MATH1231 Mathematics 1B
and
MATH1241 Higher Mathematics 1B
PAST EXAM PAPERS
AND
SOLUTIONS
Copyright 2018 School of Mathematics and Statistics, UNSW

Contents
PAST EXAM PAPERS 4
NOVEMBER 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
NOVEMBER 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
FEBRUARY 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
NOVEMBER 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
NOVEMBER 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
NOVEMBER 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
PAST HIGHER EXAM PAPERS 35
NOVEMBER 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
NOVEMBER 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
NOVEMBER 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
NOVEMBER 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
NOVEMBER 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
PAST EXAM SOLUTIONS 51
NOVEMBER 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
NOVEMBER 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
FEBRUARY 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
NOVEMBER 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
NOVEMBER 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
NOVEMBER 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
PAST HIGHER EXAM SOLUTIONS 94
NOVEMBER 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
NOVEMBER 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
NOVEMBER 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
NOVEMBER 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
NOVEMBER 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
The 2016 to 2019 exam papers will be provided on Moodle.
3
4PAST EXAM PAPERS
The four in questions in each paper were of equal value. Until 2013, Question 1 was a common
question between MATH1231 and MATH1241. Since 2013 Questions 1 and Question 2 are in
common between MATH1231 and MATH1241.
The final results for MATH1231 and MATH1241 will still be moderated so that similar performance
in the common tasks will, on average, lead to similar results for students obtaining a pass or credit.
The emphasis on topics in the syllabus can change a little from year to year.
Important Note: The algebra syllabus has changed for 2012 onwards. The changes are minimal
and effect Chapter 9 Statistics and Probability. New sections on continuous random variables and
the Normal Distribution have been added. For MATH1241 there is also a new section on the
Exponential Distribution, while joint discrete distributions has been removed.
The solutions to the examination papers contained here have been written by many members of
staff of the School of Mathematics.
While we aim to provide perfectly correct solutions, we cannot guarantee that the solutions are
error-free. Please report any serious errors to the Director of First Year Mathematics.
The format of this terms exam may be different. Details of this terms exam will be provided on
Moodle.
5MATH1231 NOVEMBER 2010
1. i) Show that the set
S =
{
x =
(
x1
x2
)
∈ R2 : x1x2 ≥ 0
}
is not a vector subspace of the vector space R2.
ii) The vectors v1, v2, v3, u and w in R
3 are defined by
v1 =

 13
−2

 , v2 =

 2−5
7

 , v3 =

 4−21
25

 , u =

31
2

 and w =

ab
c

 ,
with the set D given by D = {v1,v2,v3} and the matrix A defined in the Maple session
below.
You may use the following Maple session to assist in answering the questions below.
> with(LinearAlgebra):
> A:= <<1,3,-2>|<2,-5,7>|<4,-21,25>>;
A :=

 1 2 43 −5 −21
−2 7 25


> M:=>;
M :=

 1 2 4 a3 −5 −21 b
−2 7 25 c


> GaussianElimination(M);
 1 2 4 a0 −11 −33 b− 3a
0 0 0 b+ c− a


a) Is D a linearly independent set? Give reasons.
b) State the condition(s) for the vector w to belong to span(D).
c) Determine all possible real scalars α1, α2 and α3 such that
u = α1v1 + α2v2 + α3v3.
d) Find a basis for the kernel of A.
e) State the rank of the matrix A.
f) Find a basis for the image of A.
iii) Find each of the following integrals:
a) I1 =
∫
cosh3 x dx;
b) I2 =
∫ 1
0
x2√
4− x2 dx.
6iv) Find the general solution to the differential equation
y′′ − 5y′ + 4y = 6e2x + 12x.
2. i) For n = 0, 1, 2, . . . , let In =
∫ pi/2
0
sinn x dx. You are given that
In =
n− 1
n
In−2 for n ≥ 2.
Use this recurrence to calculate I6.
ii) Determine whether each of the following sequences converges or diverges, giving reasons
for your answers. Find the limit of each sequence that converges.
a)
{
lnn√
n
}
n = 1, 2, 3, . . .
b)
{
(−1)n
(
1− 1
n
)}
n = 1, 2, 3, . . .
iii) By using an appropriate test, determine whether each of the following series converges
or diverges:
a)
∞∑
k=1
k
k + 1
;
b)
∞∑
k=1
1
2k2 + 1
;
c)
∞∑
k=1
ek
k!
.
iv) Let A =
(
5 −8
1 −1
)
.
a) Determine the eigenvalues and corresponding eigenvectors for the matrix A.
b) Write down matrices P and D such that A = PDP−1.
c) Hence evaluate A8P .
v) Suppose that the fixed vector b in R3 is given by
b =

13
5

 .
You are given that the function T : R3 → R3, defined by
T (x) = (x · b)b for x ∈ R3
is a linear map.
Find a matrix B which transforms the vector x into T (x).
73. i) Team A has played all its matches in the home and away season and its qualification for
the finals depends on the results of three remaining matches. Teams B, C and D each
have one match against three other teams and team A will make the finals if team B
loses its last match and either of teams C or D also lose their last match. Sports analysts
have estimated that:
• team B will lose with probability 0.8;
• team C will lose with probability 0.3 and
• team D will lose with probability 0.6.
The result of any of these matches is independent of the other teams’ results and drawn
matches are not possible. Based on these data, calculate the probability that team A
will make the finals. Show all of your working.
ii) Let X be a discrete random variable with the following probability distribution:
k : 0 1 2 3 4
P (X = k) : 0.2 0.3 0.1 0.2 0.2
a) Determine E(X) and Var[(]X).
b) A random variable Y is defined by Y = −2X + 3. Determine E(Y ) and Var[(]Y ).
iii) In 1995 half of the women in Australia were taller that 164cm, and half were shorter. In
2010, clothing manufacturers need to know whether women are in general taller than in
1995, and so fifteen women were chosen at random and their heights recorded:
162 165 168 171 170 166 165 162
151 166 172 178 168 166 167
a) Assume that in 2010 Australian women were equally likely to be taller or shorter
that 164cm. What distribution could be used to model the number of women in a
random sample of 15 who were taller than 164cm?
b) Using the assumption made in part (a), determine a tail probability that measures
how unusual it would be to observe as many women taller than 164cm as was
observed in this sample.
c) From the results of part b), is there evidence that in 2010 more than half of women
in Australia are taller than 164cm?
iv) A matrix A has eigenvectors
v1 =
(
2
1
)
and v2 =
(
1
−1
)
with corresponding eigenvalues λ1 = 2 and λ2 = −3, respectively.
Determine Ab for the vector b =
(
1
1
)
.
4. i) Solve the initial value problem
(cos2 x)
dy
dx
+ y = 1 where y(0) = 3.
ii) The point P with coordinates (2, 1, 7) lies on the surface S given by z = x2 + xy2 + y4.
8a) Find the Cartesian form of the equation of the plane tangent to the surface S at the
point P .
b) Write down a vector normal to the surface S at the point P .
iii) Suppose that f(x) =
1
x
. Find the Taylor polynomial of degree 4 for f about x = 1.
iv) The position (x(t), y(t)) of a particle P at time t is given by
x(t) = t2, y(t) = t3 where t ≥ 0.
a) Find the speed, v, of the particle as a function of t.
b) Hence, or otherwise, find the length of the path travelled by the particle between
times t = 0 and t = 1.
v) A tank has a total volume of 200 litres. Initially it holds 40 litres of pure water. Brine
containing 2 grams of salt per litre is run into the tank at the rate of 3 litres per minute
and the mixture is stirred continuously so that the concentration of the dissolved salt is
uniform throughout the tank. At the same time as the brine starts to flow into the tank
the mixture is removed from the tank at the rate of 1 litre per minute.
Let t denote time, measured in minutes, from when the brine started to enter the tank
and let x(t) denote the mass of salt, in grams, present in the tank after t minutes.
Set up a first order differential equation in x and t which models this system up until
the time the tank is full.
9MATH1231 NOVEMBER 2011
1. i) Consider a set of vectors
S = {v1,v2,v3,v4} ⊂ R3
a) Can S be a spanning set for R3? Give a reason for your answer.
b) Will all such sets S be spanning sets? Give a reason for your answer.
c) Suppose S consists of the vectors
v1 =

