Cambridge A Level Mathematics - Further 9231 — 2017 May/June Paper 2 · Variant 2

*7169326746* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/22 Paper 2 May/June 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 24 printed pages and 4 blank pages. JC17 06_9231_22/2R © UCLES 2017 [Turn over

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3 1 A bullet of mass 0.08 kg is fired horizontally into a fixed vertical barrier. It enters the barrier horizontally with speed 300 m s−1 and emerges horizontally after 0.02 s. There is a constant horizontal resisting force of magnitude 1000 N. Find the speed with which the bullet emerges from the barrier. [3] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17 [Turn over

4 2 O A E D B a a 5 4a 3a A uniform smooth disc with centre O and radius a is fixed at the point D on a horizontal surface. A uniform rod of length 3a and weight W rests on the disc with its end A in contact with a rough vertical wall. The rod and the disc lie in a vertical plane that is perpendicular to the wall. The wall meets the horizontal surface at the point E such that AE = a and ED = 5 4a. A particle of weight kW is hung from the rod at B (see diagram). The coefficient of friction between the rod and the wall is 1 8 and the system is in limiting equilibrium. Find the value of k. [8] … … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17

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6 3 Two uniform small smooth spheres A and B have equal radii and masses 3m and m respectively. Sphere A is moving with speed u on a smooth horizontal surface when it collides directly with sphere B which is at rest. The coefficient of restitution between the spheres is e. (i) Find, in terms of u and e, expressions for the velocities of A and B after the collision. [3] … … … … … … … … … … … … … Sphere B continues to move until it strikes a fixed smooth vertical barrier which is perpendicular to the direction of motion of B. The coefficient of restitution between B and the barrier is 3 4. When the spheres subsequently collide, A is brought to rest. (ii) Find the value of e. [7] … … … … … … … © UCLES 2017 9231/22/M/J/17

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8 4 A B C a 2a 2a 2a Three identical uniform discs, A, B and C, each have mass m and radius a. They are joined together by uniform rods, each of which has mass 1 3m and length 2a. The discs lie in the same plane and their centres form the vertices of an equilateral triangle of side 4a. Each rod has one end rigidly attached to the circumference of a disc and the other end rigidly attached to the circumference of an adjacent disc, so that the rod lies along the line joining the centres of the two discs (see diagram). (i) Find the moment of inertia of this object about an axis l, which is perpendicular to the plane of the object and through the centre of disc A. [6] … … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17

9 The object is free to rotate about the horizontal axis l. It is released from rest in the position shown, with the centre of disc B vertically above the centre of disc A. (ii) Write down the change in the vertical position of the centre of mass of the object when the centre of disc B is vertically below the centre of disc A. Hence find the angular velocity of the object when the centre of disc B is vertically below the centre of disc A. [4] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17 [Turn over

10 5 A ! B C O a 1 3a 2 3a ag A particle of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The point A is such that OA = a and OA makes an angle ! with the upward vertical through O. The particle is held at A and then projected downwards with speed ag so that it begins to move in a vertical circle with centre O. There is a small smooth peg at the point B which is at the same horizontal level as O and at a distance 1 3a from O on the opposite side of O to A (see diagram). (i) Show that, when the string first makes contact with the peg, the speed of the particle is ag1 + 2 cos !. [2] … … … … … … … … The particle now begins to move in a vertical circle with centre B. When the particle is at the point C where angle CBO = 150Å, the tension in the string is the same as it was when the particle was at the point A. (ii) Find the value of cos !. [10] … … … … © UCLES 2017 9231/22/M/J/17

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12 6 A fair die is thrown repeatedly until a 6 is obtained. (i) Find the probability that obtaining a 6 takes no more than four throws. [2] … … … … … … … … … … … (ii) Find the least integer N such that the probability of obtaining a 6 before the Nth throw is more than 0.95. [3] … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17

13 7 A farmer grows a particular type of fruit tree. On average, the mass of fruit produced per tree has been 6.2 kg. He has developed a new kind of soil and claims that the mean mass of fruit produced per tree when growing in this new soil has increased. A random sample of 10 trees grown in the new soil is chosen. The masses, x kg, of fruit produced are summarised as follows. Σ x = 72.0 Σx2 = 542.0 Test at the 5% significance level whether the farmer’s claim is justified, assuming a normal distribution. [7] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17 [Turn over

