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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)

Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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Extract from 2011 Mathematics Admissions Test

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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

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2.0

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0.0 0.0

0.5

1.0

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1.5

2.0

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(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.

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