 12
3

 , v2 =

 33
8

 ,v3 =

 1−1
2

 , v4 =

 −11
0

 .
Determine a subset of S that forms a basis for R3.
ii) Suppose A is a fixed matrix in Mm,n. Apply the subspace theorem to show that
S = {x ∈ Rn : Ax = 0}
is a subspace of Rn.
iii) A linear map T : Rm → Rn has rank k. State the value of the nullity of T .
iv) The following MAPLE session may assist you with (b) below.
> f := (x^2-8*x-11)/((x+3)*(x^2+2)):
> convert(f, parfrac,x);
−x− 5
x2 + 2
+ 2 (x+ 3)−1
Evaluate each integral:
a) I1 =
∫
sin3 θ cos5 θ dθ
b) I2 =
∫
x2 − 8x− 11
(x+ 3)(x2 + 2)
dx
v) a) Determine whether the sequence an =
2n + lnn
1 + 3n
converges or diverges as n → ∞.
If it converges, find its limit.
b) Does the series
∞∑
n=2
(−1)n 2n+ lnn
1 + 3n
converge? Give a reason for your answer.
vi) By using an appropriate test, determine whether the series
∞∑
n=1
5n
n!
converges or diverges.
10
2. i) a) Find the general solution of y′′ + 4y = cos x.
b) If the right-hand side is changed to cos 2x, write down the form of the particular
solution you would try. (Do not attempt to find the unknown coefficient(s).)
ii) In a direct-current circuit, the total resistance z, (measured in ohms), produced by two
parallel resistors with resistances x and y ohms, is given by
1
z
=
1
x
+
1
y
.
The values of x and y are measured to be 6 ohms and 12 ohms respectively and each of
the measurements is made with an error whose absolute value is at most 0.1 ohms.
a) Use the given measurements (ignoring measurement error) to calculate the total
resistance z.
b) Show that
∂z
∂x
=
z2
x2
and
∂z
∂y
=
z2
y2
.
c) Use the total differential approximation to estimate the maximum absolute error in
the calculated value of the total resistance z, correct to three significant figures.
iii) Show that the map T : R2 → R3, defined by
T (x) =

 x1 + x2x1x2
−x2

 , for x = ( x1
x2
)
∈ R2,
is not linear.
iv) Consider the matrix
A =

 3 2 31 4 3
1 2 5

 .
The mapping T : R3 → R3 defined by
T (x) = Ax− 2x, for x ∈ R3,
is linear. (You are not required to prove this.)
a) Find the matrix B for T , such that T (x) = Bx.
b) Explain why

 −30
1

 belongs to the kernel of T .
c) Write down the rank and nullity of T .
v) Consider the matrix
A =
(
3 2
0 −1
)
.
a) Verify that A has eigenvalues 3 and −1 with eigenvectors
(
1
0
)
and
(
1
−2
)
re-
spectively.
11
b) Hence determine the solution to the system of differential equations
dx
dt
= Ax, x(0) =
(
2
1
)
.
3. i) Consider the matrix
A =
(
5 −10
2 −4
)
.
a) Find A
(
2
2
)
.
b) Find a basis for the kernel of A.
c) Find a basis for the image of A.
d) Given that A2 = A, prove that if b belongs to the image of A, then Ab = b.
e) Hence, (or otherwise), write down the eigenvalues and corresponding eigenvectors
of A.
ii) It has been determined that 80% of supporters of the Liverpool Football Club wear red
shirts on a day Liverpool plays a game. On those days 30% of non-supporters will wear
a red shirt. If 5% of all people support Liverpool, what is the exact probability a person
wearing a red shirt on a game day is a Liverpool supporter?
iii) Fred recorded the average maximum monthly temperatures at his home each month in
2009 and then again in 2010. These temperatures are shown in degrees centigrade in
the following table:
Year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2009 26 31 27 24 26 19 21 15 23 28 31 30
2010 24 30 28 26 26 22 22 19 24 29 32 26
a) What distribution could be used to model the number of times there was an increase
in maximum temperature rather than a decrease from one year to the next?
b) Calculate a tail probability that measures how unusual it would be to observe as
many months with such an increase in temperature from 2009 to 2010 as was ob-
served, if increases and decreases are equally likely.
iv) A fair coin is tossed four times and the absolute value of the difference between the
number of heads and tails is recorded as a random variable X. (Thus, if X = 4, then
either exactly 4 heads appeared, or exactly 4 tails appeared.) A probability distribution
table is partially completed below.
x 0 2 4
P (X = x) 12
a) Explain why P (X = 4) = 18
b) Find P (X = 0).
c) Find the expected value E(X) of X.
d) What is the probability that X exceeds E(X)?
12
4. i) For n ≥ 1, let In =
∫ 1
0
dx
(1 + x2)n
.
You are given that
In+1 =
1
n2n+1
+
2n− 1
2n
In, for n ≥ 1.
Use this reduction formula to calculate I2.
ii) A ship of mass m starts from rest under the force of a propeller thrust of L Newtons. It
experiences a resistance in Newtons given by kv, where v is the velocity of the ship in
metres/sec, and k is a positive constant. We assume that no other forces are in play.
Newton’s second law tells us that the force F acting on the ship may be written as
F = mdvdt .
a) By considering the forces acting on the ship, briefly explain why
dv
dt
+
k
m
v =
L
m
.
b) By treating this as a linear equation, find the integrating factor and hence solve the
equation for v as a function of t.
c) Explain why the speed of the ship cannot exceed Lk .
d) At what time will the ship reach half this speed?
iii) The Maclaurin series for tan−1 x, valid for |x| < 1, is given by
tan−1 x =
∞∑
n=0
(−1)n x
2n+1
2n+ 1
.
a) Using an appropriate convergence test, show that the Maclaurin series for tan−1 x
also converges at x = 1.
b) Hence find the value of the infinite series
1− 1
3
+
1
5
− 1
7
+ · · ·
iv) In this question you may use the Fundamental Theorem of Calculus. That is, for a
continuous function f on [a, b],
d
dt
( ∫ t
a
f(u) du
)
= f(t),
for each t ∈ (a, b).
The curve in Figure 1 is defined by the parametric equations
x(t) =
∫ t
1
cos u
u
du and y(t) =
∫ t
1
sinu
u
du, t ≥ 1,
and it has many vertical tangents. A curve in parametric form has a vertical tangent
when dxdt = 0 but
dy
dt 6= 0.
13
Figure 1
a) Find
dx
dt
and
dy
dt
.
b) Show that the curve defined above has a vertical tangent at the point where t =
t1 =
pi
2 .
c) State the next value, t2, of t, at which the curve has a vertical tangent.
d) Find, in simplest form, the length of the arc of the curve from t1 to t2.
14
MATH1231 FEBRUARY 2012
1. i) Let S = {x ∈ R2 : x1x2 ≥ 0}.
a) Show that S is closed under multiplication by a scalar.
b) Show that S is a not a subspace.
ii) Consider the set S = {v1,v2,v3,v4} of vectors in R4 given by
S =




1
−2
1
2

 ,


2
−4
2
4

 ,


1
0
5
0

 ,


3
−4
8
7



 ,
and let b =


−2
1
−10
−7

 .
a) Does b belong to span(S)? Give reasons.
b) Does S span R4? Give reasons.
c) Write down a basis for span(S).
iii) A mapping T : R3 → R2 is defined by
T (x) =
(
x1 − x3
2x1 + x2
)
.
a) Show that T is linear.
b) Write down a matrix A such that T (x) = Ax.
iv) The linear map T : R5 → R5 has matrix
A =


1 3 −3 1 4
−1 3 −9 2 −1
2 −4 14 3 −3
−1 −3 3 −1 −4
0 6 −12 3 3

 .
Using the Maple below, or otherwise, answer the following questions, giving brief reasons.
a) Find rank(T ).
b) Find nullity(T ).
c) Find a basis for ker(T ).
d) Write down conditions on b ∈ R5 for b to be in im(T ).
e) Find a basis for R5 containing as many columns of A as possible.
15
> with(LinearAlgebra):
> A:=<<1,-1,2,-1,0>|<3,3,-4,-3,6>|<-3,-9,14,3,-12>|
> <1,2,3,-1,3>|<4,-1,-3,-4,3>>:
> Id5:=IdentityMatrix(5):
> AI:=;
AI :=


1 3 −3 1 4 1 0 0 0 0
−1 3 −9 2 −1 0 1 0 0 0
2 −4 14 3 −3 0 0 1 0 0
−1 −3 3 −1 −4 0 0 0 1 0
0 6 −12 3 3 0 0 0 0 1


> ReducedRowEchelonForm(AI);

1 0 3 0 2 0 0 112 −56 −1336
0 1 −2 0 1 0 0 − 112 −16 136
0 0 0 1 −1 0 0 16 13 518
0 0 0 0 0 1 0 0 1 0
0 0 0 0 0 0 1 0 −1 −1


