SOLUTIONS FOR ADMISSIONS TEST IN
MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS
WEDNESDAY 02 NOVEMBER 2022
Mark Scheme:
Each part of Question 1 is worth 4 marks which are awarded solely for the correct answer.
Each of Questions 2–7 is worth 15 marks
x ≥ 0 and x2 + 1 = 3x or x < 0 and −x2 + 1 = −3x. The first equation√has solutions
A Either √
1
2
3 ± 5 which are both positive. The second equation has solutions 21 3 ± 13 and only
one of these is actually negative. So there are three solutions in total.
The answer is (d).
B The relationship between circle Cn of radius rn and circle Cn+1 of radius rn+1 is shown below.
rn+1 1
rn
1
The tangent is at right-angles to the radius, so there’s a right-angled triangle with hypotenuse
2
rn+1 and other sides rn and 1. Pythagoras gives rn+1 = rn2 + 1. Since r12 = 1, we have r22 = 2
and r32 = 3 and so on, up to r100
2
= 100, so the radius of C100 is 10.
The answer is (d).
C Complete the square for x and for y (x − 2k)2 − 4k 2 + (y − 2)2 − 4 + 8 = k 3 − k This is the
equation of a circle with centre (2k, 2) and r2 = k 3 + 4k 2 − k − 4 provided that expression is
positive. Factorising the cubic as (k + 4)(k − 1)(k + 1) reveals that this happens if and only if
either −4 < k < −1 or if k > 1.
The answer is (b).
4
D a1 = 8 × (3)4 and a2 = 8(8 × 34 )4 = 8 × 84 × (34 )4 and a3 = 8 × (8 × 84 × (34 )4 ) . In a similar
way,
n−1 n
an = 8 × 84 × 816 × · · · × 84 × 3(4 )
The exponent on the 8s is 1+4+16+· · ·+4n−1 = 31 (4n − 1). Then 81/3 = 2. Also use 4n = 22n .
2n
22n −1 (22n ) 62
an = 2 3 =
2
20
So for n = 10 we get 21 6(2 ) .
The answer is (e).
E We’ll calculate the square of (x + 1 + x−1 ) first;
x2 + 2x + 1 + 2x−1 + x−2
(I drew a square grid to keep track of all the cross-terms, and there’s a nice pattern which
helps). Now if we were to square this expression, the constant term independent of x would be
2(x2 )(x−2 ) + 2(2x)(2x−1 ) + 32
Most of the terms have a factor of 2 because they occur in either order. This sum is 2+8+9 = 19.
The answer is (c).
F Write α = sin 72◦ . We can substitute θ = 72◦ into the given equation, and then since
sin(360◦ ) = 0, we get the equation
0 = 16α5 − 20α3 + 5α
Either α = 0 (which is not the case because 18◦ isn’t an odd multiple of 90◦ ), or we have the
following quadratic equation for α2 ;
16(α2 )2 − 20(α2 ) + 5 = 0 (1)
q √
The quadratic formula followed by a square root, shows that α = ± 5±8 5 . One of these
values is sin(72◦ ), but what are the others? Thinking back to the point where we substituted
in θ = 72◦ , we can see that substituting θ = 36◦ × n would have given the same equation for α
for any whole number n, based on the positions of the zeroes of sin x. So these roots are sin 36◦
and sin 72◦ and the negatives of those two expressions. We want sin 72◦ which is positive and
larger than sin 36◦ so we should take the root with two + signs.
The answer is (a).
√
G This expression is true for all real n, so write n = m/ 2 and multiply through by 4 to get
m4 + 4 = (m2 + 2m + 2)(m2 − 2m + 2)
Relabelling m as n gives us an interesting way to factorise n4 + 4, so this expression is not
prime unless at least one of m2 ± 2m + 2 = 1. Completing the square(s) gives (m ± 1)2 + 1 = 1
so m = ±1. We want positive whole numbers, so we just need to check that 1 + 4 = 5 is prime
to find the only example of a prime number of this form.
The answer is (b).
H We can use laws of logarithms to write the right-hand side of the given equation as
log2 2x3 + 6x2 + 6x + 2 .
Since log2 (x) is an increasing function for x > 0, we can compare the arguments of the loga-
rithms, provided that both are positive. This gives the polynomial equation
2x3 + 7x2 + 2x + 3 = 2x3 + 6x2 + 6x + 2
which rearranges to x2 − 4x + 1 = 0, which has two real solutions. We should check that
2x3 + 7x2 + 2x + 3 is positive for these roots, but it definitely is because the roots of the
quadratic are both positive and all the coefficients of the cubic are positive.
