Question 1.
(i) What are the two properties of a valid probability density function (pdf)?
(ii) Professor X’s arrival time at his lectures is given by the continuous random variable
Y = X0 + X, where X0 is the scheduled start time and X (measured in minutes),
has the following pdf:
−0.5x
ce
,
x > 2;
fX (x) =
0,
otherwise.
Determine:
(a) The value of c.
(b) Professor X’s expected arrival time if the scheduled start time is 11am.
(c) The probability that Professor X arrives more than 10 minutes late, given that
he is more than 8 minutes late.
M1S
Answer.
(i) Given that the range of X is X, then fX (x) must satisfy,
Z ∞
fX (x) dx = 1.
fx (x) ≥ 0, ∀ x ∈ X;
−∞
(2 marks)
(range must get a mention, but it is fine to just have integral over X)
(ii) (a)
Z
∞
2
∞
ce−0.5x dx = −2e−0.5 2 = 2ce−1 ⇒ c = 0.5e.
(2 marks)
(b)
Z
∞
∞
EfX (X) =
x0.5e
dx = e −xe−0.5x 2 +
2
∞ = e 2e−1 + −2e−0.5x 2 = 4.
1−0.5x
Z
∞
e
−0.5x
dx
2
(Could also note that X = W + 2 where W ∼ Exp(0.5) and EfW (W ) = 2.)
Hence
EfY (Y ) = EfX (X0 + X) = EfX (11:00 + X)
= 11:00 + EfX (X) = 11:00 + 0:04 = 11:04am.
(3 marks)
(c)
P((X > 10) ∩ (X > 8))
P(X > 10)
=
P(X > 8)
P(X > 8)
R∞
∞
0.5e1−0.5x dx
[−e1−0.5x ]10
10
= R∞
=
[−e1−0.5x ]∞
0.5e1−0.5x dx
8
8
e−4
= −3 = e−1 .
e
P(X > 10 | X > 8) =
(3 marks)
Could also note that
P(X > 10 | X > 8) = P(W > 8 | W > 6) = P(W > 2) by lack of memory
property of exponential.
Also P(W > w) = e−0.5x , so P(W > 2) = e−1 .
M1M1
Question 2. In lectures we showed the curvature, K, of a curve in the (x, y)-plane was
K=
y 00
.
(1 + y 02 )3/2
(a) If the curve is given parametrically in the form x = f (t), y = g(t), obtain a formula
for K in terms of f , g and t.
(b) In suitable units, Prof Corti’s nose is described in polar coordinates by r = cos θ. By
defining x and y parametrically in terms of θ and using part (a), find its curvature K,
simplifying your answer as much as possible.
(c) Sketch the curve r = cos θ.
Answer. (a) We have y 0 (x) = g 0 (t)/f 0 (t). Differentiating w.r.t. x, we have
0 00
dt f 0 g 00 − f ”g 0
f g − f 00 g 0
00
=
.
y =
dx
f 02
f 03
Thus
f 0 g 00 − f 00 g 0
1
f 0 g 00 − f 00 g 0
=
.
K=
f 03
(1 + (g 0 /f 0 )2 )3/2
(f 02 + g 02 )3/2
(3 marks)
(b) We have x = r cos θ. y = r sin θ so that we can define
f (θ) = cos2 θ,
g(θ) = cos θ sin θ.
Then writing c = cos θ and s = sin θ, we have f 0 = −2sc, g 0 = (c2 − s2 ), f 00 = 2(s2 − c2 ),
g 00 = −4sc. Substituting, we have
K=
2s4 + 2c4 + 4s2 c2
(−2sc)(−4sc) − 2(s2 − c2 )(c2 − s2 )
=
= 2.
