Past Paper and Sample Solutions from Wednesday 5 November 2008

Transcription

Past Paper and Sample Solutions from Wednesday 5th November 2008
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 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.
Extract from 2008 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. The function
| = 2{3 6{2 + 5{ 7
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)
log10 >
p
(b)
log10 ( 2 )>
3
(c)
1
log10 ¶3
>
(d)
1
log10
s =
Turn Over
C. The simultaneous equations in {> |,
(cos ) { (sin ) | = 2
(sin ) { + (cos ) | = 1
are solvable
(a) for all values of in the range 0 6 ? 2;
(b) except for one value of in the range 0 6 ? 2;
(c) except for two values of in the range 0 6 ? 2;
(d) except for three values of in the range 0 6 ? 2.
D. When
1 + 3{ + 5{2 + 7{3 + · · · + 99{49
is divided by { 1 the remainder is
(a) 2000>
(b) 2500>
(c) 3000>
4
(d) 3500=
E. The highest power of { in
¾
½h
¡ 6
¢
¢3 ¡ 8
¢4 i5 h¡ 5
¡ 7
¢4 i6 3
2 5
2{ + 7 + 3{ 12
+ 3{ 12{ + { + 6
is
(a) {424 >
(b) {450 >
(c) {500 >
(d) {504 =
F. If the trapezium rule is used to estimate the integral
Z 1
i ({) d{>
0
by splitting the interval 0 6 { 6 1 into 10 intervals then an overestimate of the integral
is produced. It follows that
R1
(a) the trapezium rule with 10 intervals underestimates 0 2i ({) d{;
R1
(b) the trapezium rule with 10 intervals underestimates 0 (i ({) 1) d{;
R2
(c) the trapezium rule with 10 intervals underestimates 1 i ({ 1) d{;
R1
(d) the trapezium rule with 10 intervals underestimates 0 (1 i ({)) d{=
5
Turn Over
G. Which of the graphs below is a sketch of
|=
1
4{ {2 5
?
y
y
x
x
(a)
y
(b)
y
x
x
(c)
(d)
H. The equation
9{ 3{+1 = n
has one or more real solutions precisely when
(a) n > 9@4>
(c) n 6 1>
(b) n A 0>
6
(d) n > 5@8=
I. The function V (q) is dened for positive integers q by
V (q) = sum of the digits of q.
For example, V (723) = 7 + 2 + 3 = 12= The sum
V (1) + V (2) + V (3) + · · · + V (99)
equals
(a) 746>
(b) 862>
(c) 900>
(d) 924=
J. In the range 0 6 { ? 2 the equation
(3 + cos {)2 = 4 2 sin8 {
has
(a) 0 solutions>
(b) 1 solution>
7
(c) 2 solutions>
(d) 3 solutions=
Turn Over
(i) Find a pair of positive integers, {1 and |1 > that solve the equation
({1 )2 2 (|1 )2 = 1=
(ii) Given integers d> e> we dene two sequences {1 > {2 > {3 > = = = and |1 > |2 > |3 > = = = by setting
{q+1 = 3{q + 4|q >
|q+1 = d{q + e|q >
for q > 1=
Find positive values for d> e such that
({q+1 )2 2 (|q+1 )2 = ({q )2 2 (|q )2 =
(iii) Find a pair of integers [> \ which satisfy [ 2 2\ 2 = 1 such that [ A \ A 50=
(iv) (Using the values of d and e found in part (ii)) what is the approximate value of
{q @|q as q increases?
8
9
Turn Over
3.
;
MATHEMATICS
A
A
?
MATHEMATICS & STATISTICS
For APPLICANTS IN
MATHEMATICS & PHILOSOPHY
A
A
=
MATHEMATICS & COMPUTER SCIENCE
<
A
A
@
A
A
>
ONLY.
Computer Science applicants should turn to page 14.
(i) The graph | = i ({) of a certain function has been plotted below.
y
x
On the next three pairs of axes (A), (B), (C) are graphs of
| = i ({) >
i ({ 1) >
i ({)
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
2
| = 23{
and
2
| = 22{3{ =
Carefully label any stationary points.
(iii) Let f be a real number and dene the following integral
Z 1
2
23({3f) d{=
L (f) =
0
State the value(s) of f for which L (f) is largest. Briey explain your reasoning.
[Note you are not being asked to calculate this maximum value.]
10
|
6
p
p
p
p
p
p
p
p
p
p
p -
11
{
Turn Over
4.
<
;
@
? MATHEMATICS
MATHEMATICS & STATISTICS
ONLY.
For APPLICANTS IN
>
=
MATHEMATICS & PHILOSOPHY
Mathematics & Computer Science and Computer Science applicants should turn to
page 14.
y
Q
C
q
P
O
p
x
Let s and t be positive real numbers. Let S denote the point (s> 0) and T denote the
point (0> t) =
(i) Show that the equation of the circle F which passes through S , T and the origin R
is
{2 s{ + | 2 t| = 0=
Find the centre and area of F.
(ii) Show that
area of circle F
> =
area of triangle RS T
(iii) Find the angles RS T and RTS if
area of circle F
= 2=
area of triangle RS T
12
13
Turn Over
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 1000.
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 nth student changing the
state of every nth 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
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:
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. As y = 2x3 − 6x2 + 5x − 7 then
y ′ = 6x2 − 12x + 5.
The quadratic y ′ has discriminant 122 − 4 × 6 × 5 = 24 > 0 and hence the equation y ′ = 0 has
two distinct real roots. The answer is (c).
B. As π < 10 then
L = log10 π < 1.
So
√
√
log10 (π2 ) = 2L > L × L = L;
1
log10 π
3
= L−3 > 1;
2
√ = > 2.
log10 π
L
1