14 8 The continuous random variable X has probability density function f given by fx = T 1 4x −1 2 ≤x ≤4, 0 otherwise. (i) Find the distribution function of X. [3] … … … … … … … … … … … … The random variable Y is defined by Y = X −13. (ii) Find the probability density function of Y. [4] … … … … … … … … © UCLES 2017 9231/22/M/J/17

15 … … … … … … … … … … … … (iii) Find the median value of Y. [3] … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17 [Turn over

16 9 Two fish farmers X and Y produce a particular type of fish. Farmer X chooses a random sample of 8 of his fish and records the masses, x kg, as follows. 1.2 1.4 0.8 2.1 1.8 2.6 1.5 2.0 Farmer Y chooses a random sample of 10 of his fish and summarises the masses, y kg, as follows. Σ y = 20.2 Σ y2 = 44.6 You should assume that both distributions are normal with equal variances. Test at the 10% significance level whether the mean mass of fish produced by farmer X differs from the mean mass of fish produced by farmer Y. [10] … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17

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18 10 A random sample of 5 pairs of values x, y is given in the following table. x 1 2 4 5 8 y 7 5 8 6 4 (i) Find, showing all necessary working, the equation of the regression line of y on x. [4] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17

19 (ii) Find, showing all necessary working, the value of the product moment correlation coefficient for this sample. [3] … … … … … … … … … … … … (iii) Test, at the 10% significance level, whether there is evidence of non-zero correlation between the variables. [4] … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17 [Turn over

20 11 Answer only one of the following two alternatives. EITHER A particle P of mass 3m is attached to one end of a light elastic spring of natural length a and modulus of elasticity kmg. The other end of the spring is attached to a fixed point O on a smooth plane that is inclined to the horizontal at an angle !, where sin ! = 2 3. The system rests in equilibrium with P on the plane at the point E. The length of the spring in this position is 5 4a. (i) Find the value of k. [3] … … … … … … … … The particle P is now replaced by a particle Q of mass 2m and Q is released from rest at the point E. (ii) Show that, in the resulting motion, Q performs simple harmonic motion. State the centre and the period of the motion. [6] … … … … … … … … … … © UCLES 2017 9231/22/M/J/17

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22 (iii) Find the least tension in the spring and the maximum acceleration of Q during the motion. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17

23 OR A shop is supplied with large quantities of plant pots in packs of six. These pots can be damaged easily if they are not packed carefully. The manager of the shop is a statistician and he believes that the number of damaged pots in a pack of six has a binomial distribution. He chooses a random sample of 250 packs and records the numbers of damaged pots per pack. His results are shown in the following table. Number of damaged pots per pack (x) 0 1 2 3 4 5 6 Frequency 48 69 78 32 22 1 0 (i) Show that the mean number of damaged pots per pack in this sample is 1.656. [1] … … … … … The following table shows some of the expected frequencies, correct to 2 decimal places, using an appropriate binomial distribution. Number of damaged pots per pack (x) 0 1 2 3 4 5 6 Expected frequency 36.01 82.36 a 39.89 b 1.74 0.11 (ii) Find the values of a and b, correct to 2 decimal places [5] … … … … … … … … … © UCLES 2017 9231/22/M/J/17 [Turn over

24 … … … … … … … … … … … (iii) Use a goodness-of-fit test at the 1% significance level to determine whether the manager’s belief is justified. [8] … … … … … … … … … … … … … © UCLES 2017 9231/22/M/J/17

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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 2017 9231/22/M/J/17