16
2. i) Let
A =
(
4 −2
7 −5
)
.
a) Find the eigenvalues and corresponding eigenvectors of A.
b) Hence find the general solution to the system of linear differential equations
dy1
dt
= 4y1 − 2y2
dy2
dt
= 7y1 − 5y2.
ii) You are given the following information about two events A and B.
P (B) = 0.65, P (B|A) = 0.95, P (Bc|Ac) = 0.85.
Using a tree diagram, or otherwise, find:
a) the exact value of P (A),
b) the value of P (A|B) correct to two decimal places.
iii) A continuous random variable X has probability density function given by
f(x) =
{
α(1− x2) if 0 ≤ x ≤ 1
0 otherwise.
a) Find the value of α.
b) Find E(X).
iv) A six-sided die, which is believed to be biased, is rolled 720 times and shows a ‘6’ 100
times.
a) Write down the formula for the tail probability of getting 100 or less 6’s in 720 rolls
of a fair die.
b) Using the normal approximation to the binomial distribution, calculate the proba-
bility in (i), giving your answer to 3 decimal places.
c) Giving reasons, do you think that this die is biased?
v) You are given that a 2× 2 matrix B satisfies,
B2 = 3B − 2I.
Giving reasons, find the eigenvalues of B.
17
3. i) Consider the surface S, given by z =
√
x2 + y2, at the point
x0 =

 86
10

.
a) Determine a vector n normal to the surface S at the point x0.
b) Determine the equation of the tangent plane to the surface S at the point x0.
c) Using your result from part (ii) find an approximation to the value of
√
8.022 + 5.972.
ii) Consider the function z = xf(xy), where f is a differentiable function of one variable.
Prove that
z = x
∂z
∂x
− y ∂z
∂y
.
iii) Evaluate each integral:
a) I1 =
∫
cos3 θ sin4 θ dθ,
b) I2 =
∫
x2 + x
(x− 2)(x2 + 2) dx.
iv) Solve the differential equation defined, for x > 0, by
dy
dx
− 3
x
y =
1
x4
, y(1) = 4.
v) Consider the differential equation 2xy + (x2 + y3)
dy
dx
= 0.
a) Show that the differential equation is exact.
b) Hence solve the differential equation.
18
4. i) A tank can hold 1000 litres. Initially it holds 250 litres of pure water. Brine, which
contains 5 grams of salt per litre, is run into the tank at the rate of 6 litres per minute.
The mixture, which is stirred continuously, is run off at a rate of 1 litre per minute. Let
M(t) denote the mass of salt (measured in grams) present in the tank after t minutes,
where 0 ≤ t ≤ 150.
Determine (with reasoning) a first order differential equation for M(t).
[ Note: You are NOT being asked to solve the differential equation.]
ii) Determine the general solution of y′′ − 4y′ + 4y = x.
iii) By using an appropriate test, determine whether each of the following series converges
or diverges.
a)
∞∑
n=1
(− 1)n cos
( π
n2
)
,
b)
∞∑
n=1
sin2 3n
n2
.
iv) Determine the open interval of convergence for
∞∑
n=2
(x− 1)n
lnn
.
v) Consider the Maclaurin series
ln(1 + x) = x− x
2
2
+
x3
3
− x
4
4
+ · · · =
∞∑
n=1
(−1)n+1x
n
n
, x ∈ (−1, 1].
a) By differentiating the above series for ln(1+x), find the Maclaurin series for
1
1 + x
.
b) Hence write down the Taylor Polynomial P6(x) about 0 for
1
1 + x2
.
c) Using your result from the previous part, find a rational approximation to
1∫
0
1
1 + x2
dx.
(Your answer should be written as a fraction.)
vi) The area A of the surface of revolution about the y-axis of the curve C, described
parametrically by C = {(x(t), y(t)) ∈ R2 : a ≤ t ≤ b}, is given by
A =
b∫
a
2πx(t)
√
[x′(t)]2 + [y′(t)]2 dt .
(You are NOT required to show this.)
a) The curve C is described using polar coordinates by
r = r(θ), θ0 ≤ θ ≤ θ1 .
Use the change of variables x(θ) = r(θ) cos θ and y(θ) = r(θ) sin θ to show that the
area A of the surface of revolution of C about the y-axis is given by
A =
θ1∫
θ0
2πr cos θ
√
r2 +
(
dr
dθ
)2
dθ .
19
b) Determine the area of the surface formed when the polar curve
r = 1 + sin θ, where −π
2
≤ θ ≤ π
2
, is rotated about the y-axis.
20
MATH1231 NOVEMBER 2012
1. i) Suppose S = {v1,v2,v3,v4}, where
v1 =

 12
3

 , v2 =

 34
6

 ,v3 =

 0−2
3

 , v4 =

 32
1

 .
Find a basis for span(S).
ii) Consider the matrix
A =
(
3 1 1 0
6 2 2 1
)
.
a) Find a basis for the kernel of A.
b) Hence state the nullity(A).
iii) The vector space P2 consists of all polynomials, with real coefficients, of degree at most
2. Suppose W is the subset of P2 defined by
W = {p ∈ P2 : p′(0)p′′(0) = 0},
where ′ denotes differentiation.
a) Show that W is closed under scalar multiplication.
b) Determine, with reasons, whether or not W is a subspace of P2.
iv) a) Find constants b and c such that
3x+ 2
(x2 + 2x+ 2)(x − 1) =
1
x− 1 +
bx+ c
x2 + 2x+ 2
.
b) Hence find
J =
∫
3x+ 2
(x2 + 2x+ 2)(x − 1) dx.
v) Solve the initial value problem
dy
dx
− 1
(x− 1)y = 2, y(2) = 3, x > 1.
vi) The following MAPLE session may assist you with this question.
> x := t -> t-sin(t):
> y := t -> 1-cos(t):
> xp := diff(x(t),t):
> yp := diff(y(t),t):
> A := simplify(sqrt(xp^2 + yp^2));√
2− 2 cos(t)
> int(A*(1-cos(t)),t=0..2Pi);
32
3
21
a) Write down the integral representing the area S of the surface of revolution formed
when the curve with parametric equation
x = x(t), y = y(t), between t = t0 and t = t1, is rotated about the x-axis.
b) Use the MAPLE output above to find the area of the surface of revolution when the
curve given by
x(t) = t− sin t, y(t) = 1− cos t, 0 ≤ t ≤ 2π,
is rotated about the x-axis.
2. i) For a gas confined in a container, the ideal gas law states that the pressure P is related
to the volume V and the temperature T by
P = k
T
V
,
where k is a positive constant.
a) Find
∂P
∂V
and
∂P
∂T
.
b) The volume V is increased by 4% and the temperature T is decreased by 3%.
Use the total differential to estimate the percentage increase or decrease in the
pressure P .
ii) Determine, with reasons, whether or not the following sequences converge and find the
limit if it exists:
a) {an} where an = n√
4n2 − 1 ,
b) {bn} where bn = n!
nn
.
iii) Consider the ordinary differential equation
(2x+ 3y) dx+ (3x+ 2y) dy = 0.
a) Show that the equation is exact.
b) Find the general solution of the given equation.
iv) Consider the map T : R4 → R4 defined by
T (x) =


