Past Paper and Sample Solutions from Wednesday 5 th November 2008

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1 Past Paper and Sample Solutions from Wednesday 5 th November 2008 The Mathematics Admissions Test (MAT) is a paper based test that has been used by the University of Oxford since This extract from the 2008 test and associated sample solutions constitute the test undertaken by applicants to the Mathematics, Maths & Philosophy and Maths & Statistics undergraduate degree courses at Oxford. From 2013, the test will be used as part of the admissions process for applicants to the following courses run by the Department of Mathematics at Imperial College. UCAS code G100 G103 G125 GG31 G104 G1F3 G102 G1G3 G1GH Course title Mathematics Mathematics Mathematics (Pure Mathematics) Mathematics, Optimisation and Statistics Mathematics with a Year in Europe Mathematics with Applied Mathematics/Mathematical Physics Mathematics with Mathematical Computation Mathematics with Statistics Mathematics with Statistics for Finance IN 2013 THE ADMISSIONS TESTING SERVICE WILL BE ORGANIZING THE DISTRIBUTION AND RECEIPT OF THE MATHEMATICS TEST. SEE THIS ADMISSIONS TESTING SERVICE PAGE FOR FULL DETAILS.

2 1. For ALL APPLICANTS. For each part of the question on pages 3 7 you will be given four possible answers, just one of which is correct. Indicate for each part A J which answer (a), (b), (c), or (d) you think is correct with a tick (X) in the corresponding column in the table below. Please show any rough working in the space provided between the parts. (a) (b) (c) (d) A B C D E F G H I J 2

3 A. The function has = (a) no stationary points; (b) one stationary point; (c) two stationary points; (d) three stationary points. B. Which is the smallest of these values? (a) log 10 (b) p log10 ( 2 ) (c) 3 1 (d) log 10 1 log 10 3 Turn Over

4 C. The simultaneous equations in, are solvable (cos ) (sin ) = 2 (sin ) +(cos ) = 1 (a) for all values of in the range ; (b)exceptforonevalueof in the range ; (c) except for two values of in the range ; (d)exceptforthreevaluesof in the range D. When is divided by 1 the remainder is (a) 2000 (b) 2500 (c) 3000 (d)

5 E. The highest power of in ½ h i 5 + h i 6 ¾ 3 is (a) 424 (b) 450 (c) 500 (d) 504 F. If the trapezium rule is used to estimate the integral Z 1 0 ( ) d by splitting the interval into 10 intervals then an overestimate of the integral is produced. It follows that (a) the trapezium rule with 10 intervals underestimates R 1 0 (b) the trapezium rule with 10 intervals underestimates R 1 0 (c) the trapezium rule with 10 intervals underestimates R 2 1 (d) the trapezium rule with 10 intervals underestimates R ( ) d ; ( ( ) 1) d ; ( 1) d ; (1 ( )) d 5 Turn Over

6 G. Which of the graphs below is a sketch of y = ? y (a) y x x (b) y x x (c) (d) H. The equation = has one or more real solutions precisely when (a) > 9 4 (b) 0 (c) 6 1 (d) > 5 8 6

7 I. The function ( ) is de ned for positive integers by ( ) =sum of the digits of. For example, (723)=7+2+3=12 The sum (1) + (2) + (3) + + (99) equals (a) 746 (b) 862 (c) 900 (d) 924 J. In the range the equation (3 + cos ) 2 =4 2sin 8 has (a) 0 solutions (b) 1 solution (c) 2 solutions (d) 3 solutions 7 Turn Over

8 2. For ALL APPLICANTS. (i) Find a pair of positive integers, 1 and 1 that solve the equation ( 1 ) 2 2( 1 ) 2 =1 (ii) Given integers we de ne two sequences and by setting +1 = = + for > 1 Find positive values for such that ( +1 ) 2 2( +1 ) 2 =( ) 2 2( ) 2 (iii) Find a pair of integers which satisfy =1such that 50 (iv) (Using the values of and found in part (ii)) what is the approximate value of as increases? 8

