Past Paper and Sample Solutions from Wednesday 2nd November 2011
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 2011 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.
Extract from 2011 Mathematics Admissions Test
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)
A
B
C
D
E
F
G
H
I
J
2
(c)
(d)
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47
Sample Solutions for Extract from 2011 Mathematics Admissions Test
SOLUTIONS FOR ADMISSIONS TEST IN
MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS
WEDNESDAY 2 NOVEMBER 2011
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. Note that y = x3 − x2 − x + 1 = (x − 1)(x2 − 1) = (x − 1)2 (x + 1) so that the cubic has a repeated
root at 1 and a simple root at −1. The answer is (c).
B. Let the rectangle have sides x, y so that
2x + 2y = P,
xy = A.
Eliminating y we see that
1
x2 − P x + A = 0.
2
As x is real it follows that the discriminant is non-negative and so
2
P
− 4 × 1 × A 0 =⇒ P 2 16A.
2
The answer is (c). (This condition is in fact sufficient for x and y to be real and positive.)
C. We have xn = n3 − 9n2 + 631 so that xn > xn+1 is equivalent to
(n3 − 9n2 + 631) − ((n + 1)3 − 9(n + 1)2 + 631)
= (n3 − n3 − 3n2 − 3n − 1) − 9(n2 − n2 − 2n − 1)
= −3n2 + 15n + 8.
We note
−3x2 +15x+8 = 0 ⇐⇒ x =
−15 ±
√
√
√
225 + 96
15 − 321
15 + 321
⇐⇒ x = α =
or x = β =
.
−6
6
6
So −3x2 + 15x + 8 > 0 if and only if α < x < β. As 182 = 324 then β ≈ (15 + 18)/6 = 5 12 and so
n = 5 is the largest integer in the range α < n < β. The answer is (a).
D. In the range 0 x 2π we have
sin x 5π
1
π
when
x
;
2
6
6
sin 2x 5π
17π
1
π
13π
when
x
and when
x
.
2
12
12
12
12
One or both of these inequalities hold for x in the ranges
5π
13π
17π
π
x
and
x
12
6
12
12
which are intervals of total length 13π/12. The relevant fraction is (13π/12)/(2π) = 13/24. The
answer is (b).
1
Sample Solutions for Extract from 2011 Mathematics Admissions Test
E. Without any loss of generality, we can assume the radius of the circle to be 1. Let A be the
vertex of the angles α and β and B be the vertex of the angle γ. We see that AC has length 1/ sin α.
Applying the sine rule to the triangle ABC we find
sin γ
sin β
=
1
1/ sin α
and the answer is (b).
F. Note that
x2 + y 2 + 4x cos θ + 8y sin θ + 10 = 0
rearranges to
(x + 2 cos θ)2 + (y + 4 sin θ)2 = 4 cos2 θ + 16 sin2 θ − 10 = 12 sin2 θ − 6
and so the equation defines a circle of radius 12 sin2 θ − 6 provided sin2 θ > 12 . In the given range,
0 θ < π, we have sin θ 0 and so we need √12 < sin θ. The answer is (b).
G. For x in the interval −1 x 1, we have −1 x2 − 1 0. For t in the range −1 t 0 we
can see from the graph that f (t) = t + 1. So for −1 x 1 we have f (x2 − 1) = (x2 − 1) + 1 = x2
and hence
3 1
1
1
2
x
2
2
=
f(x − 1) dx =
x dx =
3 −1 3
−1
−1
and the answer is (d).
H. Note that log0.5 0.25 = 2 and that if x > 0 then
3
8log2 x = 23 log2 x = 2log2 x = x3
and likewise 9log3 x = x2 = 4log2 x . The given equation then rearranges to
⇐⇒
⇐⇒
⇐⇒
x = x3 − x2 − x2 + 2
x3 − 2x2 − x + 2 = 0
(x − 2)(x2 − 1) = 0
(x − 2) (x − 1) (x + 1) = 0.
So the positive roots are x = 1, 2 and the answer is (c).
I. We know, for 0 x < 2π, that sin2 x + cos2 x = 1 and that 0 sin2 x, cos2 x 1. Also, for any y
in the range 0 y 1, then y 3 < y and y 4 < y unless y = 0 or 1. So we would have
sin8 x + cos6 x < sin2 x + cos2 x = 1
unless sin2 x and cos2 x are 0 and 1 in some order. So the only solutions are x = 0, π/2, π, 3π/2 in
the given range and the answer is (b).