x2
2x3 − x1
x1 + x2 + 5x4
6x2 − 3x4

 .
You may assume (without proof) that T is linear.
Determine a matrix A such that T (x) = Ax.
v) Let A be the matrix A =
(
6 −1
2 3
)
.
a) Find the eigenvalues and eigenvectors for A.
b) Hence write down matrices P and D, where D is diagonal and P−1AP = D.
22
vi) Suppose b is a non-zero vector in R3 and consider the projection map T : R3 → R3,
defined by T (x) =
x · b
|b|2 b
a) Show that T is linear.
b) Describe geometrically the kernel of T .
3. i) Let T : R2 → R2 be the linear map which reflects a vector x =
(
x
y
)
about the line
y = x as shown in the diagram.
x
T (x)
b
b
y = x
Explain why
(
1
1
)
and
(
1
−1
)
are eigenvectors of T and give their corresponding
eigenvalues.
ii) In Sydney, 85% of workers use public transport (bus, train, or light-rail) to get to work.
On a particular day, 50% of Sydney workers use at least the bus in their travels, 40%
use at least the train, 30% use at least light-rail, 20% use at least bus and train, 10%
use at least bus and light-rail, 8% use at least train and light-rail. What percentage of
Sydney workers used all three modes of public transport to get to work on that day?
iii) Suppose X has probability density function given by
f(x) =
{
1 + 4x, x ∈ [0, a]
0, x 6∈ [0, a] for some a > 0.
a) Show that a = 12 .
b) Find, as a fraction, the value of E(X), the expected value of X.
c) Find, as a fraction, the value of Var(X), the variance of X.
iv) A six-sided die, with faces numbered 1 to 6, is suspected of being unfair so that the
number 6 will occur more frequently than should happen by chance. During 300 test
rolls of the die, the number 6 occurred 68 times.
a) Write down an expression for a tail probability that measures the chance of rolling
a 6 at least 68 times.
b) Use the normal approximation to the binomial to estimate this probability.
c) Is this evidence that the die is unfair?
v) Suppose A is a 3× 2 matrix and let AT denote its transpose.
a) Prove that the column space of the matrix AAT is a subset of the column space of
A and deduce that rank(AAT ) ≤ rank(A).
23
b) Deduce that nullity(AAT ) ≥ 1, and explain why the matrix AAT can never be the
identity.
4. i) Find the open interval of convergence I = (a, b) for the power series
∞∑
k=2
(x− 3)k
2k log k
.
ii) Determine whether each of the following series converges or diverges, stating any tests
you use.
a)
∞∑
n=1
cos(2n)
(n3 + 1)
1
2
.
b)
∞∑
n=1
n4
n!
iii) You are given that
1− x+ x2 − x3 + · · · = 1
1 + x
for − 1 < x < 1.
a) Find the Maclaurin series for log(1 + x) that is valid for −1 < x < 1.
b) Given that
∫
log x dx = x log x− x+ C, simplify the infinite series:
∞∑
n=2
(−1)nxn
(n− 1)n =
x2
1× 2 −
x3
2× 3 +
x4
3× 4 − · · ·
where −1 < x < 1.
iv) A fixed proportion k1 of a certain isotope, A, decays each hour into isotope B. Also,
a fixed proportion k2 of the isotope B decays each hour into isotope C. Let a(t) and
b(t) be the amounts of isotopes A and B respectively at time t, in hours. The rates of
change of these quantities, can be modelled by the differential equations
da
dt
= −k1a (1)
db
dt
= k1a− k2b. (2)
a) By differentiating both sides of equation (2), show that b satisfies the differential
equation
d2b
dt2
+ (k1 + k2)
db
dt
+ k1k2b = 0. (3)
b) Assuming that k1 6= k2, find the general solution of the differential equation (3).
c) Suppose now that k1 = k2. Write down the general solution of the differential
equation (3), given then b(0) = 0.
d) Under the conditions in part (c), find an expression for the time it takes for b(t) to
reach its maximum value.
24
MATH1231 NOVEMBER 2013
1. i) Let
S =



 x1x2
x3

 ∈ R3 : x31 + x32 + x33 = 0

 .
a) Prove that S is closed under scalar multiplication.
b) Show that S is not a subspace of R3.
ii) Suppose that
A =


−23 3 −5 37
−49 8 −10 73
−87 9 −15 129
−26 3 −5 40

 .
Using the following Maple output, answer the questions below.
> with(LinearAlgebra):
> A := <<-23,-49,-87,-26>|<3,8,9,3>|<-5,-10,-15,-5>
|<37,73,129,40>>;
A :=


−23 3 −5 37
−49 8 −10 73
−87 9 −15 129
−26 3 −5 40


> Eigenvectors(A); 

2
3
5
0

 ,


1 2 1 1
1 5 2 2
3 0 3 4
1 1 1 1


a) Without performing any further computations, explain why the matrix A has four
linearly independent eigenvectors.
b) Write down the general solution of
dy
dt
= Ay.
iii) A collection of discs consists of DVDs and Blu-ray discs. In the collection, 75% are
DVDs and 25% are Blu-ray discs. Among the DVDs, 60% are movies. Among the
Blu-ray discs, 90% are movies.
a) Find the probability that a disc chosen randomly from the collection is a movie.
b) Find the probability that a randomly chosen disc from the collection is a Blu-ray
disc given that it is a movie.
25
iv) Evaluate each of the following integrals:
a) I1 =
∫
5x2 − 6x+ 4
x(x− 1)(x− 2) dx.
b) I2 =
∫ pi/2
0
sin3 θ cos2 θ dθ.
v) Solve the initial value problem
dy
dx
=
sinx
y2
, y(0) = 1.
vi) Let f(x) = ln(1+4x2). The following MAPLE session may assist you with this question.
> f := x -> log(4*x^2+1):
> taylor(f(x), x = 0, 6);
4x2 − 8x4 +O(x6)
Using the MAPLE output above or otherwise, write down the values of f ′′(0) and f ′′′(0).
2. i) Suppose that f is a differentiable function of one variable, and F (x, y) is defined by
F (x, y) = f(2x− 3y2).
a) Show that F satisfies the partial differential equation
3y
∂F
∂x
+
∂F
∂y
= 0.
b) Given that F (x, 0) = cos(x) for all x, write down a formula for F (x, y).
ii) The curve C is given parametrically by x = t3, y = 2t2. Find the arc length of the curve
C between t = 0 and t = 1.
iii) a) Find the general solution y(x) for the following ordinary differential equation:
y′′ − 5y′ + 6y = 0.
b) Find the general solution of the following equation:
y′′ − 5y′ + 6y = e2x.
iv) Let T : R2 → R2 be the linear map which rotates a vector x about the origin through
π
6
anticlockwise and doubles its length, as shown in the diagram.
x
T (x)
0
pi
6
26
a) Show that T (e1) =
( √
3
1
)
, where e1 =
(
1
0
)
.
b) Find the matrix A such that T (x) = Ax for all x ∈ R2.
v) The probability density function f of a continuous random variable X is given by
f(x) =
{
1
2(2− x) for 0 ≤ x ≤ 2,
0 otherwise.
a) Find E(X) and Var(X).
b) The median of a distribution is defined to be the real number m such that P (X ≤
m) = 12 . Find the median of the above distribution.
vi) Let V be a vector space.
a) Let S be the set {v1,v2, · · · ,vn} of vectors in V .
State the condition for S to be a linearly dependent set.
b) Suppose that v1, v2 ∈ V and 0 is the zero vector of V . Prove that the set
{v1, v2, 0} is linearly dependent.
3. i) The polynomials p1, p2, p3, p4 and p5 are defined, for all x ∈ R, by
p1(x) = x
3 + 3x2 + 2x− 1
p2(x) = 2x
3 − 3x2 + 4x+ 2
p3(x) = 11x
3 − 3x2 + 22x+ 5
p4(x) = 2x
3 − 21x2 + 4x+ 3
p5(x) = 9x
3 + 18x2 + 18x+ 2
Let S = {p1, p2, p3, p4, p5}.
a) Explain why S cannot be a basis for P3(R). Using the following Maple output,
answer the questions below.
> with(LinearAlgebra):
> A := <<-1,2,3,1>|<2,4,-3,2>|<5,22,-3,11>|<3,4,-21,2>
|<2,18,18,9>|<1,0,0,0>|<0,1,0,0>|<0,0,1,0>
|<0,0,0,1>>;
A :=


−1 2 5 3 2 1 0 0 0
2 4 22 4 18 0 1 0 0
3 −3 −3 −21 18 0 0 1 0
1 2 11 2 9 0 0 0 1


> GaussianElimination(A);

−1 2 5 3 2 1 0 0 0
0 8 32 10 22 2 1 0 0
0 0 0 −634 634 94 −38 1 0
0 0 0 0 0 0 −12 0 1


27
b) Find a basis for span(S).
c) Find a basis for P3(R) which contains as many of the polynomials of S as possible.
ii) Let
A =


1 2 3 4
2 3 5 3
1 3 4 9

 .
a) Find a basis for ker(A).
b) Hence state the value of nullity(A). Give a reason.
iii) Let
C =
(
1 2
−2 5
)
.
a) Find the eigenvalue(s) of C and for each eigenvalue find the corresponding eigen-
vectors.
b) Is C diagonalisable? Give a reason for your answer.
iv) A manufacturer claims that only 20% of the one-litre size soft drinks produced by his
factory contain less than 1.05 litres. Consumer Watchdog examined a random sample of
100 bottles of one-litre soft drinks produced by this manufacturer and found that 30%
of the sample contain less than 1.05 litres. To test whether the manufacturer’s claim is
true, we shall use a binomial distribution B(n, p) to model the number, X, of bottles
containing less than 1.05 litres in a random sample of 100.
a) State the values of n and p.
b) Find an expression for P (X ≥ 30) in terms of binomial coefficients.
c) Use the normal approximation to the binomial distribution to estimate the proba-
bility in the previous part.
d) Based on this probability, is there evidence that the manufacturer has made a false
claim? (Give a reason.)
4. i) Use appropriate tests to determine whether each of the following series converges or
diverges:
a)
∞∑
k=0
2k + 1
3k
,
b)
∞∑
k=1
(−1)k√
k(k + 1)
.
ii) The Maclaurin series for sinx is
sinx =
∞∑
k=0
(−1)k x
2k+1
(2k + 1)!
, x ∈ R.
Do not prove this result.
a) Write down the Maclaurin series for sin(x2).
b) Find the Maclaurin series for cos(x).
28
c) Hence derive the identity
1√
2
=
∞∑
k=0
(−1)k π
2k
24k(2k)!
.
iii) Determine, with reasons, whether or not each of the following sequences converges and
find each limit if it exists:
a) {an} where an = lnn
n
,
b) {bn} where bn = n
√
n.
iv) A rumour is spreading through an isolated population. Modelling suggests that r(t), the
proportion of the population who have heard the rumour after t days, is governed by
a differential equation of the form
dr
dt
= kr(1− r), (4)
where k is a constant.
a) Show that the general solution to (4) is given implicitly by
r
1− r = β e
kt,
with β > 0. (You may assume that 0 < r < 1.)
b) At time t = 0, 1% of the population have heard the rumour. After 5 days, 20% of
the population have heard the rumour.
A) Find β and k. (Give k correct to 4 decimal places.)
B) Hence, estimate how long it will take until 80% of the population have heard
the rumour. (Give your answer correct to 2 decimal places.)
c) How does r(t) behave as t→∞?
29
MATH1231 NOVEMBER 2014
1. i) By expanding sin(A+B) + sin(A−B), or otherwise, find I1 =
∫
sin(5x) cos(x) dx.
ii) Evaluate the integral I2 =
∫ 1
0
x2√
4− x2 dx.
iii) Use appropriate tests to determine whether each of the following series converges or
diverges
a)
∞∑
n=3
1
n(ln n)3
,
b)
∞∑
n=1
2
2n + 3n
.
iv) Let S =



