Past Paper and Sample Solutions from Wednesday … · Past Paper and Sample Solutions from Wednesday 4th November 2009 The Mathematics Admissions Test (MAT) is a paper based test

Past Paper and Sample Solutions from Wednesday 4th November 2009

The Mathematics Admissions Test (MAT) is a paper based test that has been used by the University of

Oxford since 1996. This extract from the 2009 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 Course title

G100 Mathematics

G103 Mathematics

G125 Mathematics (Pure Mathematics)

GG31 Mathematics, Optimisation and Statistics

G104 Mathematics with a Year in Europe

G1F3 Mathematics with Applied Mathematics/Mathematical Physics

G102 Mathematics with Mathematical Computation

G1G3 Mathematics with Statistics

G1GH 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.

http://www3.imperial.ac.uk/portal/page/portallive/85B4415ABA5D0C3DE0440003BACD17D6









http://www.admissionstestingservice.org/our-services/subject-specific/mat/about-mat/

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 tablebelow. 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

Extract from 2009 Mathematics Admissions Test

A. The smallest value of

I (a) =

Z 1

0

¡x2 − a

¢2dx,

as a varies, is

(a)3

20, (b)

4

45, (c)

7

13, (d) 1.

B. The point on the circlex2 + y2 + 6x+ 8y = 75,

which is closest to the origin, is at what distance from the origin?

(a) 3, (b) 4, (c) 5, (d) 10.

3 Turn Over


C. Given a real constant c, the equation

x4 = (x− c)2

has four real solutions (including possible repeated roots) for

(a) c 6 1

4, (b) − 1

46 c 6 1

4, (c) c 6 −1

4, (d) all values of c.

D. The smallest positive integer n such that

1− 2 + 3− 4 + 5− 6 + · · ·+ (−1)n+1 n > 100,

is(a) 99, (b) 101, (c) 199, (d) 300.

4


E. In the range 0 6 x < 2π, the equation

2sin2 x + 2cos

2 x = 2

(a) has 0 solutions;(b) has 1 solution;(c) has 2 solutions;(d) holds for all values of x.

F. The equation in x3x4 − 16x3 + 18x2 + k = 0

has four real solutions(a) when −27 < k < 5;(b) when 5 < k < 27;(c) when −27 < k < −5;(d) when −5 < k < 0.

5 Turn Over


G. The graph of all those points (x, y) in the xy-plane which satisfy the equationsin y = sinx is drawn in

�10 �5 5 10

�10

�5

5

10

�10 �5 5 10

�10

�5

5

10

(a) (b)

�10 �5 5 10

�10

�5

5

10

�10 �5 5 10

�10

�5

5

10

(c) (d)

H. When the trapezium rule is used to estimate the integralZ 1

0

2x dx

by dividing the interval 0 6 x 6 1 into N subintervals the answer achieved is

(a)1

2N

½1 +

1

21/N + 1

¾, (b)

1

2N

½1 +

2

21/N − 1

¾,

(c)1

N

½1− 1

(21/N − 1)

¾, (d)

1

2N

½5

21/N + 1− 1¾.

6


I. The polynomialn2x2n+3 − 25nxn+1 + 150x7

has x2 − 1 as a factor(a) for no values of n;(b) for n = 10 only;(c) for n = 15 only;(d) for n = 10 and n = 15 only.

J. The number of pairs of positive integers x, y which solve the equation

x3 + 6x2y + 12xy2 + 8y3 = 230

is(a) 0, (b) 26, (c) 29 − 1, (d) 210 + 2.

7 Turn Over



A list of real numbers x1, x2, x3, . . . is defined by x1 = 1, x2 = 3 and then for n > 3 by

xn = 2xn−1 − xn−2 + 1.

So, for example,x3 = 2x2 − x1 + 1 = 2× 3− 1 + 1 = 6.

(i) Find the values of x4 and x5.

(ii) Find values of real constants A,B,C such that for n = 1, 2, 3,

xn = A+Bn+ Cn2. (∗)

(iii) Assuming that equation (∗) holds true for all n > 1, find the smallest n such thatxn > 800.

(iv) A second list of real numbers y1, y2, y3, . . . is defined by y1 = 1 and

yn = yn−1 + 2n

Find, explaining your reasoning, a formula for yn which holds for n > 2.

What is the approximate value of xn/yn for large values of n?

8


9 Turn Over


3.

ForAPPLICANTS IN

⎧⎪⎪⎨⎪⎪⎩MATHEMATICSMATHEMATICS & STATISTICSMATHEMATICS & PHILOSOPHYMATHEMATICS & COMPUTER SCIENCE

⎫⎪⎪⎬⎪⎪⎭ONLY.

Computer Science applicants should turn to page 14.

For a positive whole number n, the function fn (x) is defined by

fn (x) =¡x2n−1 − 1

¢2.