® IGCSE is a registered trademark. This document consists of 14 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/22 Paper 2 May/June 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 2 of 14 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 May/June 2017 © UCLES 2017 Page 3 of 14 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 May/June 2017 © UCLES 2017 Page 4 of 14 Question Answer Marks Guidance 1 0⋅08 × (300 – v) = 1000 × 0⋅02 (AEF) M1 A1 Find eqn for exit speed v from e.g. change in momentum = Ft (if 300 + v or equivalent, can allow M1 only) v = 300 – 250 = 50 [m s–1] A1 Total: 3 2 Take moments for rod about some point such as: A: RP × AP – kW × 3a cos θ = W × (3a/2) cos θ [ RP × 3a / 4 – kW × 9a / 5 = W × 9a/10 so 15 RP – 36 kW = 18 W ] O: FA × (5a/4) – kW × (3a cos θ – 5a/4) = – W × (5a/4 – (3a/2) cos θ) [ FA × 5a / 4 – kW × 11a / 20 = – W × 7a/20 so 25 FA – 11 kW = – 7 W ] P: RA × AP sin θ + FA × AP cos θ – kW × (3a – AP) cos θ = W × (3a/2 – AP) cos θ) [ RA × 3a/5 + FA × 9a/20 – kW × 27a/20 = W × 9a/20 so 12 RA + 9 FA – 27 kW = 9 W ] B: RP × (3a – AP) – RA × 3a sin θ – FA × 3a cos θ = W × (3a/2) cos θ [ RP × 9a/4 – RA × 12a/5 – FA × 9a/5 = W × 9a/10 so 45 RP – 48 RA – 36 FA = 18 W ] C: RP × (3a/2 – AP) – RA × (3a/2) sin θ – FA × (3a/2) cos θ + kW × (3a/2) cos θ = 0 [ RP × 3a/4 – RA × 6a/5 – FA × 9a/10 + kW × 9a/10 = 0 so 15 RP – 24 RA – 18 FA + 18 kW = 0 ] F: RP cos θ × (3a – AP) cos θ – RP sin θ × AP sin θ – FA × 3a cos θ = W × (3a/2) cos θ [ (3/5)RP × 27a/20 – (4/5)RP × 3a/5 – FA × 9a/5 = W × 9a/10 so 81 RP – 48 RP – 180 FA = 90 W ] M1 A1 FA here denotes friction on rod measured in downward dirn; P denotes point of contact of rod and disc; θ denotes angle between rod and horizontal. [AP = 3a/4, sin θ = 4/5, cos θ = 3/5, tan θ = 4/3, 3a – AP = 9a/4, 3a/2 – AP = 3a/4] See note below on solving question without introducing RP (C denotes mid-point of AB) (F is vertically below B, on AO extended)

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 14 Question Answer Marks Guidance Find two more indep. eqns, e.g. resolution of forces on rod: Horizontally: RA = RP sin θ [= 4RP /5] Vertically: FA + (k + 1)W = RP cos θ [= 3RP /5] Along AB: RA cos θ = FA sin θ + (k + 1)W sin θ Normal to AB: RP = RA sin θ + FA cos θ + (k + 1)W cos θ B1 B1 A second moment eqn. may be used instead of a resolution Count as 2 eqns if used with moments about P (so RP absent) FA = +RA / 8 or – RA / 8 as appropriate B1 Relate FA and RA (may be implied; and must be consistent with friction taken down or up in above eqns) [sin θ = 4/5, cos θ = 3/5, tan θ = 4/3] M1 Eliminate θ from all reqd. independent eqns. for forces Find either value of k from reqd. independent eqns. for forces FA ↓: [RP = 6W, FA = 3W/5, RA = 24W/5], k = 2 FA ↑: [RP = 30W/17, FA = 3W/17, RA = 24W/17], k = 4/17 M1 A1 (or 0⋅235) Total: 8

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 14 Question Answer Marks Guidance 3(i) 3mvA + mvB = 3mu, vB – vA = eu (AEF) M1 Use momentum and Newton’s law (M0 if inconsistent LHS signs; allow 3vA + vB = 3u) vA = ¼ (3 – e) u, vB = ¾ (1 + e) u A1, A1 Combine to find velocities of A and B after colln. (signs must be consistent with chosen direction) Total: 3 3(ii) vB′ = – ¾ vB [= – (9/16) (1 + e)u] (AEF) B1 Relate velocity vB′ of B after colln. with wall to vB [3mVA +] mVB = 3mvA + mvB′ [VB = 3 (9 – 7e) u/16] M1 Use momentum (allow m omitted and VA = 0) VB [– VA ] = – e(vB′ – vA) [VB = e (21 + 5e) u/16] M1 Use Newton’s law EITHER: [4VA =] (3 – e) vA + (1 + e) vB′ = 0 ¼(3 – e) 2 – (9/16)(1 + e) 2 = 0 (AEF) (M1 A1) Eliminate VB with VA = 0 and substitute for vA and vB′ OR: 3 (9 – 7e) = e (21 + 5e) (M1 A1) 5e2 + 42e – 27 = 0, e = 3/5 or 0⋅6 M1 A1 Form and solve quadratic for e, rejecting root –9 Total: 7