The answer is (c).
I The game is fair; the distribution of the number of heads that Alice gets is the same as the
distribution of the number of tails that Bob gets. All we have to work out is the probability of
a draw, and then subtract that from 1 and divide by 2.
For a draw, both players would need to see the same number of their target side of the coin.
There are 2 2 2 2 2 2
5 5 5 5 5 5
+ + + + +
0 1 2 3 4 5
1
ways that might happen, each with probability 1024 . Adding that up gives the probability of a
63
draw to be 256 , so the probability that Alice wins is 12 − 512
63
= 193
512
.
Alternatively, count the ways that Alice wins (a bit slower).
The answer is (a).
J Write y = mx + c. A repeated root becomes a point (x, y) where the line y = mx + c is tangent
to the curve x2 + y 2 = 1 and also tangent to the curve (x − 3)2 + (y − 1)2 = 1. These are circles.
There are four lines that are tangent to both circles.
The answer is (e).
2
(i) (x2 +N y 2 )2 −19(2xy)2 = x4 +2N x2 y 2 +N 2 y 4 −76x2 y 2 and we have x4 −38x2 y 2 +192 y 4 = z 2 . So
comparing coefficients we see that if 192 = N 2 and 2N − 76 = −38 then the printed expression
is equal to z 2 . This happens if N = 19. 2 marks
(ii) If x = 13 and y = 3 then z = −2.
Using the relationship in part (i), we get (13 + 19 × 32 )2 − 19(2 × 13 × 3) = (−2)2 . So
3402 − 19 × 782 = 4. 3 marks
(iii) Now divide both sides by 4 to get 1702 −19×392 = 1. This is not the only solution; for example
we could use the relationship in (i) to generate the solution (1702 +19×392 )2 −19(2×170×39)2 =
1. 3 marks
(iv) Suppose x2 − 25y 2 = 1. The left-hand side factorises to (x − 5y)(x + 5y). Both of these brackets
are whole numbers, so they must both be 1 or both be −1. But then adding be 2x = ±2 and
we want x > 1. So there are no solutions. 3 marks
(v) Modify the relationship in part (i) to show that if x2 −17y 2 = z then (x2 +17y 2 )−17(2xy)2 = z 2 .
Now note that x = 4 and y = 1 would give 42 − 17 × 1 = −1 so z = −1. Use these values of
x and y in the relationship to generate the solution (42 + 17 × 12 )2 − 17(2 × 4 × 1) = 1, or in
other words 333 − 17 × 82 = 1. 4 marks
3
(i) Sketch.
y
1.8
1.6
1.4
1.2
√ (0, 1) √
(− 2, 1) 1 ( 2, 1)
0.8
y = (x2 − 1)2
0.6
0.4
0.2
(−1, 0) (1, 0)
x
−1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 1.2 1.4
−0.2
−0.4
y = (x2 − 1)3
−0.6
−0.8
−1
(0, −1)
3 marks
(ii) (x2 − 1)2m ≥ 0 so the integral can’t be zero unless a = 0. 1 mark
R1 Ra
(iii) Note first that 0 (x2 − 1)2m−1 dx < 0 and then 1 (x2 − 1)2m−1 dx is positive and increases
without bound as a increases. So at some point there is a value of am such that the integral
from 1 to a balances out the negative contribution from the integral from 0 to 1. 2 marks
Ra a3 √
(iv) 0 1 (x2 − 1) dx = 0 so 31 − a1 = 0. Now a1 > 0 so a1 = 3. 3 marks
Ra 7 5 √
(v) Note that 0 (x2 − 1)3 dx = a7 − 3a5 + a3 − a. Check that this is negative when a = 2 and
√
positive when a = 3, so there’s a change of sign and therefore a solution a2 somewhere in
between. To do the check, it’s perhaps convenient to write the integral as
a
5a6 − 21a4 + 35a2 − 35
35
3 marks
√
(vi) Between 0 and 2 the integrand
√ is very close to zero if m is large;√it’s negative between 0 and
1 then
√ positive from 1 to 2. We’re told that the integral up to 2 is negative. But then for
x > 2 the integrand quickly grows, √ and the integral will quickly cancel that negative value.
So the value of am will be close to 2 for large m. 3 marks
4
(i) Sketch.
