Past Paper and Sample Solutions from Wednesday 31 October 2007

Transcription

Past Paper and Sample Solutions from Wednesday 31st October 2007
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 2007 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 2007 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)
A. Let r and s be integers. Then
6r+s × 12r−s
8r × 9r+2s
is an integer if
(a) r + s 6 0,
(b) s 6 0,
(c) r 6 0,
(d) r > s.
B. The greatest value which the function
¢2
¡
f (x) = 3 sin2 (10x + 11) − 7
takes, as x varies over all real values, equals
(a)
− 9,
(b) 16,
3
(c) 49,
(d) 100.
Turn Over
C. The number of solutions x to the equation
7 sin x + 2 cos2 x = 5,
in the range 0 6 x < 2π, is
(a) 1,
(b) 2,
(c) 3,
(d) 4.
D. The point on the circle
(x − 5)2 + (y − 4)2 = 4
which is closest to the circle
(x − 1)2 + (y − 1)2 = 1
is
(a)
(3.4, 2.8) ,
(b)
(c)
(3, 4) ,
4
(5, 2) ,
(d)
(3.8, 2.4) .
E. If x and n are integers then
(1 − x)n (2 − x)2n (3 − x)3n (4 − x)4n (5 − x)5n
is
(a) negative when n > 5 and x < 5,
(b) negative when n is odd and x > 5,
(c) negative when n is a multiple of 3 and x > 5,
(d) negative when n is even and x < 5.
F. The equation
8x + 4 = 4x + 2x+2
has
(a) no real solutions;
(b) one real solution;
(c) two real solutions;
(d) three real solutions.
5
Turn Over
G. On which of the axes below is a sketch of the graph
¡ ¢
y = 2−x sin2 x2 ?
(a)
(b)
(c)
(d)
H. Given a function f (x) , you are told that
Z 1
Z 2
3f (x) dx +
2f (x) dx = 7,
0
1
Z 2
Z 2
f (x) dx +
f (x) dx = 1.
0
It follows that
R2
0
1
f (x) dx equals
(a)
− 1,
(b) 0,
(c)
6
1
,
2
(d) 2.
I. Given that a and b are positive and
4 (log10 a)2 + (log10 b)2 = 1,
then the greatest possible value of a is
(a)
1
,
10
(b) 1,
(c)
√
10,
√
(d) 10 2 .
J. The inequality
¢ ¡
¢ ¡
¢
¡
¢
¡
(n + 1) + n4 + 2 + n9 + 3 + n16 + 4 + · · · + n10000 + 100 > k
is true for all n > 1. It follows that
(a) k < 1300,
(b) k 2 < 101,
(c) k > 10110000 ,
(d) k < 5150.
7
Turn Over
Let
fn (x) = (2 + (−2)n ) x2 + (n + 3) x + n2
where n is a positive integer and x is any real number.
(i) Write down f3 (x) .
Find the maximum value of f3 (x).
For what values of n does fn (x) have a maximum value (as x varies)?
[Note you are not being asked to calculate the value of this maximum.]
(ii) Write down f1 (x).
Calculate f1 (f1 (x)) and f1 (f1 (f1 (x))).
Find an expression, simplified as much as possible, for
f1 (f1 (f1 (· · · f1 (x))))
where f1 is applied k times. [Here k is a positive integer.]
(iii) Write down f2 (x) .
The function
f2 (f2 (f2 (· · · f2 (x)))) ,
where f2 is applied k times, is a polynomial in x. What is the degree of this polynomial?
8
9
Turn Over
3.
⎧
MATHEMATICS
⎪
⎪
⎨
MATHEMATICS & STATISTICS
For APPLICANTS IN
MATHEMATICS & PHILOSOPHY
⎪
⎪
⎩
MATHEMATICS & COMPUTER SCIENCE
Computer Science applicants should turn to page 14.
Let
I (c) =
Z
0
where c is a real number.
1
⎫
⎪
⎪
⎬
⎪
⎪
⎭
ONLY.