(i) On the axes provided opposite, sketch the graph of y = f2 (x) labelling where thegraph meets the axes.

(ii) On the same axes sketch the graph of y = fn (x) where n is a large positive integer.

(iii) Determine Z 1

0

fn (x) dx.

(iv) The positive constants A and B are such thatZ 1

0

fn (x) dx 6 1−A

n+Bfor all n > 1.

Show that(3n− 1) (n+B) > A (4n− 1)n,

and explain why A 6 3/4.

(v) When A = 3/4, what is the smallest possible value of B?

10


6

- x

y

p p

p ppp

p ppp

p p

p

p

p

p

11 Turn Over


4.

For APPLICANTS IN

⎧⎨⎩ MATHEMATICSMATHEMATICS & STATISTICSMATHEMATICS & PHILOSOPHY

⎫⎬⎭ ONLY.

Mathematics & Computer Science and Computer Science applicants should turn topage 14.

As shown in the diagram below: C is the parabola with equation y = x2; P is the point(0, 1); Q is the point (a, a2) on C; L is the normal to C which passes through Q.

PC L

��,�2�

Q

�2 �1 1 2

0.5

1.0

1.5

2.0

2.5

(i) Find the equation of L.

(ii) For what values of a does L pass through P?

(iii) Determine |QP |2 as a function of a, where |QP | denotes the distance from P to Q.

(iv) Find the values of a for which |QP | is smallest.

(v) Find a point R, in the xy-plane but not on C, such that |RQ| is smallest for a uniquevalue of a. Briefly justify your answer.

12


13 Turn Over



Given an n × n grid of squares, where n > 1, a tour is a path drawn within the gridsuch that:

• along its way the path moves, horizontally or vertically, from the centre of onesquare to the centre of an adjacent square;

• the path starts and finishes in the same square;

• the path visits the centre of every other square just once.For example, below is a tour drawn in a 6× 6 grid of squares which starts and finishesin the top-left square.

For parts (i)-(iv) it is assumed that n is even.

(i) With the aid of a diagram, show how a tour, which starts and finishes in the top-leftsquare, can be drawn in any n× n grid.

(ii) Is a tour still possible if the start/finish point is changed to the centre of a differentsquare? Justify your answer.

Suppose now that a robot is programmed to move along a tour of an n × n grid. Therobot understands two commands:

• command R which turns the robot clockwise through a right angle;

• command F which moves the robot forward to the centre of the next square.

The robot has a program, a list of commands, which it performs in the given order tocomplete a tour; say that, in total, command R appears r times in the program andcommand F appears f times.

(iii) Initially the robot is in the top-left square pointing to the right. Assuming the firstcommand is an F , what is the value of f? Explain also why r + 1 is a multiple of 4.

(iv) Must the results of part (iii) still hold if the robot starts and finishes at the centreof a different square? Explain your reasoning.

(v) Show that a tour of an n× n grid is not possible when n is odd.

14


SOLUTIONS FOR ADMISSIONS TEST INMATHEMATICS, JOINT SCHOOLS AND COMPUTER SCIENCE

WEDNESDAY 4 NOVEMBER 2009

Mark Scheme:

Each part of Question 1 is worth four marks which are awarded solely for the correct answer.

Each of Questions 2-7 is worth 15 marks

QUESTION 1:

A. We have

I (a) =

∫ 1

0

(x2 − a

)2dx

=

∫ 1

0

(x4 − 2ax2 + a2

)dx

=1

5− 2a3+ a2

=

(a− 1

3

)2+

(1

5− 19

)

=

(a− 1

3

)2+4

45.

So the smallest value of I (a) is 4/45, achieved when a = 1/3. The answer is (b).

B. Completing the square we rewrite x2 + y2 + 6x+ 8y = 75 as

(x+ 3)2 + (y + 4)2 = 100.

Hence the circle has centre (−3,−4) and radius 10. The radius from the centre, back through theorigin, meets the circle at (3, 4) which is at a distance 5 from the origin. The answer is (c).

C. Taking square roots of both sides of x4 = (x− c)2 , we see

x2 = x− c, or x2 = c− x.

The first quadratic has discriminant 1 − 4c and the second quadratic has discriminant 1 + 4c. Inorder that the original equation has four real solutions both of these quadratic must have two realsolutions and so nonnegative discriminants. Hence c � 1/4 and c � −1/4. The answer is (b).

D. Summing the first few terms of the series

1− 2 + 3− 4 + 5− 6 + · · ·+ (−1)n+1 n,

we see that we get the sequence

1,−1, 2,−2, 3,−3, 4,−4, . . .

The first time that this equals or exceeds 100 is when n = 199. The answer is (c).