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 14 Question Answer Marks Guidance 4(i) Idiscs = ½ ma2 + 2 × {½ ma2 + m (4a)2 } [= (½ + 2 × {33/2)} ma2 = 67 ma2/2] M1 A1 Find MI of discs about axis l IAB or IAC = ⅓(⅓m) a2 + (⅓m) (2a)2 (AEF) [= 13 ma2/9] M1 A1 Find MI of e.g. rod joining one of A,B or A,C about axis l (M1 for finding MI of any of the 3 rods) IBC = ⅓(⅓m) a2 + (⅓m) (2a√3)2 (AEF) [= 37 ma2/9] A1 Find MI of rod joining B,C about axis l I = (67/2 + 37/9 + 2 × 13/9) ma2 = 81 ma2/2 A1 Combine to find MI of object about axis l Total: 6 4(ii) h = 4a B1 Find or state vertical change h of centre of mass ½ I ω2 = 4 mgh, ω2 = 64g /81a M1 A1 FT Find angular velocity ω when B below A by energy (FT on I) ω = (8/9) √(g/a) or 0⋅889 √(g/a) or 2⋅81/√a A1 (requires some simplification for this A1) Total: 4 5(i) ½mv1 2 = ½mu2 + mga cos α v1 2 = ag + 2 ag cos α, v1 = √ (ag (1 + 2 cos α)) AG M1 A1 Verify v1 for string horizontal by consvn of energy (A0 if no m) Total: 2

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 8 of 14 Question Answer Marks Guidance 5(ii) TA + mg cos α = m (√ag)2 /a, TA = mg (1 – cos α) ½mv2 2 = ½mu2 + mga cos α – mg ⅔ a cos 60° M1 A1 Find tension TA at A from F = ma radially Find v2 2 at C by consvn. of energy (A0 if no m) or ½mv1 2 – mg ⅔ a cos 60° M1 A1 v2 2 = ag + 2ag cos α – ⅔ ag = ag (⅓ + 2 cos α) A1 TC + mg cos 60° = m v2 2 / ⅔ a [= 3m v2 2 / 2 a] [TC = 3mg cos α] M1 A1 Find tension TC at C from F = ma radially mg (1 – cos α) = 3mg (⅓ + 2 cos α)/2 – ½mg M1 A1 Find cos α from TA = TC and substituting for v2 2 1 – cos α = 3 cos α, cos α = ¼ A1 Total: 10

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 9 of 14 Question Answer Marks Guidance 6(i) P(X ≤ 4) = 1 – q4 M1 Find prob. of score of 6 on no more than 4 throws = 671/1296 or 0⋅518 A1 Set q = 5/6 and evaluate Total: 2 6(ii) 1 – qN – 1 > 0⋅95 M1 Formulate condition for N (1 – qN is M0) (5/6) N – 1 < 0⋅05, N – 1 > log 0⋅05 / log 5/6 M1 Set q = 5/6, rearrange and take logs (any base) to give bound N – 1 > 16⋅4[3], Nmin = 18 A1 Find Nmin (N – 1 < 16⋅4 or N – 1 = 16⋅4 earns M1 M1 A0) Total: 3 7 x = 7⋅2 B1 Find sample mean s2 = (542 – 722 /10) / 9 [ = 118/45 or 2⋅622 or 1⋅6192 ] M1 Estimate population variance (allow biased here: 2⋅36 or 1⋅5362) H0: µ = 6⋅2, H1: µ > 6⋅2 (AEF) B1 State hypotheses (B0 forx …) t9, 0.95 = 1⋅83[3] B1 State or use correct tabular t-value t = (x – 6⋅2) / (s/√10) = 1⋅95 [Accept H1:] M1 A1 Find value of t (or can comparex with 6⋅2 + 0⋅939 = 7⋅14 ) Consistent conclusion Claim (of mean mass increased) is justified (AEF) B1 FT (FT on both t-values) Total: 7