¡
¢
(x − c)2 + c2 dx
(i) Sketch y = (x − 1)2 + 1 for the values −1 6 x 6 3 on the axes below and show on
your graph the area represented by the integral I (1) .
(ii) Without explicitly calculating I (c) , explain why I (c) > 0 for any value of c.
(iii) Calculate I (c) .
(iv) What is the minimum value of I (c) (as c varies)?
(v) What is the maximum value of I (sin θ) as θ varies?
y
6
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
- x
10
11
Turn Over
4.
⎧
⎫
⎨ MATHEMATICS
⎬
MATHEMATICS & STATISTICS
For APPLICANTS IN
ONLY.
⎩
⎭
MATHEMATICS & PHILOSOPHY
Mathematics & Computer Science and Computer Science applicants should turn to
page 14.
In the diagram below is sketched the circle with centre (1, 1) and radius 1 and a line
L. The line L is tangential to the circle at Q; further L meets the y-axis at R and the
x-axis at P in such a way that the angle OP Q equals θ where 0 < θ < π/2.
(i) Show that the co-ordinates of Q are
(1 + sin θ, 1 + cos θ) ,
and that the gradient of P QR is − tan θ.
Write down the equation of the line P QR and so find the co-ordinates of P.
(ii) The region bounded by the circle, the x-axis and P Q has area A (θ); the region
bounded by the circle, the y-axis and QR has area B (θ). (See diagram.)
Explain why
A (θ) = B (π/2 − θ)
for any θ.
Calculate A (π/4) .
(iii) Show that
A
³π ´
3
=
√
π
3− .
3
12
13
Turn Over
Let f (n) be a function defined, for any integer n > 0, as follows:
⎧
1
if n = 0,
⎨
2
f (n) =
(f (n/2)) if n > 0 and n is even,
⎩
2f (n − 1) if n > 0 and n is odd.
(i) What is the value of f (5)?
The recursion depth of f (n) is defined to be the number of other integers m such that
the value of f (m) is calculated whilst computing the value of f (n) . For example, the
recursion depth of f (4) is 3, because the values of f (2) , f (1) , and f (0) need to be
calculated on the way to computing the value of f (4).
(ii) What is the recursion depth of f (5)?
Now let g (n) be a function, defined for all integers n > 0, as follows:
⎧
0
if n = 0,
⎨
1 + g (n/2) if n > 0 and n is even,
g (n) =
⎩
1 + g (n − 1) if n > 0 and n is odd.
(iii) What is g (5)?
¡ ¢
(iv) What is g 2k , where k > 0 is an integer? Briefly explain your answer.
¢
¡
(v) What is g 2l + 2k where l > k > 0 are integers? Briefly explain your answer.
(vi) Explain briefly why the value of g (n) is equal to the recursion depth of f (n).
14
Sample Solutions for Extract from 2007 Mathematics Admissions Test
SOLUTIONS FOR ADMISSIONS TEST IN
MATHEMATICS, JOINT SCHOOLS AND COMPUTER SCIENCE
WEDNESDAY 31 OCTOBER 2007
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: Separating out the powers of 2 and 3 we have
6r+s × 12r−s
= 2(r+s+2r−2s−3r) × 3(r+s+r−s−2r−4s) = 2−s × 3−4s
8r × 9r+2s
which is an integer if s 0. The answer is (b).
B:
• sin (10x + 11) takes values between −1 and 1 as x varies;
• sin2 (10x + 11) takes values between 0 and 1 as x varies;
• 3 sin2 (10x + 11) takes values between 0 and 3 as x varies;
• 3 sin2 (10x + 11) − 7 takes values between −7 and −4 as x varies;
2
• 3 sin2 (10x + 11) − 7 takes values between 16 and 49 as x varies.
The answer is (c).