1

Sample Solutions for Extract from 2009 Mathematics Admissions Test

E. As 0 � sin2 x � 1 and 0 � cos2 x � 1 then

1 � 2sin2 x� 2, 1 � 2cos

2 x� 2.

So2sin

2 x + 2cos2 x = 2

is only possible when sin2 x = 0 = cos2 x, but this is true for no value of x as cos2 x+ sin2 x = 1 forall x. The solution is (a).

F. Settingy = 3x4 − 16x3 + 18x2 + k

we seey′ = 12x3 − 48x2 + 36x = 12x (x− 1) (x− 3)

and so y has turning points at x = 0, 1, 2 which are respectively minimum, maximum, minimumbecause of the shape of a quartic. There will be four real solutions when both minima are below thex-axis and the maximum is above the x-axis. Now

y (0) = k, y (1) = 5 + k, y (3) = 32 (27− 48 + 18) + k = k − 27.

Hence we needk < 0, k > −5, k < 27

each to be true. The answer is (d).

G. Thinking about the graph/periodicity of sine, we know that

sin (x+ 2π) = sin x

and so all the lines y = x+ 2nπ where n is an integer will be part of the graph of sin y = sin x. Butalso we have

sin (π − x) = sin xand so all the lines y = (2n+ 1)π − x where n is an integer will also be part of sin y = sin x. Theanswer is (c).

H. Using the trapezium rule with N subintervals to calculate∫ 102x dx we get

h

2[y0 + 2y1 + · · ·+ 2yN−1 + yN ]

=1

2N

[1 + 2× 21/N + · · ·+ 2(N−1)/N + 2

]

=1

2N

[2(1 + 21/N + · · ·+ 2(N−1)/N

)+ 1]

=1

2N

2((21/N

)N − 1)

21/N − 1 + 1

=1

2N

{1 +

2

21/N − 1

}.

The answer is (b).

2


I. In order for the polynomial

p (x) = n2x2n+3 − 25nxn+1 + 150x7

to have x2− 1 = (x− 1) (x+ 1) as a factor, the polynomial must be zero at both x = 1 and x = −1.Now

p (1) = n2 − 25n+ 150 = (n− 10) (n− 15)is zero when n = 10 or n = 15. And

p (−1) = −n2 + (−1)n 25n− 150 ={− (n− 10) (n− 15) n is even− (n+ 10) (n+ 15) n is odd

is zero when n = 10 or n = −15. Only n = 10 meets both requirements and so the answer is (b)

J. Factorising the given equation we see it now reads as

(x+ 2y)3 = 230

and sox+ 2y = 210,

where x and y are both positive integers. The solutions (x, y) are then of the form

(210 − 2, 1

),

(210 − 4, 2

), . . . . . .

(4, 29 − 2

),

(2, 29 − 1

)

and so the answer is (c).

3


2. (i) Using the given recurrence relation

x4 = 2x3 − x2 + 1 = 2× 6− 3 + 1 = 10;x5 = 2x4 − x3 + 1 = 2× 10− 6 + 1 = 15.

(ii) Setting n = 1, n = 2, n = 3 into A+Bn+Cn2 and using the known values of x1, x2, x3, we seethat

We haveA+B + C = 1, A+ 2B + 4C = 3, A+ 3B + 9C = 6.

So, subtractingB + 3C = 2, B + 5C = 3,

giving

C =1

2, B =

1

2, A = 0.

(iii) If n (n+ 1) /2 � 800 thenn (n+ 1) � 1600 = 402.

Now n (n+ 1) is an increasing function, and by inspection 39× 40 < 402 < 40× 41, so n = 40 is theleast possible n.

(iv) For for n � 2,

yn = 1 + 4 + 6 + · · ·+ 2n

= 1 +

(n− 12

)(4 + 2n) [AP formula]

= 1 + (n− 1) (2 + n)= n2 + n− 1.

Asyn = n

2 + n− 1 = 2xn − 1then xn/yn is approximately 1/2 for large values of n because xn and yn both become large.

4


3. (i) We have f2 (x) = (x3 − 1)2. Note that f2 (x) = 0 only when x = 1 and f2 (0) = 1; so the graph

crosses the axes only at (1, 0) and (0, 1) .

-2 -1 0 1 2

1

2

3

4

Now differentiating f2 (x) = x6 − 2x3 + 1 gives

f ′2 (x) = 6x5 − 6x2 = 6x2

(x3 − 1

)

is zero only at x = 0 and x = 1. As f2 (x) is large for large positive and negative values, and asf2 (x) � 0 = f2 (1) then f2 (x) has a minimum at x = 1 and an inflexion point at x = 0.