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10 3. For APPLICANTS IN MATHEMATICS MATHEMATICS & STATISTICS MATHEMATICS & PHILOSOPHY MATHEMATICS & COMPUTER SCIENCE Computer Science applicants should turn to page 14. (i) The graph = ( ) of a certain function has been plotted below. y ONLY. x Onthenextthreepairsofaxes(A),(B),(C)aregraphsof = ( ) ( 1) ( ) in some order. Say which axes correspond to which graphs. y y y x x x (A) (B) (C) (ii) Sketch, on the axes opposite, graphs of both of the following functions Carefully label any stationary points. =2 2 and =2 2 2 (iii) Let be a real number and de ne the following integral ( ) = Z ( )2 d State the value(s) of for which ( ) is largest. Brie y explain your reasoning. [Note you are not being asked to calculate this maximum value.] 10

11 6 p p p p p p p p p p p - 11 Turn Over

12 4. MATHEMATICS For APPLICANTS IN MATHEMATICS & STATISTICS ONLY. MATHEMATICS & PHILOSOPHY Mathematics & Computer Science and Computer Science applicants should turn to page 14. y Q q C O p P x Let and be positive real numbers. Let denote the point ( 0) and denote the point (0 ) (i) Show that the equation of the circle which passes through, and the origin is =0 Find the centre and area of. (ii) Show that area of circle area of triangle > (iii) Find the angles and if area of circle area of triangle =2 12

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14 5. For ALL APPLICANTS. The Millennium school has 1000 students and 1000 student lockers. The lockers are in a line in a long corridor and are numbered from 1 to Initially all the lockers are closed (but unlocked). The rst student walks along the corridor and opens every locker. The second student then walks along the corridor and closes every second locker, i.e. closes lockers 2, 4, 6, etc. At that point there are 500 lockers that are open and 500 that are closed. The third student then walks along the corridor, changing the state of every third locker. Thus s/he closes locker 3 (which had been left open by the rst student), opens locker 6 (closed by the second student), closes locker 9, etc. All the remaining students now walk by in order, with the th student changing the state of every th locker, and this continues until all 1000 students have walked along the corridor. (i) How many lockers are closed immediately after the third student has walked along the corridor? Explain your reasoning. (ii) How many lockers are closed immediately after the fourth student has walked along the corridor? Explain your reasoning. (iii) At the end (after all 1000 students have passed), what is the state of locker 100? Explain your reasoning. (iv) After the hundredth student has walked along the corridor, what is the state of locker 1000? Explain your reasoning. 14

15 Sample Solutions for Extract from 2008 Mathematics Admissions Test SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, JOINT SCHOOLS AND COMPUTER SCIENCE WEDNESDAY 5 NOVEMBER 2008 Mark Scheme: EachpartofQuestion1isworthfourmarkswhichareawardedsolelyforthecorrectanswer. EachofQuestions2-7isworth15marks QUESTION 1: A.Asy=2x 3 6x 2 +5x 7then y =6x 2 12x+5. Thequadraticy hasdiscriminant =24>0andhencetheequationy =0has two distinct real roots. The answer is(c). B.Asπ<10then So log10 (π 2 )= 2L> L L=L; Theansweris(a). L=log 10 π<1. ( ) 3 1 =L 3 >1; log 10 π 1 log 10 π = 2 L >2. C. Wewillwritec=cosθands=sinθforeaseofnotation. Eliminatingyfromthesimultaneous equations cx sy=2, sx+cy=1; weget and similarly eliminating x we find 2c+s=c(cx sy)+s(sx+cy)= ( c 2 +s 2) x=x c 2s=( s)(cx sy)+c(sx+cy)= ( s 2 +c 2) y=y. Hencetheequationscaresolvableforanyvalueofθ. Theansweris(a). D.Bytheremaindertheoremwhenapolynomialp(x)isdividedbyx 1thentheremainder isp(1). Sotherequiredremainderhereis = 50 2 (1+99)=2500 as the series is an arithmetic progression. The answer is(b). 1