xy
z

 ∈ R3 : z2 = x2 + y2

.
a) Prove that S is closed under scalar multiplication.
b) Prove that S is not a subspace of R3.
v) Let A =
(
0 −1
2 3
)
.
a) Find the eigenvalues and eigenvectors for the matrix A.
b) Write down an invertible matrix M and a diagonal matrix D such that
D =M−1AM.
vi) Let
A =


1 2 2 −1
2 4 10 −44
3 −3 −3 24
1 2 1 6


Using the MAPLE output below, find a basis for ker(A).
> with(LinearAlgebra):
> A := <<1,2,3,1>|<2,4,-3,2>|<2,10,-3,1>|<-1,-44,24,6>>;

1 2 2 −1
2 4 10 −44
3 −3 −3 24
1 2 1 6


> ReducedRowEchelonForm(A); 

1 0 0 5
0 1 0 4
0 0 1 −7
0 0 0 0


30
2. i) Consider the initial value problem
dy
dx
+(2+
1
x
)y =
2
x
, with y(1) = 0, defined for x > 0.
a) Show that an integrating factor for this equation is xe2x.
b) Hence solve the initial value problem.
ii) Find the general solution to
d2y
dx2
+ 2
dy
dx
+ 2y = 20e2x.
iii) Consider the MAPLE session:
> a:=n->n^n/n!*(x-1)^n;
a := n→ n
n(x− 1)n
n!
> a(n+1);
(n+ 1)(n+1)(x− 1)(n+1)
(n+ 1)!
> limit(a(n+1)/a(n),n=infinity);
ex− e
Using MAPLE session above, or otherwise, find the open interval of convergence I =
(a, b) for the power series
∞∑
n=1
nn(x− 1)n
n!
.
iv) Consider the set S consisting of the vectors v1 =

 11
2

 ,v2 =

 23
−1

 from R3 and
let u =

 1−1
12

 .
a) Find scalars λ and µ such u = λv1 + µv2.
b) A linear transformation T : R3 → R3 has v1,v2 as eigenvectors with eigenvalues 2
and −1, respectively.
α) Find T (u) as a linear combination of v1,v2.
β) Denote T (T (u)) by T 2(u), T (T (T (u))) by T 3(u), and so on. Express T n(u) as
a linear combination of v1,v2, where n is a positive integer.
31
v) The two most popular soft drinks in Old South Wales are AppleAde and BananAde.
Assume that no-one in Old South Wales likes both of these drinks equally (that is,
everyone has a preference for one or the other). Past statistics show that 50 % of the
population prefer AppleAde.
Last month the manufacturer advertised AppleAde on television for a week. After that,
a survey was conducted by taking a random sample of 100 people. Of the 100 people
sampled, 60 preferred AppleAde and 40 preferred BananAde.
a) Assuming that the advertising had no effect on people’s preferences, write down
an expression for the tail probability that 60 or more people preferred AppleAde in
a sample of 100.
b) Use the normal approximation to the binomial to calculate the tail probability in
(a), giving your answer to 3 decimal places.
c) Giving reasons, is there evidence that the advertising campaign increased the per-
centage of the population that prefer AppleAde?
3. i) Let P1 denote the space of all polynomials of degree less or equal to 1.
Prove that function T : R3 → P1 defined by
T (v) = 2a+ (b− c)x, for all v =

ab
c

 ,
is a linear transformation.
ii) Let S be the set of vectors {v1,v2,v3,v4,v5} given by
S =




21
12
−6
−31
−12

 ,


11
20
−14
−29
−21

 ,


18
−72
60
50
78

 ,


29
−52
46
21
57

 ,


16
−17
−31
28
−21




.
Using the following MAPLE output, answer the questions below. Give reasons.
> with(LinearAlgebra):
> M := <<21, 12, -6, -31, -12>|<11, 20, -14, -29, -21>
|<18, -72, 60, 50, 78>|<29, -52, 46, 21, 57>
|<16, -17, -31, 28, -21>>;


21 11 18 29 16
12 20 −72 −52 −17
−6 −14 60 46 −31
−31 −29 50 21 28
−12 −21 78 57 −21


32
> GaussianElimination(M);


21 11 18 29 16
0 967 −5767 −4807 −1837
0 0 0 0 −3778
0 0 0 0 0
0 0 0 0 0


a) Find a basis for span(S).
b) Write down the dimension of span(S).
iii) Employment data at a large company reveal that 40 % of the employees are thirty years
of age or younger; 60 % are older than thirty. All employees belong to exactly one of
the types: full-time, part-time or casual. Among those who are thirty years of age or
younger, 25% are full-time; 15% are part-time; the others are casual. Among those who
are older than thirty, 45% are full-time; 20% are part-time; the others are casual.
a) What proportion of the employees are full-time?
b) What is the probability that a randomly chosen employee is older than thirty given
that the employment type of the employee is full-time?
iv) The discrete random variable X can only take the values −2,−1, 1, 2, 5.
The probability distribution for X is given in the following table:
x −2 −1 1 2 5
P (X = x) 0.3 4c 0.35 6c 0.05
Another random variable Y is defined by Y = X2.
a) Find the value of c.
b) Calculate P (Y = 4).
v) The density function f for a random variable X is defined by
f(x) =
{
3x2 for 0 < x ≤ 1
0 otherwise.
a) Find the cumulative probability density function F corresponding to f .
b) Calculate E(X) for the probability density function f .
c) Calculate Var(X).
vi) Let B = {v1,v2,v3} be a set of three non-zero vectors in R3.
a) State the definition for the set B to be a linearly independent set.
33
b) Prove that if B is an orthogonal set then B is linearly independent.
c) Hence explain why any orthogonal set of 3 non-zero vectors in R3 forms a basis for
R
3.
4. i) A right circular cone with base radius r cm and a perpendicular height h cm is formed
by rotating the line segment given by
y =
r
h
x, 0 ≤ x ≤ h,
about the x-axis.
a) Use calculus to show that the surface area S of the curved surface of the cone is
given by
S = πr
√
r2 + h2.
b) Find the partial derivatives
∂S
∂r
and
∂S
∂h
.
c) If the values of r and h are measured to be 3.0 and 4.0 cms respectively and each
of these measurements is made with an error whose absolute value is at most 0.05
cm, then use the total differential approximation of S to estimate the maximum
absolute error in the measured value of the total surface area S.
ii) During the winter the daytime temperature in the Physics Theatre is maintained at
20◦C. The heating is turned off at 10pm and turned on again at 6am. On a certain
day, the temperature inside the Theatre at 11pm was found to be 18◦C. The outside
temperature was found to be constant throughout the night at 10◦C. Let P (t) be the
temperature (in ◦C) in the Physics Theatre at time t (in hours), from 10pm. The rate
of cooling of the air inside the Theatre can be modelled by the differential equation
dP
dt
= −k(P − 10),
where k is a positive constant. (Do not prove this.)
a) By first solving the differential equation, show that k = ln(5/4) .
b) Find, to two decimal places, the temperature inside the Physics Theatre when the
heating was turned on at 6am.
34
iii) The nth term of the sequence {an} is given by
an :=