(ii) Consider fn (x) = (x2n−1 − 1)2 when n is large and positive. Then fn (x) still crosses the axes

at (1, 0) and (0, 1) only, is large for large negative and positive values and takes only nonnegativevalues. For −1 < x < 1 we have that x2n−1 − 1 is very close to −1 and so fn (x) is close to 1 in therange −1 < x < 1. For large values of x it is now the case that fn (x) increases even more rapidlythan f2 (x) does.

(iii) Note ∫ 1

0

(x2n−1 − 1

)2dx =

∫ 1

0

(x4n−2 − 2x2n−1 + 1

)dx =

1

4n− 1 −1

n+ 1.

(iv) So ∫ 1

0

fn (x) dx �1

4n− 1 −1

n+ 1 � 1− A

n+B⇐⇒ 3n− 1

(4n− 1)n �A

n+B

Rearranging this gives

(3n− 1) (n+B) � A (4n− 1)n.3n2 − n−B + 3Bn � 4An2 −An

Comparing coefficients of n2 (the dominant term) we need 3 � 4A so that A � 3/4. Or one mightexpand the brackets to get

(3− 4A)n2 + (3B +A− 1)n−B � 0 for n � 1;

if 4A > 3 the LHS is an "upside-down" parabola and so negative for large n which is a contradiction.Hence A � 3/4.

(v) When A = 3/4, we need

4 (3n− 1) (n+B) � 3 (4n− 1)n for n � 1;

12n2 + 12nB − 4n− 4B � 12n2 − 3n for n � 1;

(12B − 1)n− 4B � 0 for n � 1.

The graph of the LHS against n is a line with gradient 12B − 1; so we need 12B − 1 � 0 for anon-decreasing line and also (12B − 1) − 4B � 0 for the first n = 1 case to hold. So B � 1/8 and1/8 is the smallest value.

5


4. (i) By differentiating, we know that the gradient at Q is 2a and so the normal has gradient−1/ (2a) when a �= 0. Hence the line L has equation

y − a2 = −12a(x− a) .

If a = 0 the equation is x = 0. The equation can be better written as

x+ 2ay = 2a3 + a

which also includes the a = 0 case.

(ii) So L passes through (0, 1) if

0 + 2a = 2a3 + a =⇒ a(2a2 − 1

)= 0 =⇒ a =

±1√2, or a = 0.

(iii) As Q = (a, a2) and P = (0, 1) then the distance |PQ|2 is determined as

|PQ|2 = a2 +(a2 − 1

)2= a4 − a2 + 1.

(iv) This can be differentiated, and set to zero, so that

4a3 − 2a = 0 =⇒ a = 0,1√2,−1√2

though 0 is a local maximum. [Note these were the three values determined in part (ii).] Or bycompleting the square we have

a4 − a2 + 1 =(a2 − 1

2

)2+3

4

which we can see is minimal when a2 = 1/2.

(v) We need to find a point R, in the xy-plane, not on C, such that |RQ| is smallest for a uniquevalue of a. There are many points where this is the case but we are only asked to find one suchpoint..Now R = (0,−1) is such a point and it is relatively easy to justify this: the point (0, 0) on Cis distance 1 away; no other point in the upper half-plane, and hence no other point on C, is as closeas (0, 0) . [The same argument can be used for any point on the negative y-axis.]

6


5. (i) A tour of an even by even grid can be produced by generalizing the sketch below.

By moving right along the first row, dropping a row, left to column two, dropping a row, and so, andthen finally coming back all the way along the bottom row and coming up the leftmost column, wecomplete a tour. As n is even then this general E-shapes has n/2 "arms" and so it is the case thathaving descended all the rows the tour returns left along the bottom row.

(ii) Given it is possible to draw a tour which goes through all the squares, our new starting pointis somewhere on the original tour and so a new tour can begin from there and proceed along theoriginal tour to the top-left corner and then catch up on the part of the original tour that it hadmissed.

(iii) We have f = n2 as the robot has to move through each of the squares of the grid withoutrepetition and moves into a new square with each F command. As it starts in a corner going right itmust return to that corner by moving up and so to go from travelling right to up means that robothas turned 270 degrees over all, or one right angle short of a number of whole turns.Equivalentlyr + 1 clockwise quarter turns have led to a whole number of turns in all and so r + 1 must be amultiple of 4 (there being four quarter turns in a whole turn).

(iv) As the robot still travels through all the squares we still have f = n2. But if, for example, therobot set off going right along one of the flat parts of the E-like tour above then it would overall turnthrough 360 degrees and, instead, r would be a multiple of 4

(v) Suppose now that n is odd. Then f = n2 also is odd. But as any tour begins and ends in thesame place, for every move to the right there must be one to the left, and for every up there mustbe a down i.e. f must be even. Hence no tour is possible for odd n.

7


Documents

Past Paper and Sample Solutions from Wednesday … · Past Paper and Sample Solutions from Wednesday 4th November 2009 The Mathematics Admissions Test (MAT) is a paper based test