16 Sample Solutions for Extract from 2008 Mathematics Admissions Test E.Thehighestpowerofxin(2x 6 +7) 3 isx 18 andin(3x 8 12) 4 isx 32 sothehighestpowerin [ ] 5 is(x 32 ) 5 =x 160. Thehighestpowerofxin(3x 5 12x 2 ) 5 isx 25 andin(x 7 +6) 4 isx 28, sothatthehighestpowerofxin[...] 6 is(x 28 ) 6 =x 168. Thusthehighestpowerofxin{...} 3 is(x 168 ) 3 =x 504. Theansweris(d). F. Suppose that, when the trapezium rule is used to estimate the integral 1 f(x)dx, an 0 overestimate of E is produced. If the same number of intervals are used in the following calculations then: (a)toestimate 1 2f(x) dxanoverestimateof2ewillbeproduced,astherelevantgraphs 0 havebeenstretchedbyafactorof2andallareasdoubled; (b) to estimate 1 (f(x) 1)dx an overestimate of E will be produced, as the relevant 0 graphshavebeentranslateddownby1andallareasremainthesame; (c)toestimate 2 f(x 1)dxanoverestimateofEwillbeproduced,astherelevantgraphs 1 havebeentranslaterightby1andallareasremainthesame; (d)toestimate 1 (1 f(x))dxanunderestimateofe willbeproduced, astherelevant 0 graphs have been reflected in the x-axis turning the overestimate to an underestimate and translatedupby1,whichchangesnothingwithregardtoareas. Theansweris(d). G. As 4x x 2 5 = (x 2) 2 1, theny =(4x x 2 5) 1 is alwaysnegative andhas a minimumvalueatx=2. Theansweris(c). H.Ifwesety=3 x thentheequation9 x 3 x+1 =knowreads This has solutions y 2 3y k=0. y= 3± 9+4k 2 whicharerealwhenk 9/4. Asy=3 x thenwefurtherneedthaty>0forxtobereal,but thisisnotaproblemasthelargerrootisclearlypositive. Theansweris(a). I.Wehave S(1)+S(2)+S(3)+ +S(99)=S(00)+S(01)+ +S(99) andinthe100two-digitnumbers00,...,99therearetwenty0s,twenty1s,...,twenty9s.so S(1)+S(2)+S(3)+ +S(99)=20 ( )= (0+9)=900 andtheansweris(c). J.Notethat (3+cosx) 2 (3 1) 2 =4; 4 2sin 8 x 4. Sotheequationwillholdonlywehncosx= 1andsinx=0. Intherange0 x<2π this onlyoccursatx=π. Theansweris(b). 2

17 Sample Solutions for Extract from 2008 Mathematics Admissions Test 2. (i)afairlyobviouspair(x 1,y 1 )thatsatisfy(x 1 ) 2 2(y 1 ) 2 =1iisx 1 =3andy 1 =2. (ii) Note (x n+1 ) 2 2(y n+1 ) 2 =(x n ) 2 2(y n ) 2 (3x n +4y n ) 2 2(ax n +by n ) 2 =(x n ) 2 2(y n ) 2 ( 8 2a 2) (x n ) 2 +(24 4ab)x n y n + ( 18 2b 2) (y n ) 2 =0 Inordertohave(x n+1 ) 2 2(y n+1 ) 2 =(x n ) 2 2(y n ) 2 weneed 2a 2 =8, 4ab=24, 2b 2 =18. Wefurtherrequirethata,b>0. Weseethata=2andb=3solveallthreeequations. (iii)startingwithx 1 =3,y 1 =2wefind: SoX=99andY =70issuchapair. x 1 =3, y 1 =2; x 2 = =17, y 2 = =12; x 3 = =99, y 3 = =70. (iv)forthegeneratedsequences,(x n ),(y n ),wehave (x n ) 2 2(y n ) 2 =1 foreachn. Alsotheintegersx n andy n aregettingincreasinglylargerbecauseofhowtheyaredefinedin (ii). So ( ) 2 xn 2= forlargen, y n (y n ) andx n /y n 2asx n andy n arebothpositive. 3