Alternatively, and writing s = sin θ and c = cos θ to ease the notation, we see
s8 + c6 = 1
⇐⇒ s8 + (1 − s2 )3 = 1
⇐⇒ s8 − s6 + 3s4 − 3s2 = 0
⇐⇒ s2 (s6 − s4 + 3s2 − 3) = 0
⇐⇒ s2 (s2 − 1)(s4 + 3) = 0
to see that we have roots when s = 0, 1, −1 at x = 0, π, π/2, 3π/2 as before.
2
Sample Solutions for Extract from 2011 Mathematics Admissions Test
J. We note
f(1) = 1,
f (2) = f(1) = 1,
f (3) = f(1)2 − 2 = −1,
f (4) = f (2) = 1,
f(5) = f (2)2 − 2 = −1,
f (6) = f(3) = −1,
f (7) = f (3)2 − 2 = −1,
f (8) = f (4) = 1.
The function f can only take values 1 and −1 because of the nature of the two rules defining f ;
specifically if we set f(n) = ±1 into either of the two rules we can only achieve ±1 for further values
of f. Moreover f(n) = 1 is only possible when the first rule f (2n) = f (n) has been repeatedly applied
to determine f (n); any use of the second rule would lead to f (n) = −1. Repeated application of only
the first rule will determine f(n) when n is a power of 2; for such n we have f (n) = 1 and otherwise
f(n) = −1.
Amongst the numbers 1 n 100 the powers of 2 are 1, 2, 4, 8, 16, 32, 64. So there are 7 powers of
2 and 93 other numbers. Hence
f(1) + f (2) + f(3) + · · · + f(100) = 1 × 7 + (−1) × 93 = −86
and the answer is (a).
3
Sample Solutions for Extract from 2011 Mathematics Admissions Test
3
2. (i) [2 marks] Multiplying x = 2x + 1 by x we get the first equation. Multiplying by x again we
get
x5 = x2 + 2x3 = x2 + 2 (2x + 1) = 2 + 4x + x2 .
(ii) [5 marks] If xk = Ak + Bk x + Ck x2 then
xk+1 = Ak x + Bk x2 + Ck x3
= Ak x + Bk x2 + Ck (2x + 1)
= Ck + (Ak + 2Ck ) x + Bk x2 .
Comparing the coefficients of 1, x, x2 we then get
Ak+1 = Ck ,
Bk+1 = Ak + 2Ck,
Ck+1 = Bk .
(iii) [5 marks] With Dk = Ak + Ck − Bk we have
Dk+1 =
=
=
=
Ak+1 + Ck+1 − Bk+1
Ck + Bk − (Ak + 2Ck )
Bk − Ak − Ck
−Dk .
[using the previous part]
As 1 = 1 + 0x + 0x2 then A0 = 1, B0 = 0, C0 = 0 and D0 = 1. So Dk = (−1)k and
Ak + Ck = Bk + (−1)k .
(iv) [3 marks] Let Fk = Ak+1 + Ck+1 . Then
Fk + Fk+1 =
=
=
=
=
=
(Ak+1 + Ck+1 ) + (Ak+2 + Ck+2 )
Bk+1 + (−1)k+1 + Bk+2 + (−1)k+2
Bk+1 + Bk+2
Ck+2 + Ck+3
Ak+3 + Ck+3
Fk+2 .
4
Sample Solutions for Extract from 2011 Mathematics Admissions Test
3
3. (i) [2 marks] As y = m (x − a) and y = x − x touch at x = b then their gradients agree and so,
differentiating, m = 3b2 − 1.
(ii) [3 marks] As they also meet when x = b, then the y-coordinates of the graphs also agree so that
b3 − b = m (b − a) = 3b2 − 1 (b − a) .
Solving for a we have
a= b−
b3 − b
3b2 − 1
=
3b3 − b − b3 + b
2b3
=
.
3b2 − 1
3b2 − 1
(iii) [3 marks] If a is a large negative number then the line y = m (x − a) will be almost horizontal
√
meaning that the line will be tangential very close to the cubic’s (local) maximum where x = −1/ √3.