C: Using the identity sin2 x + cos2 x = 1 we see
7 sin x + 2 cos2 x = 5
⇐⇒ 2 sin2 x − 7 sin x + 3 = 0
⇐⇒ (2 sin x − 1) (sin x − 3) = 0
Now sin x = 3 has no solutions, and in the range 0 x < 2π we note sin x takes the value 1/2 twice (at π/6 and at
5π/6). The answer is (b).
2
2
D: The circle with equation (x − 5) + (y − 4) = 4 has centre (5, 4) and radius 2.
2
2
The circle with equation (x − 1) + (y − 1) = 1 has centre (1, 1) and radius 1.
2
2
The vector from the first circle’s centre to the second circle’s centre is (−4, −3) which has length (−4) + (−3) = 5.
So the point on the first circle, closest to the second is
(5, 4) +
2
(−4, −3) = (5, 4) + (−1.6, −1.2) = (3.4, 2.8) .
5
The answer is (a).
1
E: Let
fn (x) = (1 − x)n (2 − x)2n (3 − x)3n (4 − x)4n (5 − x)5n .
• If x = 4 then fn (x) = 0 and so (a) and (d) are false.
• If n = 6 then each exponent in fn (x) is even and so (c) is false.
• If x > 5 then each bracket is negative, and if n is odd then
fn (x) = (negative) (positive) (negative) (positive) (negative) < 0.
The answer is (b).
F: If we set y = 2x then the equation 8x + 4 = 4x + 2x+2 can be rewritten as
y 3 + 4 = y 2 + 4y
⇐⇒ y 3 − y 2 − 4y + 4 = 0
⇐⇒ (y − 1) y 2 − 4 = 0
So y = 1, 2, −2 are the possible values for y. But as y = 2x > 0 then only positive values for y will lead to real values
for x. Hence y = 1, 2 and x = 0, 1 are the only possible x-values. The answer is (c).
G: If y = 2−x sin2 x2 then note that y > 0, which discounts (b). Also y (0) = 0 which discounts
2 (d). Finally the
2
−x
points where graph (c) meets the x-axis arise regularly — this is not the case with y = 2 sin x where y = 0 at
√
√ √
x = π, 2π, 3π, . . . The answer is (a).
H: If we set
A=
1
f (x) dx,
B=
0
then we have the equations
3A + 2B = 7,
2
f (x) dx
1
(A + B) + B = 1.
Solving these simultaneous equations we find A = 3 and B = −1. Hence
2
f (x) dx = A + B = 3 − 1 = 2.
0
The answer is (d).
I: Note that a is largest when log10 a is largest. As 4 (log10 a)2 + (log10 b)2 = 1 then (log10 a)2 is largest when b = 1
and log10 b = 0. So
√
1
=⇒ a = 10.
4 (log10 a)2 = 1 =⇒ log10 a =
2
The answer is (c).
J: Note that
(n + 1) + n4 + 2 + n9 + 3 + n16 + 4 + · · · + n10000 + 100
increases as n increases. So the inequality will hold for all n 1 if it holds for n = 1. So we need
(1 + 1) + (1 + 2) + (1 + 3) + (1 + 4) + · · · + (1 + 100) > k
⇐⇒ 2 + 3 + 4 + · · · + 101 > k
100
(2 + 101) > k
⇐⇒
2
⇐⇒ 5150 > k.
The answer is (d).
2
QUESTION 2: We have
fn (x) = (2 + (−2)n ) x2 + (n + 3) x + n2 .
(i) So
2
2
1
7
1
21
3
2
= −6
x−
−
= −6 x −
+ .
f3 (x) = −6x + 6x + 9 = −6 x − x −
2
2
4
2
2
2
So the maximum is
21
2
= 10.5 achieved at x = 1/2.
For any n, fn (x) is a quadratic in x which has a maximum when the lead coefficient is negative. If 2 + (−2)n < 0
then n is an odd number greater than 1.
