2a, λ = 8W/3 or 2⋅67W M1 A1 Find modulus λ using Hooke’s Law 3

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 7 of 15 Question Answer Marks Guidance 5(i) AABCD = 24a2, AEFGH = 8a2, AFrame = 16a2 MABCD = 3m/2 and mEFGH = m/2 B1 Use areas to find masses MABCD and mEFGH IABCD = ⅓ MABCD ((3a)2 + (2a)2) [= (13/3) MABCD a2 = (13/2) ma2] B1 Find MI of ABCD about axis at centre Q IEFGH = ⅓ mEFGH ((2a)2 + a2) [= (5/3) mEFGH a2 = (5/6) ma2] B1 Find MI of EFGH about axis at centre Q IABCD – IEFGH + m × (6a)2 [= (121/2 – 113/6) ma2 or (17/3 + 36) ma2 = (125/3) ma2] M1 A1 Find MI of frame about axis at O Result also follows from combining four rectangular parts IObject = (11m/12) × (4a)2 [= (44/3) ma2] B1 Find MI of small object about axis at O I = (13/2 – 5/6 + 36 + 44/3) ma2 = (169/3) ma2 AG A1 Combine to verify MI of system about axis at O 7 5(ii) [–] I d2θ/dt2 or [–] Iα = mg × 6a sin θ + (11/12)mg × 4a sin θ or (23/12)mg × (116/23)a sin θ [ = (29/3) mga sin θ ] M1 A1 Use eqn of circular motion to relate d2θ/dt2 to sin θ, where θ is angle of QO with vertical d2θ/dt2 or α = – (29g / 169a) θ or – (0⋅172 g / a) θ M1 A1 Approximate sin θ by θ to show SHM (M0 if wrong sign or cos θ ≈ θ used) T = 2π / √(29g/169a) = 26π √(a/29g) or 15⋅2√(a/g) or 4⋅80√a (AEF) A1 Find period T from T = 2π/ω (A1 requires some simplification) 5

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 8 of 15 Question Answer Marks Guidance 6(i) p = (1/6)2 or 1/36 B1 Find (or imply) probability p of pair of sixes in one throw 1/p = 36 B1 Find mean value of X 6(ii) P(X = 12) = p (1 – p)11 = 0⋅0204 M1 A1 Find prob. of needing exactly 12 throws 6(iii) P(X > 12) = (1 – p)12 = 0⋅713 M1 A1 Find prob. of needing more than 12 throws 2 Question Answer Marks Guidance 7(i) F(x) = ∫ f(x) dx = – e-0.2x + c = 1 – e-0.2x (x ≥ 0) M1 State, or integrate and use F(0) = 0 or F(x) → 1 as x → ∞ and F(x) = 0 (x < 0 or otherwise) A1 to find, F(x) (A0 if case x < 0 omitted) 2 7(ii) P(X > 2) = 1 – F(2) = e-0.4 = 0⋅670 M1 A1 Find P(X > 2): (M0 for F(2)) 3 7(iii) 1 – e-0.2m = ½ , e0.2m = 2 M1 Find median value m from F(m) or 1 – F(m) = ½ m = 5 ln 2 or 3⋅47 M1 A1 3

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 9 of 15 Question Answer Marks Guidance 8 H0: No association or method is independent of gender (AEF) B1 State (at least) null hypothesis Ei: 9⋅0 19⋅8 31⋅2 6⋅0 13⋅2 20⋅8 M1 A1 Find expected values Ei (A0 if rounded to integers) χ2 = 0⋅111 + 3⋅073 + 1⋅482 + 0⋅167 + 4⋅610 + 2⋅223 (to 2 d.p.) = 11⋅7 (or 12⋅3) M1 A1 Find value of χ2 from Σ (Ei – Oi)2 / Ei [or Σ Oi 2/Ei – n ] (allow 12⋅3 if integer values of Ei used) χ2, 0.99 2 = 9⋅21 B1 State or use correct tabular χ2 value Reject H0 if χ2 > tabular value (AEF) M1 Valid method for reaching conclusion 11⋅7 [± 0⋅1] > 9⋅21 so there is an association A1 Correct conclusion, from correct values 8 Question Answer Marks Guidance 9(i) r = Sxy / √(Sxx Syy) with e.g. Sxy = 212⋅62 – 35⋅9 × 36⋅8/8 = 47⋅48 (or 5⋅935) Sxx = 216⋅47 – 35⋅92/8 = 55⋅37 (or 6⋅921) Syy = 244⋅96 – 36⋅82/8 = 75⋅68 (or 9⋅46) (all to 3 s.f.) M1 A1 Find correlation coefficient r (Insufficient working loses first A1) r = 0⋅733 *A1 3