[4s2 c2 + (c2 − s2 )2 ]3/2
(c2 + s2 )3
(5 marks)
(c) Constant curvature means a circle. Symmetry about θ = 0, and passing through (0, 1)
when θ = 0 and (0, 0) when θ = 21 π means a circle of radius 12 , centre ( 12 , 0). (2 marks)
Total : 10
[Notes for markers: In the first part, some will miss a factor of f 0 . There will be some
algebraic errors - use your judgement how much credit to give. If all markers agree, you
can redistribute marks between the parts. Indeed, you may to a large extent do as you
choose, but of course you must be consistent across all the scripts. Remember they only
have 15 minutes for this question. 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 3.
(a) For all real numbers r and s determine the type of the conic x21 + rx1 x2 + s = 0.
(You can use any method you like. You may need to consider several cases depending on the values of r and s.)
(b) Find the volume of the parallelepiped in R3 built on the vectors (1, −1, 1), (1, 1, 1)
and (1, 2, 4).
(c) Find the area of the triangle in R3 with corners (1, −1, 1), (1, 1, 1) and (1, 2, 4).
Answer.
(a) The characteristic polynomial t2 − t − r2 /2 has two real roots:
√
√
λ1 = (1 + 1 + r2 )/2 > 0 and λ2 = (1 − 1 + r2 )/2 ≤ 0;
moreover, λ2 = 0 if and only if r = 0. By lectures, after an appropriate rotation,
the equation takes the form
λ1 y12 + λ2 y22 + s = 0.
If r 6= 0 and s 6= 0, this conic is a hyperbola. If r = 0 and s > 0, the set of solutions
is empty; if r = 0 and s = 0, this is a double line; if r = 0 and s < 0, this is a pair
of parallel lines. If r 6= 0 and s = 0, this is a pair of lines meeting at a point.
(6 marks) [Subtract one mark for each missing degenerate case.]
(b)


1 −1 1 det  1
1 1  = 6
1
2 4 (2 marks)
(c)
1
|(0, 2, 0).(0, 3, 3)| = 3.
2
(2 marks)
M1F
Name (IN CAPITAL LETTERS!): . . . . . . . . . . . . . . . . . . . . . . . . . . . . TID:
CID: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Personal tutor: . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Question 4.
(a) Show that, for all integers n, 5n3 + 7n5 ≡ 0 (mod 12).
(b) Let a, b be integers > 0. Prove by any means that
hcf (a, b) = hcf a + b, lcm (a, b)
where, as usual, hcf and lcm are the highest common factor and the least common
multiple.
Answer.
(a) (5 marks) By the uniqueness part in the Chinese remainder theorem, is is
sufficient to show that A = 5n3 + 7n5 is ≡ 0 (mod 3) and (mod 4). We verify this
as follows:
(
0 if n ≡ 0 (mod 3)
A ≡ n3 (−1 + n2 ) (mod 3) ≡
0 if n ≡ ±1 (mod 3)
and, similarly
A ≡ n3 (1 − n2 )
(
0 if n ≡ 0, 2 (mod 4)
(mod 4) ≡
0 if n ≡ ±1 (mod 4)
(b) (5 marks) As usual write d = hcf (a, b), a = da0 , b = db0 with hcf (a0 , b0 ) = 1
and lcm (a, b) = da0 b0 . From:
a + b = d(a0 + b0 ) and lcm (a, b) = da0 b0
it follows that hcf a + b, lcm (a, b) = d hcf (a0 + b0 , a0 b0 ). Thus, to conclude, it is
enough to show that:
hcf (a0 + b0 , a0 b0 ) = 1
Suppose for a contradiction that p is a prime and p divides both a0 b0 and a0 + b0 .
Then, because p is prime and divides a0 b0 , then it divides a0 or b0 . In the first case
it also divides b0 = (a0 + b0 ) − a0 , and in the second case it also divides a0 . In both
cases, p divides both a0 and b0 , a contradiction since hcf (a0 , b0 ) = 1.1
1
This is a very hard question. I don’t expect many students will have answered. In part (b),
consider giving 2 marks for showing that one side divides the other.