(ii) Setting n = 1 we have f1 (x) = 4x + 1. So
f1 (f1 (x)) = 4 (4x + 1) + 1 = 16x + 5
f1 (f1 (f1 (x))) = 4 (16x + 5) + 1 = 64x + 21
More generally
4k − 1
.
f1k (x) = 4k x + 1 + 4 + · · · + 4k−1 = 4k x +
3
(iii) Setting n = 2 we have f2 (x) = 6x2 + 5x + 4. So f2k (x) is a polynomial of degree 2k .
QUESTION 3: (i)
(ii) As x − c2 + c2 0 for all x then I (c) 0.
(iii)
I (c) =
0
1
2
2
(x − c) + c
(x − c)3
dx =
3
1
0
+ c2 =
(1 − c)3 c3
1
+
+ c2 = 2c2 − c + .
3
3
3
(iv) Completing the square
2
2
5
5
c 1
1
1
2
=2
+
=2 c−
+ .
I (c) = 2 c − +
c−
2 6
4
48
4
24
So the minimum is 5/24.
(v) If c can only vary between ±1 then the maximum is at I (−1) as −1 is furthest from 1/4. In this case
2
5
50
5
150 + 10
160
10
5
+
=
+
=
=
= .
I (−1) = 2
4
24
16 24
48
48
3
3
QUESTION 4: (i) Let C = (1, 1) denote the centre of the circle. then CQ makes angle θ with the vertical and is of
length 1. So
−−→
Q = C + CQ = (1, 1) + (sin θ, cos θ) = (1 + sin θ, 1 + cos θ) .
The gradient of the line is
−RO
= − tan θ
PO
by looking at the triangle OP R. So using the formula y − yQ = m (x − xQ ) we have
y − 1 − cos θ = − tan θ (x − 1 − sin θ) .
At P we have y = 0 and so we have
x = cot θ (1 + cos θ) + 1 + sin θ =
cos θ + cos2 θ + sin θ + sin2 θ
= 1 + cot θ + csc θ.
sin θ
(ii) If we consider the diagram with π/2 − θ as the angle OP R rather than θ, then this is just a reflection of the
θ-diagram in the y = x line. Hence, comparing areas,
A (θ) = B (π/2 − θ) .
So when θ = π/4 we have, dividing up the triangle
But A (π/4) = B (π/4) and Pπ/4
1
A (π/4) + B (π/4) + 3π/4 + 1 = Pπ/4 Rπ/4
2
√
√
= Rπ/4 = 1 + 1 + 2 = 2 + 2. Hence
2A (π/4) + 3π/4 + 1 =
giving
√ 2
√
1
2+ 2 =3+2 2
2
√
3π
2−
.
8
(iii) Let D = (1, 0). When θ = π/3 we can calculate A (π/3) as the area of the congruent right-angled triangles DCP
and P CQ minus 1/3 of the circle. So
π
2
π √
π
1
1 × Pπ/3 − 1 × 1 − = 1 + √ + √ − 1 − = 3 − .
A (π/3) = 2
2
3
3
3
3
3
A (π/4) = 1 +
QUESTION 5: (i)
f (5) = 2f (4) = 2 (f (2))2 = 2
2
2 f (1)2
= 2 22
= 32.
(ii) As we had to calculate f (4) , f (2) , f (1) , f (0) on the way then f (5) has recursion depth 4.
(iii)
g (5) = 1 + g (4) = 1 + 1 + g (2) = 1 + 1 + 1 + g (1) = 1 + 1 + 1 + 1 + g (0) = 4.
(iv) For any natural number k
g 2k = 1 + g 2k−1 = · · · = k + g 20 = k + g (1) = k + 1.
(v) For natural numbers l > k 0
g 2l + 2k = k + g 2l−k + 1 = k + 1 + g 2l−k = k + 1 + l − k + 1 = l + 2.
(vi) In the definition of g (n) a further 1 is added to previously calculated values at each stage whether n is even or
odd; as g (0) = 0 then g (n) is a measure of the number of previously calculated values, i.e. g (n) equals the recursion
depth.
4

Past Paper and Sample Solutions from Wednesday 31 October 2007

Transcription

Similar documents