The answer is (a).
C. We will write c = cos θ and s = sin θ for ease of notation. Eliminating y from the simultaneous equations
cx − sy = 2,
sx + cy = 1;
we get
2c + s = c (cx − sy) + s (sx + cy) = c2 + s2 x = x
and similarly eliminating x we find
c − 2s = (−s) (cx − sy) + c (sx + cy) = s2 + c2 y = y.
Hence the equations care solvable for any value of θ. The answer is (a).
D. By the remainder theorem when a polynomial p (x) is divided by x − 1 then the remainder
is p (1). So the required remainder here is
1 + 3 + 5 + 7 + · · · + 99 =
50
(1 + 99) = 2500
2
as the series is an arithmetic progression. The answer is (b).
1
3
4
E. The highest power of x in (2x6 + 7) is x18 and in (3x8 − 12) is x32 so the highest power in
5
5
4
[· · · ]5 is (x32 ) = x160 . The highest power of x in (3x5 − 12x2 ) is x25 and in (x7 + 6) is x28 ,
6
6
so that the highest power of x in [. . .] is (x28 ) = x168 . Thus the highest power of x in {. . .}3
3
is (x168 ) = x504 . The answer is (d).
1
F. Suppose that, when the trapezium rule is used to estimate the integral 0 f (x) dx, an
overestimate of E is produced. If the same number of intervals are used in the following
calculations then: 1
(a) to estimate 0 2f (x) dx an overestimate of 2E will be produced, as the relevant graphs
have been stretched by a factor of 2 and all areas doubled;
1
(b) to estimate 0 (f (x) − 1) dx an overestimate of E will be produced, as the relevant
graphs have been translated
down by 1 and all areas remain the same;
2
(c) to estimate 1 f (x − 1) dx an overestimate of E will be produced, as the relevant graphs
have been translateright by 1 and all areas remain the same;
1
(d) to estimate 0 (1 − f (x)) dx an underestimate of E will be produced, as the relevant
graphs have been reflected in the x-axis — turning the overestimate to an underestimate — and
translated up by 1, which changes nothing with regard to areas. The answer is (d).
G. As 4x − x2 − 5 = − (x − 2)2 − 1, then y = (4x − x2 − 5)
minimum value at x = 2. The answer is (c).
−1
is always negative and has a
H. If we set y = 3x then the equation 9x − 3x+1 = k now reads
y 2 − 3y − k = 0.
This has solutions
√
9 + 4k
y=
2
x
which are real when k −9/4. As y = 3 then we further need that y > 0 for x to be real, but
this is not a problem as the larger root is clearly positive. The answer is (a).
3±
I. We have
S (1) + S (2) + S (3) + · · · + S (99) = S (00) + S (01) + · · · + S (99)
and in the 100 two-digit numbers 00, . . . , 99 there are twenty 0s, twenty 1s, . . . , twenty 9s. So
S (1) + S (2) + S (3) + · · · + S (99) = 20 × (0 + 1 + · · · + 9) = 20 ×
10
(0 + 9) = 900
2
and the answer is (c).
J. Note that
(3 + cos x)2 (3 − 1)2 = 4;
4 − 2 sin8 x 4.
So the equation will hold only wehn cos x = −1 and sin x = 0. In the range 0 x < 2π this
only occurs at x = π. The answer is (b).
2
2. (i) A fairly obvious pair (x1 , y1 ) that satisfy (x1 )2 − 2 (y1 )2 = 1 iis x1 = 3 and y1 = 2.
(ii) Note
(xn+1 )2 − 2 (yn+1 )2 = (xn )2 − 2 (yn )2
⇐⇒ (3xn + 4yn )2 − 2 (axn + byn )2 = (xn )2 − 2 (yn )2
⇐⇒ 8 − 2a2 (xn)2 + (24 − 4ab) xn yn + 18 − 2b2 (yn )2 = 0
In order to have (xn+1 )2 − 2 (yn+1 )2 = (xn )2 − 2 (yn )2 we need
2a2 = 8,
4ab = 24,
2b2 = 18.
We further require that a, b > 0. We see that a = 2 and b = 3 solve all three equations.
(iii) Starting with x1 = 3, y1 = 2 we find:
x1 = 3,
y1 = 2;
x2 = 3 × 3 + 4 × 2 = 17,
y2 = 2 × 3 + 3 × 2 = 12;
x3 = 3 × 17 + 4 × 12 = 99, y3 = 2 × 17 + 3 × 12 = 70.
So X = 99 and Y = 70 is such a pair.
(iv) For the generated sequences, (xn ) , (yn ) , we have
(xn)2 − 2 (yn )2 = 1 for each n.
Also the integers xn and yn are getting increasingly larger because of how they are defined in
(ii). So
2
1
xn
−2 =
≈ 0 for large n,
yn
(yn )2
√
and xn /yn ≈ 2 as xn and yn are both positive.
3
3. (i) (A) is −f (x) ; (B) is f (−x) ; (C) is f (x − 1) .
2
2
2
2
(ii) As 22x−x = 2 × 2−(x−1) then the graph of y = 22x−x is the graph of y = 2−x translated
to the right by 1 and stretched parallel to the y-axis by a factor of 2.
2.0
1.5
1.0
0.5
-2
1
-1
−x2
Plots of y = 2
2
2
and y = 2
2
2x−x2
3
.
(ii) c = 12 . The graph of 2−(x−c) is the graph of 2−x translated c to the right. The integral
I (c) represents the area under the graph between 0 x 1. As the graph is symmetric/even
and decreasing away from 0 then this area is maximised by having the apex half way along the
interval 0 x 1, i.e. at x = 1/2 which occurs when c = 12 .
4
4. (i) We can complete the squares in x2 − px + y 2 − qy = 0 to get
q 2 p2 + q 2
p 2 + y−
=
x−
2
2
4
(1)
which is the equation of the circle with centre: (p/2, q/2) and area: π (p2 + q 2 ) /4. Either by
checking the original question, or the rearranged one, we can see that