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 10 of 14 Question Answer Marks Guidance 8(i) F(x) = ∫ f(x) dx = x2/8 – x/4 [+ c] M1 Find or state distribution function F(x) for 2 ⩽ x ⩽ 4 using F(2) = 0 or F(4) = 1 to find c if necessary = x2/8 – x/4 or {(x – 1)2 – 1}/8 (AEF) A1 State F(x) for other values of x F(x) = 0 (x < 2), F(x) = 1 (x > 4) A1 Total: 3 8(ii) EITHER: G(y) = P(Y < y) = P((X – 1)3 < y) = P(X < 1 + y1/3) = F(1 + y1/3) = (1 + y1/3)2 /8 – (1 + y1/3)/4 or (y 2/3 – 1)/8 (M1 A1) Find or state G(y) for 2 ⩽x ⩽ 4 from Y = (X – 1)3 (allow < or ⩽ throughout) OR: Use x = 1 + y1/3 to find f(x) = ¼ y1/3 and dx/dy = ⅓ y –2/3 (M1 A1) Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy g(y) [= G′(y)] = (1/12) y –1/3 or 1 / (12 y1/3) A1 Find g(y) in simplified form for 1 ⩽ y ⩽ 27 [g(y) = 0 otherwise] A1 State corresponding range of y for G(y) or g(y) Total: 4 8(iii) (m2/3 – 1)/8 = ½ M1 Find median value m of Y from G(m) = ½ m2/3 = 5, m = √125 or 5√5 or 11⋅2 M1 A1 Total: 3

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 11 of 14 Question Answer Marks Guidance 9 H0: µX = µY , H1: µX ≠ µY (AEF) B1 State hypotheses (B0 forx …) x = 13⋅4/8 or 1⋅67[5], y = 2⋅02 (all to 3 s.f.) B1 Find sample means (values to 3 s.f. throughout) sX 2 = (24⋅7 – 13⋅42/8) / 7 = 451/1400 or 0⋅3221 or 0⋅56782 and sY 2 = (44⋅6 – 20⋅22/10) / 9 = 949/2250 or 0⋅4218 or 0⋅64942 M1 Estimate or imply popln. variances (allow biased here: 0⋅2819 or 0⋅53092) (allow biased here: 0⋅3796 or 0⋅61612) s2 = (7 sX 2 + 9 sY 2) / 16 (AEF) or (24⋅7 – 13⋅42/8 + 44⋅6 – 20⋅22/10) / 16 M1 A1 Estimate (pooled) common variance (note sX 2 and sY 2 not needed explicitly) = 6051/16 000 or 0⋅3782 or 0⋅61502 A1 t16, 0.95 = 1⋅746 *B1 State or use correct tabular t value [–] t = (y –x) / s √(1/8 + 1/10) = 1⋅18 M1 A1 Find value of t (or can comparey –x = 0⋅345 with 0⋅509) t < 1⋅75 so mean masses are the same (AEF) DB1 FT Correct conclusion (FT on t, dep *B1) SR: Ζ = (y –x) / √(sX 2/8 + sY 2/10) = 0⋅345 / √(0⋅078) = 1⋅20 (B1) SR: Implicitly taking sX 2, sY 2 as unequal popln. variances (may also earn first B1 B1 M1) Ζ < 1⋅645 so mean masses are the same (AEF) (B1FT) Comparison with Ζ0.95 and conclusion (FT on Ζ) (can earn at most 5/10) Total: 10