n1/n/2, for n odd,
1/2, for n even.
a) Explain briefly why
1
2
≤ an ≤ n
1/n
2
, for n ≥ 1.
b) Prove that
lim
n→∞n
1/n = 1.
c) Hence find
lim
n→∞ an.
Give reasons for your answer.
d) Does the series
∞∑
n=2
(−1)nan
converge or diverge? Name any test that you use.
iv) Let f(x) be a function satisfying
f(0) = 0, f ′(0) = 1, f ′′(0) = −2 and |f ′′′(x)| ≤ 6, for 0 ≤ x ≤ 1.
a) Write down the second Taylor polynomial p2(x) of f(x) about x = 0.
b) The function f(x) is approximated by p2(x), for x in the interval
0 ≤ x ≤ 1.
Using Taylor’s Theorem with Lagrange remainder formula, show that
1
8
≤ f(1/2) ≤ 3
8
.
c) Using (a), or otherwise, find
lim
x→0+
x− f(x)
x2
.
35
PAST HIGHER EXAM PAPERS
36
MATH1241 NOVEMBER 2010
2. i) A surface with height z = F (x, y) is defined by
z = 1 +
25
x2 + y2
for 1 ≤ x2 + y2 ≤ 25.
Find the equation of the plane tangent to the surface at the point (1, 2, 6) on the surface.
ii) A function u(x, t) is defined implicitly by the equation
u(x, t) = f(x+ tu(x, t)),
where f is a differentiable function of one variable. Show that u(x, t) is a solution of the
partial differential equation
∂u
∂t
= u
∂u
∂x
.
iii) Find the integral
I =
∫
6x2 − 11x+ 9
(x− 1)(x2 − 2x+ 2) dx.
iv) Let P3(R) denote the vector space of polynomials of degree 3 or less with real coefficients
and let S be given by
S = {p ∈ P3(R) : (x− 3)p′(x)− 2p(x) = 0 for all x ∈ R} .
Show that S is a subspace of P3(R).
v) On the basis of the health records of a particular group of people, an insurance company
accepted 60% of the group for a 10 year life insurance policy and rejected the others.
Ten years later the company examined the survival rates for the whole group and found
that 80% of those accepted for the policy had survived the 10 years, while 40% of those
rejected had survived the 10 years. If a person from the original group did survive 10
years, what is the probability that they had been refused a life insurance policy?
vi) Suppose that the vectors vj ∈ Rn, for j = 1, 2, . . . , n, are mutually orthogonal, i.e.
vj · vk = 0 for j 6= k and j, k = 1, 2, . . . , n.
Let x be an arbitrary vector in Rn.
a) Given that the set G = {v1,v2,v3, . . . ,vn} is a basis for Rn, find an expression for
x in terms of the vectors in the basis G.
b) Suppose that the linear map T : Rn → R has the property that
T (vj) = γj(vj · vj) for j = 1, 2, . . . , n,
where the γj ∈ R are given constants.
Find an expression for T (x) in terms of x, the constants γj and the vectors vj .
3. i) Let A =
(
5 −8
1 −1
)
.
a) Determine the eigenvalues and corresponding eigenvectors for the matrix A.
37
b) Write down matrices P and D such that A = PDP−1.
c) Hence evaluate A8P .
ii) According to the Bureau of Meteorology, Sydney experienced rain on 103 out of the 365
days in 2009. Let us make the simple assumptions that
• the probability p of rain on any given day in Sydney is constant; (5)
• the weather on any given day is independent of the weather on
other days. (6)
a) Under assumptions (1) and (2), use the stated data to estimate the probability p
that a given day in Sydney is rainy.
b) Under assumptions (1) and (2), estimate the probability that precisely 5 out of 10
given days are rainy in Sydney.
c) Suppose that Sydney will experience rain on 1 January, 2011. Using the assumptions
(1) and (2), estimate the probability that the 6th of January 2011 is the first dry
day of the new year.
iii) Let R[R] denote the vector space of real-valued functions defined on R. Let S be the
subspace of R[R] that is spanned by the ordered basis B = (e−x, sin(x), cos(x)). Define
the map T : S → S by
T (f) = f ′ − 2f where f ′(x) = df
dx
.
a) Show that the map T is linear.
b) Calculate the matrix C which represents T with respect to the basis B.
c) What is the rank of the matrix C found in part (b)?
d) From part (c), what can be deduced about the solutions y ∈ S of
y′ − 2y = g
where g is a given function in S?
e) Use your results from parts (b) and (d) to find all non-zero solutions y ∈ S satisfying
y′ − 2y = e−x .
4. i) Find the general solution of the ordinary differential equation
(x2 − xy)dy
dx
− 2y2 + 5xy = 3x2.
ii) Suppose that ak =
k
ln(k!)
for k = 2, 3, 4, . . .. In the questions below you
may use the inequality
ln(k!) − ln(k) + k − 1 < k ln(k) < ln(k!) + k − 1 for k = 2, 3, 4, . . ..
a) Find lim
k→∞
ak, clearly stating your reasons.
38
b) Does the series
∞∑
k=2
ak converge? Give reasons for your answer.
iii) Determine the radius of convergence of the power series
∞∑
n=1
1
2n
√
n
(x− 1)n.
iv) a) Write down the first three non-zero terms in the Taylor series of sinhx about x = 0,
including an expression for the Lagrange form of the remainder.
b) Hence, or otherwise, write down the first three non-zero terms in the Taylor series,
about x = 0, of the function
Shi(x) =
∫ x
0
sinh t
t
dt.
v) A curve C in the xy-plane is defined parametrically by
x(t) = a cos t, y(t) = b sin t, for 0 ≤ t ≤ π, where 0 < a < b.
Find the surface area of the ellipsoid defined by rotating the curve C about the x-axis.
vi) A tank has a total volume of 200 litres. Initially it holds 40 litres of pure water. Brine
containing 2 grams of salt per litre is run into the tank at the rate of 3 litres per minute
and the mixture is stirred continuously so that the concentration of the dissolved salt is
uniform throughout the tank. At the same time as the brine starts to flow into the tank
the mixture is removed from the tank at the rate of 1 litre per minute.
Let t denote time, measured in minutes, from when the brine started to enter the tank
and let x(t) denote the mass of salt, in grams, present in the tank after t minutes.
Set up a first order differential equation in x and t which models this system up until
the time the tank is full.
39
MATH1241 NOVEMBER 2011
2. i) Let f : R2 → R and let (x0, y0) be a point in R2.
a) Carefully write down the definition of the derivative fy(x0, y0).
b) If f(x, y) := x1/3y2/3 find the value fy(0, 0).
ii) In a direct-current circuit, the total resistance z, (measured in ohms), produced by two
parallel resistors with resistances x and y ohms, is given by
1
z
=
1
x
+
1
y
.
The values of x and y are measured to be 6 ohms and 12 ohms respectively and each of
the measurements is made with an error whose absolute value is at most 0.1 ohms.
a) Use the given measurements (ignoring the measurement error) to calculate the total
resistance z.
b) Show that
∂z
∂x
=
z2
x2
and
∂z
∂y
=
z2
y2
.
c) Use the total differential approximation to estimate the maximum absolute error in
the calculated value of the total resistance z, correct to three significant figures.
iii) Consider the vectors
v1 =