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 10 of 15 Question Answer Marks Guidance 9(ii) H0: ρ = 0, H1: ρ > 0 B1 State both hypotheses (B0 for r …) EITHER: r8, 1% = 0⋅789 (*B1 State or use correct tabular one-tail r-value Accept H0 if |r| < tab. r-value (AEF) M1) State or imply valid method for conclusion OR: tr = r√((n-2) / (1 – r2)) = 2⋅64, t6,0⋅99 = 3⋅143 (*B1 Accept H0 if | tr | < tab. t-value (AEF) M1) No positive correlation (AEF) DA1 4 9(iii) r14, 1% = 0⋅661, r15, 1% = 0⋅641 so nmin = 15 M1 A1 Find nmin from relevant two-tail tabular value[s] SC: Award B1 for stating 15 without any justification 2

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 11 of 15 Question Answer Marks Guidance 10(i) x = 0⋅91, y = 1⋅205 sX 2 = 19⋅56 / 49 [= 489/1225 or 0⋅3992] and sY 2 = 30⋅25 / 59 [= 121/236 or 0⋅5127] B1 M1 Find both sample means Estimate both population variances (allow biased here: 0⋅3912 and 0⋅5042) EITHER: s2 = sX 2/50 + sY 2/60 (M1 Estimate or imply combined variance = 0⋅01653 or 0⋅12862 (to 3 s.f. throughout) A1 [±] (y –x) ± z s M1 Find confidence interval for difference Y – X or X – Y z0.95 = 1⋅645 [±] 0⋅295 ± 0⋅211 (allow 0⋅212) A1 Use appropriate tabular value Evaluate confidence interval (either form) or [±] [0⋅084, 0⋅506] (allow [±] [0⋅083, 0⋅507]) A1) OR: Assume equal [population] variances s2 = (49 sX 2 + 59 sY 2) / 108 or (19⋅56 + 30⋅25) / 108 (B1 State assumption Find or imply pooled estimate of common variance (note sX 2 and sY 2 not needed explicitly so first M1 may be implied by result) = 4981/10800 or 0⋅461 or 0⋅6792 B1 [±] (y –x) ± z s √(1/50 + 1/60) M1 Find confidence interval for difference Y – X or X – Y z0.95 = 1⋅645 A1 Use appropriate tabular z-value (or t-value from calculator) [±] 0⋅295 ± 0⋅214 or [±] [0⋅081, 0⋅509] A1) Evaluate confidence interval (either form) 7

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 12 of 15 Question Answer Marks Guidance 10(ii) z = (1⋅205 – 0⋅91) /s = 0⋅295 /s = 2⋅29[4] [or 2⋅26[9]] (to 3 s.f.) M1 A1 Find value of z (either sign) (z) = 0⋅989[1] [or 0⋅988[4]] A1 Find Φ(z) 100 × (1 – 0⋅989) × 2 = 2⋅2 [or 2⋅3] (to 1 d.p.) M1 A1 Find limiting value for α, based on two-tail test (M0 for basing on one-tail test) α < (or ⩽) 2⋅2 [or 2⋅3] A1 Find set of possible values of α (Treat α instead of α % as misread) 6 Question Answer Marks Guidance 11A(i) ½mv2 = ½mu2 – mga (cos α + cos β ) M1 A1 Find v2 at A′ from conservation of energy (A0 if no m) mv2/a = mg cos β u2 = ag cos β + 2ag (1/16 + cos β) B1 Use F = ma radially at A′ with RA ′ = 0 Use cos α = 1/16 and eliminate v2 to verify u2 = (1/8) ag (1 + 24 cos β) AG M1 A1 5