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 12 of 14 Question Answer Marks Guidance 10(i) Σ x = 20, Σ y = 30, Σ xy = 111, Σ x2 = 110, Σ y2 = 190 Sxy = 111 – 20 × 30/5 = – 9 or – 1⋅8 Sxx = 110 – 202/5 = 30 or 6 [Syy = 190 – 302/5 = 10 or 2 ] b = Sxy / Sxx = – 9/30 = – 3/10 or – 0⋅3 M1 A1 Find reqd. values (y – 6) = b (x – 4), y = – 0⋅3x + 7⋅2 M1 A1 Find gradient b in y –y = b (x –x) and hence eqn. of regression line (may be implied by writing y = a + bx and finding a, b) Total: 4 10(ii) r = Sxy / √(Sxx Syy) = – 9 / √(30 × 10) M1 A1 Find correlation coefficient r = – 0⋅520 *A1 Total: 3 10(iii) H0: ρ = 0, H1: ρ ≠ 0 B1 State both hypotheses (B0 for r …) r5, 10% = 0⋅805 *B1 State or use correct tabular two-tail r-value Accept H0 if |r| < tab. value (AEF) M1 State or imply valid method for conclusion No [non-zero] correlation (AEF) DA1 Correct conclusion (dep *A1, *B1) Total: 4

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 13 of 14 Question Answer Marks Guidance 11(a)(i) T = 3mg sin α [= 2mg] B1 Find T by resolving forces along plane on P T = kmg (5a/4 – a) / a [= ¼ kmg] B1 Find T using Hooke’s Law k = 8 B1 Combine using sin α = ⅔ to find k Total: 3 11(a)(ii) EITHER: ± 2m d2OQ/dt2 = 2mg sin α – kmg (OQ – a) / a (M1 A1 Apply Newton’s law at general point (e.g. below E) d2OQ/dt2 = (4g/a) (7a/6 – OQ) A1 Substitute values of k and sin α d2x/dt2 = – (4g/a) x where x = OQ – 7a/6 A1) Derive standard SHM form (requires minus sign) OR: 2mg sin α = kmg (e – a) / a, e = 7a/6 (M1 Find new equilibrium distance e from O ± 2m d2x/dt2 = 2mg sin α – kmg (e + x – a) / a M1 A1 Apply Newton’s law at general point (e.g. below E) d2x/dt2 = – (4g/a) x A1) Derive standard SHM form (requires minus sign) Centre is 7a/6 (or 1⋅17 a) from O B1 State centre of motion Period is π√(a/g) or 0⋅993√a B1 State period in simplified form, allowing g = 10 Total: 6 11(a)(iii) x0 = 5a/4 – e = a/12 B1 Find amplitude x0 of motion Tmin = kmg (5a/4 – 2x0 – a) / a = 2 mg/3 M1 A1 Find least tension (d2x/dt2)max = [±] (4g/a) x0 = [±] ⅓ g M1 A1 Find maximum acceleration (accepting either sign) Total: 5

9231/22 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 14 of 14 Question Answer Marks Guidance 11(b)(i) x = (1/250) Σ x f(x) = 414/250 = 1⋅656 AG B1 Verify given mean Total: 1 11(b)(ii) p = x /6 = 0⋅276, q = 0⋅724 M1 A1 Use 250 6Ci q6-i pi and find p and q a = 250 6C2 q4 p2 = 78⋅49 ± 0⋅01 (to 2 d.p.) A2 Find either exp. value b = 250 6C4 q2 p4 = 11⋅41 ± 0⋅01 (to 2 d.p.) A1 Find other exp. value (deduct single A1 if either value given to only 1 d.p.) Total: 5 11(b)(iii) H0: Distribution fits data or distribution is binomial (AEF) B1 State (at least) null hypothesis in full Combine values consistent with all exp. values ⩾ 5 Oi: 48 69 78 32 23 Ei: 36⋅01 82⋅36 78⋅49 39⋅89 13⋅26 (± 0⋅01) M1FT A1 (FT for M1 but not A1 on values of a, b) χ2 = 3⋅992 + 2⋅167 + 0⋅003 + 1⋅561 + 7⋅154 M1 Find χ2 = 14⋅9 A1 No. n of cells: 7 6 5 4 3 χn-2, 0.99 2: 15⋅09 13⋅28 11⋅34 9⋅210 6⋅635 B1FT State or use consistent tabular value χn-2, 0.99 2 (to 3 s.f.) [FT on number, n, of cells used to find χ2] Accept H1 if χ2 > tabular value (AEF) 14⋅9 [± 0⋅1] > 11⋅34 so distn. doesn’t fit [data] M1 State or imply valid method for conclusion Conclusion (requires both values correct) or manager’s belief not justified (AEF) A1 Total: 8