 1−1
−3

 , v2 =

21
2

 , v3 =

72
3

 , w =

ab
c

 and u =

56
0

 .
a) Does {v1,v2,v3} span R3? If so, give reasons; if not, then find condition(s) on a, b, c
such that w belongs to span(v1,v2,v3).
b) Determine whether or not u is in span(v1,v2,v3).
iv) One semester, 65% of the students studying MATH1231/41 attended all their tutorials,
and of these students, 45% were awarded credit (CR) or higher in the final exam. In
contrast, only 30% of the students who did not attend all their tutorials were awarded
credit (CR) or higher in the final exam.
a) What percentage of the students studying MATH1231/41 were not awarded credit
(CR) or higher in the final exam?
b) Of the students who were awarded credit (CR) or higher in the final exam, what
percentage, correct to 1 decimal place, attended all their tutorials ?
3. i) Consider the ordinary differential equation (ODE)
dy
dx
=
xy − y2
x2
.
a) Determine the general solution y = y(x) to the above ODE.
b) Find the particular solution that passes through the point (1, 1).
ii) a) Determine whether the following series converges or diverges.
∞∑
k=2
ln(k!)
k3
.
40
b) Does the following series converge absolutely or conditionally? Give reasons for your
answers. ∞∑
k=2
(−1)k log k√
k2 + 1
;
iii) Consider the ODE
dy
dx
= ex
2
y, y(0) = 1.
a) Write down the Maclaurin series for ex
2
.
b) Hence, or otherwise, find a solution (involving a series) to the above initial value
problem.
c) Carefully determine the values of x for which the solution is defined.
iv) A sequence of positive numbers {ak}∞k=1 is defined recursively by
ak+1 =
√
1 + ak, a1 = 1.
a) Prove that ak is strictly increasing.
b) Prove that ak is bounded above by 3.
c) Hence, explain why there is a number L such that
lim
k→∞
ak = L.
d) Determine the value of L.
4. i) Let A =
(
22 −100
5 −23
)
.
a) Determine the eigenvalues and corresponding eigenvectors for the matrix A.
b) Write down matrices P and D such that A = PDP−1.
c) Hence evaluate AnP , for any positive integer n.
ii) A discrete random variable X has the probability distribution given by pk =
c
2k
for
k = 0, 1, 2, 3, 4, where c is a constant.
a) Find the value of the constant c.
b) Calculate P (X = 2).
c) Calculate P ((X − 2)2 < 4).
iii) Let V be the vector space M22(R) of all 2× 2 real matrices.
Define the linear map T : M22 → M22 by T (A) = AT , where AT denotes the transpose
of the matrix A.
a) Show that for any non-zero n× n matrix C, if CT = λC then λ = ±1.
b) Find the matrix representation B for T with respect to the ordered basis {v1,v2,v3,v4}
given by {(
1 0
0 0
)
,
(
0 1
0 0
)
,
(
0 0
1 0
)
,
(
0 0
0 1
)}
.
c) State the rank of T .
d) Explain why the eigenvalues of T are ±1.
e) Find bases for each of the eigenspaces of T .
41
MATH1241 NOVEMBER 2012
2. i) For any given differentiable function f : R→ R, let u be defined by
u(x, y) = f(s), where s =
x
y
.
Determine the partial derivatives
∂u
∂x
and
∂u
∂y
.
ii) Consider the function u(x, y) = ln(x2 + y2). Show that u satisfies Laplace’s equation
∂2u
∂x2
+
∂2u
∂y2
= 0.
iii) a) Show that the area A of the planar region between the lines x = −a
2
and x =
a
2
bounded by the ellipse
x2
a2
+
y2
b2
= 1,
where a, b > 0, has the form Kab, and find the value of the constant K.
b) If the ellipse is deformed in such a way that a is increased by 4% and b is decreased by
1%, use the total differential approximation to show that the approximate relative
change of area
∆A
A
is independent of a and b and calculate its value.
iv) Consider the following vectors in R4:
v1 =


1
2
1
3

 , v2 =


2
3
4
1

 , v3 =


0
1
−2
5

 , v4 =


5
7
11
0

 , a =


a
b
c
d

 .
Let S be the set of vectors {v1,v2,v3,v4}.
a) Find a subset of S that is linearly independent, containing as many vectors from S
as possible.
b) Give conditions on a such that a ∈ span(S).
c) Find all ways of representing a as a linear combination of v1, v2, and v4.
d) Describe span(S) geometrically.
v) Suppose that a continuous random variable T has an exponential distribution with
parameter λ.
a) Find P (4− 2T ≥ 0).
b) Find E(T 2).
3. i) A six-sided die, with faces numbered 1 to 6, is suspected of being unfair so that the
number 6 will occur more frequently than should happen by chance. During 300 test
rolls of the die, the number 6 occurred 68 times.
a) Write down an expression for a tail probability that measures the chance of getting
a 6 at least 68 times.
42
b) Use the normal approximation to the binomial to estimate this probability.
c) Is this evidence that the die is unfair?
ii) We roll a die successively, until we roll six consecutive sixes, in which case we stop rolling
the die.
a) What is the probability that we stop after the thirteenth throw?
b) What is the probability that we roll the die at least ten times?
iii) Let A be a 3× 2 matrix and let AT denote its transpose.
a) Prove that the column space of the matrix AAT is a subset of the column space of
A and deduce that rank(AAT ) ≤ rank(A).
b) Is it possible for AAT to be the identity matrix? Give examples or reasons for your
answer.
iv) Let M be a 3× 3 matrix and MT be its transpose. Assume that M is orthogonal, that
is, MTM = I. Let m1, m2 and m3 be the three columns of M .
a) Explain why detM = ±1.
b) Explain why m1, m2 and m3 are mutually perpendicular vectors of length 1.
c) Show that for any square matrix A,
(Av) ·w = v · (ATw)
for all v,w ∈ R3, where · denotes the dot product.
d) Deduce that |Mv| = |v| for all v ∈ R3.
e) Explain why M must have at least one real eigenvalue and at least one real eigen-
vector v ∈ R3.
f) Show that if u ∈ R3 is an eigenvector of M , then the corresponding eigenvalue is 1
or −1.
4. i) a) Write down the general solution to the second order differential equation
d2y
dx2
+
6
dy
dx
+ 9y = 0, y(0) = 0.
b) For what value of x does the solution have a maximum?
ii) a) Consider the solution of the differential equation
dy
dx
=
xy
x2 − a2 , a > 0 (7)
which passes through the point (x0, y0) = (2a, a).
Find the equation of the normal to this solution at (x0, y0).
b) Obtain the general solution of the differential equation (7) and interpret it geomet-
rically for |x| < a and |x| > a.
iii) a) Sketch the graph of the function
f(x) =
x
x2 − 99
for x ≥ 0.
43
b) Carefully investigate the (conditional) convergence of the series
∞∑
n=0
(−1)nn
n2 − 99 ,
justifying the steps in your proof.
iv) a) State the Maclaurin series of f(t) = 1− cos t.
b) Let g(t) =
{
1−cos t
t t 6= 0
0 t = 0.
Find the Maclaurin series of the function
Cin(x) =
∫ x
0
g(t)dt,
and state where the series converges.
v) Consider the sequence defined by a1 =
1
4 , and, for n ≥ 1,
an+1 = 2an(1− an).
a) Show that the sequence is bounded above by 12 .
b) Show that the sequence is strictly increasing.
c) Explain why the sequence is convergent.
d) Find lim
n→∞ an.
44
MATH1241 NOVEMBER 2013
3. i) Suppose that a continuous random variable T has an exponential distribution with
parameter λ, λ > 0.
a) Find Var(8− 2T ).
b) Find P (λT ≥ 1).
ii) An industrial machine produces large numbers of light bulbs of which 2% are defective.
The defects occur randomly during production. The bulbs are packaged into boxes of
100.
By approximating the binomial distribution with the normal distribution, estimate the
probability that a box contains more than 3 defective bulbs.
iii) An urn contains a red balls and b black balls, where a, b ≥ 1. Draw a ball randomly from
the urn, replace it with c, c ≥ 1, balls of the same colour, and repeat this procedure n,
n ≥ 1, times. Let Ri denote the event that a red ball is drawn on the ith draw.
a) Briefly explain why P (R1) =
a
a+ b
.
b) Show that P (R2) = P (R1).
iv) A real n × n matrix A has n distinct eigenvalues λ1, . . . , λn and can be expressed as
M−1DM, whereM is an invertible matrix and D is the diagonal matrix with λ1, . . . , λn
as its diagonal entries.
a) Show that Ak =M−1DkM .
b) Explain why Dk is a diagonal matrix with diagonal entries λk1 , . . . , λ
k
n.
c) Write the characteristic polynomial p(x) of A as
p(x) = xn + an−1xn−1 + · · ·+ a1x+ a0 .
By replacing the variable x by the matrix D, and a0 with a0I, where I is the n× n
identity matrix, prove that p(D) = 0 where 0 is the n× n zero matrix.
d) Hence or otherwise, prove that p(A) = 0.
e) Use the result in (d) to show that the matrix A =
(
2 1
4 −1
)
satisfies A2 = A+6I.
4. i) Using the substitition y(x) = xv(x), solve the differential equation
x2
dy
dx
= 4x2 + xy + y2.
ii) Find the interval of convergence of the power series, (making sure to check the end
points),
∞∑
n=1
2nn2
n3 + 1
(x− 1)n.
iii) A certain chemical reaction creates a substance S from two other chemicals A and B.
Let X be the amount in grams of the substance S at a given time t. It can be shown
that
dX
dt
= k(k1 −X)(k2 −X),
45
where the constant k depends only on the reaction conditions, while the constants k1
and k2 depend on the amount of the original chemicals A and B present at the start of
the reaction.
a) The reaction is run with initial conditions set so that X(0) = 0 and k1 = k2 = 2g.
If there is 1 gram of S present after 1 hour, show it will take a total of 9 hours to
create 1.8 g of S.
b) Chemical A is a lot cheaper than chemical B, so the reaction is re-run under the
same conditions, that is, with the same value of k, but with an excess of chemical
A available. If we take X(0) = 0, k1 = 20g and k2 = 2g, how long will it now take
to create 1.8 g of S? (Give your answer correct to 2 decimal places.)
iv) Let f : [a, b]→ [a, b] be a continuous function such that f(x) ≥ x for all x ∈ [a, b].
Let c1 ∈ [a, b) and define a sequence {ck}∞k=1 ⊂ [a, b] by
ck+1 = f(ck),
for k ≥ 1.
a) Prove that the sequence is convergent.
b) Let L = lim
n→∞ cn. Prove that f(L) = L.
46
MATH1241 NOVEMBER 2014
1. i) By expanding sin(A+B) + sin(A−B), or otherwise, find I1 =
∫
sin(5x) cos(x) dx.
ii) Evaluate the integral I2 =
∫ 1
0
x2√
4− x2 dx.
iii) Use appropriate tests to determine whether each of the following series converges or
diverges
a)
∞∑
n=3
1
n(ln n)3
,
b)
∞∑
n=1
2
2n + 3n
.
iv) Let S =



xy
z

 ∈ R3 : z2 = x2 + y2

.
a) Prove that S is closed under scalar multiplication.
b) Prove that S is not a subspace of R3.
v) Let A =
(
0 −1
2 3
)
.
a) Find the eigenvalues and eigenvectors for the matrix A.
b) Write down an invertible matrix M and a diagonal matrix D such that
D =M−1AM.
vi) Let
A =