Alternatively one could argue that a = 2b3 / (3b2 − 1) ≪ 0 only when 3b2 − 1 ≈ 0 and so b ≈ −1/ 3
as b < 0.
(iv) [2 marks] If we expand (x − b)2 (x − c) and compare the coefficients of x2 we get 0 = −2b − c
and so c = −2b.
(v) [5 marks] We can see that as a increases then the tangent line rises and so the area of R increases.
So this area is greatest when a = −1. In this case b = a = −1 and c = 2. Hence the largest area
achieved by R is
2
2
3
3x + 2 − x3 dx
2 (x + 1) − x − x dx =
−1
−1
2
x4
3x2
+ 2x −
=
2
4
−1
1
27
3
= (6 + 4 − 4) −
−2−
= .
2
4
4
5
Sample Solutions for Extract from 2011 Mathematics Admissions Test
4.
y
y
2.0
2.0
1.5
1.5
1.0
1.0
0.5
0.5
0.0
0.0
0.5
1.0
1.5
Figure A
2.0
x
0.0
0.0
0.5
1.0
Figure B
1.5
2.0
x
(i) [4 marks] The largest value of x + y will be achieved
where x + y = k is tangential
√at the√point
to the boundary of Q. By symmetry, this occurs at 1/ 2, 1/ 2 . So the largest value of x + y on
√
Q is 2.
(ii) [6 marks] Again we can see that the maximum
√value√of xy on Q is when xy = k is tangential to
Q’s boundary. This takes place once more at 1/ 2, 1/ 2 where xy = 1/2. As x2 + y 2 also takes
its largest value (of 1) there then the maximum of
x2 + y 2 + 4xy = 1 +
4
= 3.
2
Further xy takes a minimum value of 0 on Q (on the axes). At (1, 0) and (0, 1) we also have x2 + y 2
taking its maximum value (of 1) and so we have
x2 + y 2 − 6xy = 1 − 6 × 0 = 1
as the largest value of x2 + y 2 − 6xy achieved at points (x, y) in Q.
(iii) [5 marks] As x2 + y 2 − 4x − 2y = k can have the squares completed to become
(x − 2)2 + (y − 1)2 = k + 5
we see that the curve is the circle with centre (2, 1) and radius
√
k + 5.
So the function x2 + y 2 − 4x − 2y increases as we move away from (2, 1) and the closest point to (2, 1)
in Q is √15 (2, 1). At that point
2
2
x + y − 4x − 2y =
2
√
5
or we can note
k=
2
+
1
√
5
2
√
8
2
10
− √ − √ =1− √ =1−2 5
5
5
5
2
√
√
5 − 1 − 5 = 1 − 2 5.
6
Sample Solutions for Extract from 2011 Mathematics Admissions Test
5. (i) [2 marks] All solutions in a semi-grid of size 4 are as follows:
RRR,
RRU,
RUR,
RU U,
U RR,
U RU,
U U R,
U U U.
(ii) [2 marks] Write R for a right-move and U for an up-move. A solution in a semi-grid of size n is
then a sequence of n − 1 letters, U or R.
(iii) [2 marks] Each solution consists of n − 1 letters, and each letter can be chosen independently;
thus there are 2n−1 solutions in total.
(iv) [3 marks] In a semi-grid of size 4 there are now ten solutions: R, U and the eight solutions from
part (ii).
In a semi-grid of size 5: the same ten solutions.
(v) [6 marks] The modified definitions ensure that the goal squares are actually the squares on the
diagonals of semi-grids of sizes 2, 4, 6, . . . and so on, up to n − 1 (if n is odd) or n (if n is even).
Then, by using the result from part (iii) of this question, the total number of paths from the original
location to a goal square can be obtained as follows. If n = 2k then the sum is
k
22−1 + 24−1 + · · · + 22k−1 =
1 2i
2 .
2 i=1
And in fact if n = 2k + 1 then we arrive at the same sum.
As this is the sum of a geometric progression, we obtain
2 k
1 4 4k − 1
=
4 −1 .
2
4−1
3
This can also be written as
2
3
22k − 1 or
2
3
(2n−1 − 1) when n is odd and
7
2
3
(2n − 1) when n is even.