1.5 y
(4, 1)
1
0.5
x
5 10 15 20
−0.5
The turning point is at (4, 1). 3 marks
√ √
(ii) Note that 4x + 1 − x − 1 = 4x + 1 − 4x+1 4
− 34 . So this is a combination of a stretch and
two translations. First squash by a factor of 4 parallel to the x-axis, then translate 14 units to
the left and 34 units down.
The turning point is at 43 , 14 .
2 marks
(iii) There are three cases depending on whether x is to the left√or to the right of each of A and
B. If x < −1 then we want (1 − x)(−1 − x) = 1 so x = − 2. If −1 < x <√1 then we want
(1 − x)(1 + x) = 1 so x =
√ 0. If x > 1 then we want (x − 1)(x + 1) = 1 so x = 2. So the three
points are (0, 0) and (± 2, 0). 2 marks
(iv) We have ((x − 1)2 + y 2 )((x + 1)2 + y 2 ) = 1. Expand and collect terms for a quadratic in y 2 ;
y 4 + 2(x2 + 1)y 2 + x4 − 2x2 = 0
with (positive, real) solution
√
q
y= 4x2 + 1 − x2 − 1
4 marks
2 2
(v) This is the equation from
√ (ii)
with x replaced by x and with y replaced with y . The turning
point will now be at 23 , 12 . 2 marks
(vi) Sketch.
y
1
x
−1 1
−1
2 marks
5
(i) Pick 1, 3, 5, 7, 9, 11, 13. 1 mark
(ii) Pick 1, 3, 5, . . . , 2k+1. The items in odd positions are untouched when first picked. For example,
when door 2l + 1 is first picked, no door numbered greater than 2l − 1 has been picked before
that, so the item behind door 2l + 1 is what it originally was. Thus each opened door results
in a new item for Alice and k + 1 doors are opened. 2 marks
(iii) Pick 1, 1, 4, 4, 7, 7, 10, 10, 13, 13. Any permutation of this sequence works. Other sequences may
work. 1 mark
(iv) Pick 1, 1, 4, 4, . . . , 3l + 1, 3l + 1, . . . , 3k + 1, 3k + 1. (Permutations allowed.) Group the items as,
(a1 , a2 ), (a3 , a4 , a5 ), (a6 , a7 , a8 ), . . . , (a3k , a3k+1 ).
Observe that the first time any door from each group is picked, the items are in their original
position. Except at the boundaries, the door that is picked is always the one in the middle, i.e.
a door of the form 3l + 1. Thus Alice wins at least 2 items from each group, and there are k + 1
groups. 2 marks
(v) (a) Pick 1, 1, 3. Many permutations work: (1, 3, 1), (3, 3, 1), (3, 1, 3). But (1, 3, 3) or (3, 1, 1)
don’t work. Anything involving 2 also doesn’t work. 1 mark
(b) Pick 1, 1, 5, 5, 3. Note that 1 and 5 have to be picked at least once before picking 3; any
ordering of (1, 1, 5, 5, 3) that respects this property will work. No other sequence works; this
can be checked by (somewhat painful) exhaustive enumeration, reducing cases by branch
and bound. 1 mark
(vi) Pick 1, 1, 5, 5, 9, 9, 13, 13, 3, 7, 11. Note that 1 and 5 have to be picked at least once each before
picking 3; 5 and 9 have to be picked at least once each before picking 7, and so on. Any
permutation that respects these conditions works just fine as the proof below shows.
1 mark
(vii) Pick 1, 1, 5, 5, . . . , 4l + 1, 4l + 1, . . . , 4k + 1, 4k + 1, then followed by 3, 7, . . . , 4l + 3, . . . , 4k − 1.
This is a total of 3k + 2 door choices. Group items as below.
(a1 , a2 ), (a3 ), (a4 , a5 , a6 ), (a7 ), . . . , (a4l−1 ), (a4l , a4l+1 , a4l+2 ) . . . , (a4k−1 ), (a4k , a4k+1 ).
Note that Alice is guaranteed to get at least two items from the groups that have 2 or 3 elements
each and picks up all the items behind doors numbered 4l + 3 for some l. 2 marks
(viii) No. Since we only need to show that Alice can’t guarantee winning all items, it is OK to assume
that the host knows Alice’s entire sequence (which may be longer than 6). The host can always
deny Alice one of items a3 or a4 . Whenever Alice picks either door 1 or 2, the host will always
swap items behind doors 1 and 2; likewise for doors 5 and 6. Now suppose for the first time
Alice picks door 3 or 4. Without loss of generality this is door 3; then the host can permanently
deny item a4 to Alice. Depending on whether the next pick among {3, 4} is 3 or 4, the host
can always guarantee that item a4 is behind the door that Alice won’t pick next. For noticing
the importance of 3 and 4; for proper justification 4 marks
6
(i) No. Suppose A, B, C, don’t have the same opinion, e.g. A is △ and the other two are □.
Then after one round B will have △, but A and C will have □; then after the second round C
will have △, but A and B will have □; and after the third round we are back to the starting
configuration and this repeats indefinitely. 2 marks
(ii) Yes. Label the layers, 1, 2, 3, 4 from the bottom. Nodes in layer 1 never change their opinion.
Nodes in layer 2 may change their opinion after the first round but never subsequently, and so
on. So after three rounds, no opinions will change in the network. 3 marks
(iii) The network below works.
△A
D1 △ D2 △
C1 △ C2 △ C3 △ C4 △
B1 B2 B3 B4 B5 B6 B7 B8
As argued in the previous part, we have 4 layers, and the layers remain static after rounds 0,
1, 2, 3 respectively from the bottom. The node C1 changes its opinion only if B1 and B2 are
both □, the same for C2 , wrt B3 and B4 , etc. So if not all Bi are □, at least one of the Ci ’s
remains △ forever. Likewise, if one of the Ci ’s is △ forever, then one of the Di ’s remains △
forever. In which case A would retain its △ opinion. However, if all Bi were □, then after one
round all Ci would become □; after two rounds all Di would become □ and then finally after
three rounds A would become □ and the network would remain stable. 3 marks
(iv) No. Consider the combined network below.
N1 + N2
X△
A B C
△ □ □
Note that after one round, X and C will not have changed their opinion, but A and B will
both have flipped their opinions. After the second round, we return to the starting position.
This continues indefinitely. 3 marks
(v) (a) 2n . 1 mark
(b) Simulate the influencer network dynamics for 2n +1 rounds. Since there are only 2n possible
configurations, at least two assignments are equal. If the network is stable, the period of the
system must be 1, so the assignment at time 2n and 2n + 1 must be the same. Otherwise,
the system is not stable. 3 marks
7
(i) Not all are achievable. We cannot have both the odd tokens in decreasing order and the even
tokens in increasing order. So the valid sequences (531642) and (642531) are not achievable.
2 marks
(ii) There are 8. All the even tokens must be together and likewise all the odd ones. We have
either evens first or second (2 choices), the evens could be increasing or decreasing (2 choices),
the odds could be increasing or decreasing (2 choices), multiplying together gives 23 = 8 valid
sequences. 3 marks
(iii) The data operator can achieve 6 sequences. All sequences where the odd tokens are increasing
are possible. When the odds are first, we just pass all odd tokens to output; depending on
whether the evens are increasing or decreasing, we either first use pushL or pushR on all evens,
and then eventually pop them all. When the odds are second, we always use pushL on the odds
at the start, for evens if they are increasing use pass directly, otherwise, use pushR on them.
Eventually all (remaining) tokens are in storage in the right order and then just pop everything.
When odds are decreasing, they must be all pushed in using pushR, if the evens are increasing
and appear first, they are directly sent to output using pass, if they appear second, then they
are pushed in using pushL and eventually everything is output using pop.
Making both odds and evens in decreasing order is not possible, as in order to make sequences
reverse the input order they must be stored using pushR; however, in this case they must be
interleaved in the storage unit, which makes the sequence invalid. 4 marks
(iv) There are a total of 3! · 23 = 48 such sequences. Call the three groups of tokens G0 , G1 , G2 ,
where Gi is all tokens that are of the form 3k + i. The groups can appear in any order, giving
3! choices, and then within each group we have two choices for each (increasing or decreasing).
3 marks
(v) There are only 3! = 6 possible achievable sequences. Note that the three groups defined above
G0 , G1 , G2 can’t be mixed in the storage unit. So one of the group needs to go straight to the
output channel using pass operations. This will appear as the first group in the output channel
and has to be in increasing order. For the other two groups, one will all have to be stored using
pushL operations and the other by pushR operations to avoid mixing. Thus, the second group
that is output will have to be in decreasing order and the last group in increasing order again.
So we no longer have any choice on increasing/decreasing order within the groups, but we can
decide which group goes first (use pass directly on them), which goes second (use pushR and
then pop after the first group is finished), and which goes last (use pushL and then pop after
the first two groups are finished). 3 marks