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 13 of 15 Question Answer Marks Guidance 11A(ii) ½mw2 = ½mu2 – mga (1 + cos 2β ) B1 Find w2 at B′ from conservation of energy (B0 if no m) mw2/a = mg cos 2β u2 = ag cos 2β + 2ag (1 + cos 2β) B1 Use F = ma radially at B′ with RB ′ = 0 Eliminate w2 to find u2 = ag (2 + 3 cos 2β) M1 1 + 24 cos β = 8 (2 + 3 cos 2β) = 16 + 48 cos2 β – 24 16 cos2 β – 8 cos β – 3 = 0 M1 Combine eqns for u2 Formulate and solve quadratic to find cos β cos β = ¾ [rejecting – ¼] M1 A1 6 11A(iii) u2 = (19/8) ag or 2⋅375 ag R = mu2/a + mg cos α = (19/8 + 1/16) mg B1 Find u2 using value of cos β Use F = ma radially at A to find reaction R at A = (39/16) mg or 2⋅44 mg M1 A1 3

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 14 of 15 Question Answer Marks Guidance 11B(i) e.g. Sxy = 218⋅72 – 45⋅3 × 40⋅5/9 = 14⋅87 or 1⋅652 Sxx = 245⋅59 – 45⋅32/9 = 17⋅58 or 1⋅953 or Syy = 195⋅11 – 40⋅52/9 = 12⋅86 or 1⋅429 Find reqd. values EITHER: b1 = Sxy / Sxx = 14⋅87 /17⋅58 = 0⋅84585 = 0⋅846 (y – 40⋅5/9) = b1 (x – 45⋅3/9) (y – 4⋅5) = 0⋅846 (x – 5⋅033) (M1 A1 Find gradient in y –y = b1 (x –x) to 3 s.f. Find eqn. of regression line to 3 s.f. y = 0⋅846x + 0⋅24257 = 0⋅846x + 0⋅243 M1 A1 x = 4⋅68 A1) Find x when y = 4⋅2 OR: b2 = Sxy / Syy = 14⋅87 /12⋅86 = 1⋅1563 = 1⋅16 (x – 45⋅3/9) = b2 (y – 40⋅5/9) (x – 5⋅033) = 1⋅16 (y – 4⋅5) (M1 A1 Find gradient in x –x = b2 (y –y) to 3 s.f. Find eqn. of regression line to 3 s.f. x = 1⋅16y – 0⋅170[04] M1A1 x = 4⋅69 A1) Find x when y = 4⋅2 (A0 for x = 4⋅70) 5

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 15 of 15 Question Answer Marks Guidance 11B(ii) H0: µx – µy = 0⋅3, H1: µx – µy > 0⋅3 (AEF) B1 State hypotheses (B0 forx … ) d = 4⋅8 / 9 or 8/15 or 0⋅533 (where d = x – y) B1 Find sample mean s2 = (3⋅26 – 4⋅82/9) / 8 = 7/80 or 0⋅0875 or 0⋅2962 M1 A1 Estimate population variance (allow biased here: 7/90 or 0⋅0778 or 0⋅2792) t = (d – 0⋅3) / (s/√9) = 2⋅37 M1 A1 Find value of t t8, 0.975 = 2⋅306 or 2⋅31 B1 State or use correct tabular t-value (or can compared with 0⋅3 + t8, 0.975 s/√ 9 = 0⋅527) Reject H0 if t > tabular value (AEF) 2⋅37 [± 0⋅1] > 2⋅31 so accept belief M1 Valid method for reaching conclusion Correct conclusion, from correct values [of increase of more than 0⋅3] (AEF) A1 SC: Wrong (hypothesis) test can earn only B1 for hypotheses 9