0
at (0, 0) ,

p2 − p2 + 0 = 0 at (p, 0) ,
x2 − px + y 2 − qy =

0 + q 2 − q 2 = 0 at (0, q) .
(ii) The area of OP Q is pq/2. So
area of circle C
=
area of triangle OP Q
Note
π (p2 + q 2 )
4
1
pq
2
=
π (p2 + q 2 )
.
2pq
π (p2 + q2 )
π ⇐⇒ p2 + q 2 2pq ⇐⇒ (p − q)2 0,
2pq
proving the required inequality.
(iii) Rearranging
π (p2 + q 2 )
= 2π ⇐⇒ p2 + q 2 = 4pq ⇐⇒
2pq
2
p
p
−4
+ 1 = 0,
q
q
which is a quadratic equation in p/q, and so
√
√
4 ± 16 − 4
p
=
= 2 ± 3.
q
2
Now p/q = tan OQP, q/p = tan OP Q and so
√ √
{tan OQP, tan OP Q} = 2 − 3, 2 + 3
with the order depending on whether p < q or p > q.
√ √ [It happens that arctan 2 − 3 = π/12 and arctan 2 + 3 = 5π/12, but appreciation of
this was not expected.]
5
5. (i) After the first/second/third students have gone by the doors look like:
Locker
1 2
Student 1 O O
Student 2 O C
Student 3 O C
3
O
O
C
4
O
C
C
5
O
O
O
6
O
C
O
7
O
O
O
8
O
C
C
9
O
O
C
10
O
C
C
11
O
O
O
12
O
C
O
13
O
O
O
14
O
C
C
We can see that the lockers now repeat in a pattern OCCCOO every 6 lockers. As 1000 =
166 × 6 + 4 we have 166 repeats of this pattern and 4 remaining lockers that go OCCC. So there
are 166 × 3 = 498 closed lockers amongst the complete cycles and 3 further in the incomplete
cycle. That is, there are 501 closed lockers in all.
(ii) After the fourth student has gone by we have the following:
Locker
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Student 3 O C C C O O O C C C O O O C
Student 4 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 contains 5 closed and 7 open doors. Now 1000 = 83 × 12 + 4 and so we have 83 × 5 = 415 closed
lockers amongst the complete cycles and 2 further amongst the incomplete cycle OCCO. In all
then there are 417 closed lockers.
(iii) Locker 100 starts off closed (as all lockers do) and then its state is altered by every nth
student where n is a factor of 100, i.e. by students 1, 2, 4, 5, 10, 20, 25, 50, 100. So 9 students
change the state and as this is odd then overall the state will have been changed to open.
(iv) Locker 1000 starts off closed (as all lockers do) and then its state is altered by every nth
student where n divides 1000 and n 100, i.e. by 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100. So 11
students change the door’s state and as this is odd then overall the state will again have been
changed to open.
6

Past Paper and Sample Solutions from Wednesday 5 November 2008

Transcription

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