18 Sample Solutions for Extract from 2008 Mathematics Admissions Test 3. (i)(a)is f(x); (B)isf( x); (C)isf(x 1). (ii)as2 2x x2 =2 2 (x 1)2 thenthegraphofy=2 2x x2 isthegraphofy=2 x2 translated totherightby1andstretchedparalleltothey-axisbyafactorof Plotsofy=2 x2 andy=2 2x x2. (ii)c= 1 2. Thegraphof2 (x c)2 isthegraphof2 x2 translatedctotheright. Theintegral I(c)representstheareaunderthegraphbetween0 x 1. Asthegraphissymmetric/even anddecreasingawayfrom0thenthisareaismaximisedbyhavingtheapexhalfwayalongthe interval0 x 1,i.e. atx=1/2whichoccurswhenc=

19 Sample Solutions for Extract from 2008 Mathematics Admissions Test 4. (i)wecancompletethesquaresinx 2 px+y 2 qy=0toget ( x p ) 2+ ( y q ) 2= p 2 +q whichis theequationofthecirclewithcentre: (p/2,q/2) andarea: π(p 2 +q 2 )/4. Eitherby checking the original question, or the rearranged one, we can see that 0 at (0,0), x 2 px+y 2 qy= p 2 p 2 +0=0 at (p,0), 0+q 2 q 2 =0 at (0,q). (ii)theareaofopqispq/2. So Note areaofcirclec areaoftriangleopq = π(p 2 +q 2 ) 2pq proving the required inequality. ( π(p 2 +q 2 ) 4 )/( ) 1 2 pq = π(p2 +q 2 ). 2pq π p 2 +q 2 2pq (p q) 2 0, (1) (iii) Rearranging π(p 2 +q 2 ) 2pq =2π p 2 +q 2 =4pq ( ) 2 ( ) p p 4 +1=0, q q whichisaquadraticequationinp/q,andso p q = 4± 16 4 =2± 3. 2 Nowp/q=tanOQP,q/p=tanOPQandso {tanoqp,tanopq}= withtheorderdependingonwhetherp<qorp>q. { 2 3,2+ } 3 [It happens that arctan ( 2 3 ) = π/12 and arctan ( 2+ 3 ) = 5π/12, but appreciation of this was not expected.] 5

20 Sample Solutions for Extract from 2008 Mathematics Admissions Test 5. (i) After the first/second/third students have gone by the doors look like: Locker Student1 O O O O O O O O O O O O O O Student2 O C O C O C O C O C O C O C Student3 O C C C O O O C C C O O O C We can see that the lockers now repeat in a pattern OCCCOO every 6 lockers. As 1000 = wehave166repeatsofthispatternand4remaininglockersthatgoOCCC.Sothere are166 3=498closedlockersamongstthecompletecyclesand3furtherintheincomplete cycle. Thatis,thereare501closedlockersinall. (ii)afterthefourthstudenthasgonebywehavethefollowing: Locker Student3 O C C C O O O C C C O O O C Student4 O C C O O O O O C C O C O C with the pattern repeating every 12 lockers in the form OCCOOOOOCCOC. Each cycle contains5closedand7opendoors. Now1000= andsowehave83 5=415closed lockers amongst the complete cycles and 2 further amongst the incomplete cycle OCCO. In all then there are 417 closed lockers. (iii)locker100startsoffclosed(asalllockersdo)andthenitsstate isalteredbyeverynth studentwhere n is afactorof 100, i.e. bystudents 1,2,4,5,10,20,25,50,100. So9students changethestateandasthisisoddthenoverallthestatewillhavebeenchangedtoopen. (iv)locker1000startsoffclosed(asalllockersdo)andthenitsstateisalteredbyeverynth student where n divides 1000 and n 100, i.e. by 1,2,4,5,8,10,20,25,40,50,100. So 11 studentschangethedoor sstateandasthisisoddthenoverallthestatewillagainhavebeen changed to open. 6

MATHEMATICS ADMISSIONS TEST

MATHEMATICS ADMISSIONS TEST For candidates applying for Mathematics, Computer Science or one of their joint degrees at OXFORD UNIVERSITY and/or IMPERIAL COLLEGE LONDON Wednesday 2 November 2016 Time Allowed: