M1F
Name (IN CAPITAL LETTERS!): . . . . . . . . . . . . . . . . . . . . . . . . . . . . TID:
CID: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Personal tutor: . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Question 1.
(a) Prove that if x, y, z are real numbers such that x + y + z = 0, then xy + yz +
zx ≤ 0.
(b) For each of the following statements, either prove it or give a couterexample
to show that it is false.
(i) The product of two rational numbers is alway rational;
(ii) The product of two irrational numbers is alway irrational;
(iii) The product of two irrational numbers is always rational;
(iv) The product of a non-zero rational and an irrational is always irrational.
Answer.
(a) ≤ (x + y + z)2 = x2 + y 2 + z 2 + 2(xy + yz + zx) ≥ +(xy + yz + zx).
4 marks
(b)
(i) True: it is one of the axioms of Q that there is a product on it.
1 mark
√ 2
√
but xy =√ ( 2) = 2 is not
(ii) False: x = y = 2 6∈ Q are both irrational
√
irrational. (If you want x 6= y take x = 2 and y = 2 2; but note that I
didn’t ask you for them to be distinct.)
1 mark
√
√
(iii) False: take x = 2 and y = √3. Both x and y are irrational (as shown in
class) and their product xy = 6 is also irrational (as shown in class).
2 marks
(iv) True. Suppose that x is irrational and y 6= 0 is rational. I claim that the
product z = xy is irrational. Indeed supposing for a contradiction that z were
rational, we would have that
xy
z
x=
= ∈Q
y
y
(indeed, because of the axioms of Q, y1 ∈ Q and then, by the axioms of Q
again, the product z × y1 ∈ Q) which contradicts the assumption that x is
irrational.
2 marks
M1S
Question 2.
You have two coins which look identical. One is fair, and the other comes up Heads with
probability 0.6. Assume that successive flips are independent for both coins. You flip one of
the coins five times and obtain the sequence HHT T H. You flip the other coin three times
and obtain the sequence HT H. Which of the coins is more likely to be the biased one?
Answer.
(i) Let A1 be the event that you obtain the sequence HHT T H and A2 be the event that
you obtain the sequence HT H. Let B be the event that the coin is biased.
We need to compare P(B | A1 ) and P(B | A2 ).
From Bayes we have:
P(A1 | B)P(B)
P(B | A1 ) =
P(A1 )
We have
P(A1 | B) = (0.6)3 (0.4)2 ,
P(A1 | B C ) = (0.5)5 ,
P(B) = 0.5
and
P(A1 ) = P(A1 | B)P(B) + P(A1 | B C )P(B C )
= (0.6)3 (0.4)2 (0.5) + (0.5)5 (0.5)
= (0.6)3 (0.4)2 (0.5) + (0.5)6
Giving
(0.6)3 (0.4)2
(0.6)3 (0.4)2 (0.5)
=
(0.6)3 (0.4)2 (0.5) + (0.5)6
(0.6)3 (0.4)2 + (0.5)5
(4 marks) (2 marks for theorem of total prob, 2 for correct formulation of
Bayes theorem)
P(B | A1 ) =
P(A2 | B) = (0.6)2 (0.4),
P(A2 | B C ) = (0.5)3 ,
P(B) = 0.5
and
P(A2 ) = P(A2 | B)P(B) + P(A2 | B C )P(B C )
= (0.6)2 (0.4)(0.5) + (0.5)3 (0.5)
= (0.6)2 (0.4)(0.5) + (0.5)4
Giving
P(B | A2 ) =
(0.6)2 (0.4)(0.5)
(0.6)2 (0.4)
=
(0.6)2 (0.4)(0.5) + (0.5)4
(0.6)2 (0.4) + (0.5)3
(4 marks)
Note
P(B | A1 ) =
(0.24)(0.6)2 (0.4)
(0.24) (0.6)2 (0.4) +
(0.5)2
(0.5)3
0.24
=
(0.6)2 (0.4)
< P(B | A2 )
0.25
(0.6)2 (0.4) + 0.24
(0.5)3
So the coin whose outcome is HT H is more likely to be the biased one. (2 marks)
M1M1
Question 3. After a Monday test, a maths lecturer (of your choice) mysteriously falls
from the top of the Queen’s Tower. After a time t > 0 he/she has fallen a distance x,
where
h
√ i
1
x = log cosh t k ,
k
and k is a positive constant related to air resistance (we have assumed gravity g = 1.)
(a) Find the first two terms in an approximation for x if k is small.
(b) If k is large (perhaps the lecturer had a parachute), show that
x ' U (t − t0 )
where U and t0 are constants you should find in terms of k.
Answer. (a) We have cosh ξ = 12 (eξ + e−ξ ) = 1 + 21 ξ 2 + 1/(24)ξ 4 + O(ξ 6 )
and so
1
1
1 2 4
x = log 1 + 12 kt2 + 24
k t + O(k 3 t6 ) = log(1 + u),
say.
k
k
Now
log(1 + u) = u − 21 u2 + O(u3 )
(1 mark)
(1 mark)
so
x=
1h
k
1 2
kt
2
+
1 2 4
k t
24
−
1
2
1 2
kt
2
+
1 2 4 2
k t
24
i
+ . . . = 12 t2 −
1
kt4
12
+ ...
(4 marks)
(b) When ξ is large, cosh ξ = 21 eξ + something exponentially small. It follows that
log cosh ξ ' log( 12 ) + log(eξ ) = − log 2 + ξ. So when k is large,
x'
√
1
(− log 2 + t k)
k
=⇒
1
U0 = √
k
log 2
and t0 = √
k
(4 marks)
Total 10
[Note: Strictly speaking, rather than k small or large we should consider kt2 being small
or large. Those of you who did A-level mechanics should have expected the 12 (g)t2 term
in part (a). In part (b), U0 is called the terminal velocity, as it is the limit of your speed
as t → ∞. For a human falling through air this is about 100-200 mph, which usually is
“terminal”. For smaller creatures such as mice, k is larger so that U0 is much smaller.
Small animals may survive a fall of any height.]
[Notes for markers: Give marks both for method as well as numerical exactness. I expect
many will forget the ui2 term in the log expansion You may award partial credit for good
attempts and redistribute marks between the parts as you choose, provided all markers
agree. Indeed, to a large extent, you may do as you choose, but of course you must be
consistent across all the scripts. The students will eventually see a copy of this sheet.
Don’t forget to initial your work to help in the “Meet your Marker” sessions.]
M1GLA
Name (IN CAPITAL LETTERS!): . . . . . . . . . . . . . . . . . . . . . . . . . . . . TID:
CID: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Personal tutor: . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Question 4.
For the following matrices A find the smallest integer m > 0 such that Am = I (the
identity matrix) or show that no such m exists:
−1 −1
−1 −1
−1 1
−1
0
0 −1
,
,
,
,
.
1
1
1
0
−1 0
1 −1
1
0
Answer.
(1) no such m exists, because A2 = 0, hence Am 6= I for any m; (2) m
= 3; (3) m
= 3;
1 0
(4) no such m exists, because one shows by induction that Am = ±
6= 0;
−m 1
(5) m = 4.
(10 marks) [2 marks for each case, if a complete proof is given in cases 1 and 4]