1 2 2 −1
2 4 10 −44
3 −3 −3 24
1 2 1 6


Using the MAPLE output below, find a basis for ker(A).
> with(LinearAlgebra):
> A := <<1,2,3,1>|<2,4,-3,2>|<2,10,-3,1>|<-1,-44,24,6>>;

1 2 2 −1
2 4 10 −44
3 −3 −3 24
1 2 1 6


> ReducedRowEchelonForm(A); 

1 0 0 5
0 1 0 4
0 0 1 −7
0 0 0 0


47
2. i) Consider the initial value problem
dy
dx
+(2+
1
x
)y =
2
x
, with y(1) = 0, defined for x > 0.
a) Show that an integrating factor for this equation is xe2x.
b) Hence solve the initial value problem.
ii) Find the general solution to
d2y
dx2
+ 2
dy
dx
+ 2y = 20e2x.
iii) Consider the MAPLE session:
> a:=n->n^n/n!*(x-1)^n;
a := n→ n
n(x− 1)n
n!
> a(n+1);
(n+ 1)(n+1)(x− 1)(n+1)
(n+ 1)!
> limit(a(n+1)/a(n),n=infinity);
ex− e
Using MAPLE session above, or otherwise, find the open interval of convergence I =
(a, b) for the power series
∞∑
n=1
nn(x− 1)n
n!
.
iv) Consider the set S consisting of the vectors v1 =

 11
2

 ,v2 =

 23
−1

 from R3 and
let u =

 1−1
12

 .
a) Find scalars λ and µ such u = λv1 + µv2.
b) A linear transformation T : R3 → R3 has v1,v2 as eigenvectors with eigenvalues 2
and −1, respectively.
α) Find T (u) as a linear combination of v1,v2.
β) Denote T (T (u)) by T 2(u), T (T (T (u))) by T 3(u), and so on. Express T n(u) as
a linear combination of v1,v2, where n is a positive integer.
48
v) The two most popular soft drinks in Old South Wales are AppleAde and BananAde.
Assume that no-one in Old South Wales likes both of these drinks equally (that is,
everyone has a preference for one or the other). Past statistics show that 50 % of the
population prefer AppleAde.
Last month the manufacturer advertised AppleAde on television for a week. After that,
a survey was conducted by taking a random sample of 100 people. Of the 100 people
sampled, 60 preferred AppleAde and 40 preferred BananAde.
a) Assuming that the advertising had no effect on people’s preferences, write down
an expression for the tail probability that 60 or more people preferred AppleAde in
a sample of 100.
b) Use the normal approximation to the binomial to calculate the tail probability in
(a), giving your answer to 3 decimal places.
c) Giving reasons, is there evidence that the advertising campaign increased the per-
centage of the population that prefer AppleAde?
3. i) Prove that the function T : P(R)→ R2 defined by
T (p) =
(
p(0)
p(1)
)
, for all polynomials p ∈ P(R) ,
is a linear transformation.
ii) Given a square matrix A, the matrix exponential eA is defined by
eA =
∞∑
k=0
1
k!
Ak = I +A+
1
2!
A2 +
1
3!
A3 + · · · .
(You may assume without proof that this series always converges.)
Let N =
(
0 1
0 0
)
.
a) Find N2.
b) Hence or otherwise, calculate the matrix exponential eN .
c) Prove that if a matrix P is idempotent (that is, P 2 = P ), then eP = I + (e− 1)P .
iii) Let R[R] denote the vector space of real-valued functions defined on R. Let S be the
subspace of R[R] that is spanned by the ordered basis B = {cos(x), sin(x)}. Define the
linear map T : S → S by
T (f) = f − 2f ′ where f ′ = df
dx
.
a) Calculate the matrix C that represents T with respect to the basis B.
b) State the rank of the matrix C found in part (a).
c) From part b), what can be deduced about the solutions y ∈ S of
y − 2y′ = g
where g is a given function in S?
49
d) Using parts a) and c), or otherwise, find all non-zero solutions y ∈ S satisfying
y − 2y′ = cos x .
A company finds that on average 0.05 of its customers return a particular product as
faulty within the six year warranty period. Assuming that the number of returned
products follows an exponential distribution, find, to 3 decimal places, the probability
that for a given product the first claim occurs within the warranty period.
iv) Two (not necessarily distinct) vectors v1, v2 are chosen at random from the (finite) set
of vectors
S =
{(
x
y
)
∈ R2 : x, y ∈ {−1, 0, 1}
}
.
Define A to be the event that v1 and v2 are linearly independent.
a) Show that P (A) = 1627 .
b) Define the discrete random variable X = dim(span {v1,v2}).
Copy and complete the following table for the probability distribution pk = P (X =
k).
k 0 1 2
pk
c) Calculate E(X).
d) If the vector b is chosen randomly from S, state the value of P (B|A), where B is
the event that b ∈ span {v1,v2}.
4. i) Show that the equation of the tangent plane to the paraboloid S given by
x2
a2
+
y2
b2
=
z
c
at the point P (x0, y0, z0) on S is
x0x
a2
+
y0y
b2
=
z + z0
2c
.
ii) Current observations of the Universe suggest that the reciprocal, u, of the matter density
satisfies a differential equation that can be written as(
du
dt
)2
= k2(2u+ ǫu2),
where k is a positive constant, ǫ = ±1 and t is time. Which of the two values ǫ takes is
still a matter of debate, but u is never negative. The radius of the universe is a positive
fractional power of u, so if u→ 0 the universe is crushed to a point.
Take time t = 0 to be the present and assume that u(0) = 1. Measurements tell us that
du
dt
is currently positive.
a) Solve the differential equation in the ǫ = 1 case and show that in this case there is
no value of t > 0 for which u = 0. (In this case the universe grows without bound.
50
b) Repeat the previous part with ǫ = −1 and show that in this case there is a t > 0
for which u = 0. (In this case the universe is said to undergo a Big Crunch).
iii) The following MAPLE session may assist you with this question.
> x:=-8/3*t^3 + 12*t^2 +2:
> y:=(t+1)^2*(t-3)^2:
> factor( diff(x,t)^2 + diff(y,t)^2 );
16(t− 3)2(t2 + 1)2
Let C be the curve given parametrically by(
−8
3
t3 + 12t2 + 2, (t+ 1)2(t− 3)2
)
.
a) A parametric curve
(
x(t), y(t)
)
has vertical tangent when
dx
dt
= 0 and
dy
dt
6= 0. Show
that the curve C has exactly one vertical tangent.
b) Find the arc length of the curve C from t = 0 to t = 3.
iv) Suppose that y satisfies the initial value problem
dy
dx
+ y2 = cos(x) with y(0) = 0 .
Using implicit differentiation, or otherwise, find the first two non-zero terms of the
Maclaurin series of y.
v) Let c be a positive real number and define the sequence {an} for n ≥ 1 by
an = n(c
1/n − 1).
Find, with reasons, lim
n→∞ an.
51
PAST EXAM SOLUTIONS
52
MATH1231 November 2010 Solutions
1. i) Take x =
(
2
2
)
, y =
(−3
−1
)
. Both vectors lie in S, but their sum, x+y =
(−1
1
)
, does
NOT lie in S. Hence S is not closed under vector addition.
ii) a) No, since 2v1 − 3v2 + v3 = 0.
b) w ∈ span(D) if and only if b+ c− a = 0.
c) We obtain u from w by putting a = 3, b = 1, c = 2. This gives the system
 1 2 4 30 −11 −33 −8
0 0 0 0

 .
Put α3 = λ, then backsubstitution gives α2 = −3λ+ 811 , α1 = 2λ+ 1711 .
d) Write the solution to (c) as
α1α2
α3

 = λ

 2−3
1

+

17118
11
0

 .
Then the kernel of A is the homogenous solution which is span



 2−3
1



 . This
vector is a basis for the kernel.
e) rank(A) = 3−nullity(A) = 2.
f)



 13
−2

 ,

 2−5
7



.
iii) a) I1 =
∫
cosh3 x dx
=
∫
cosh x(1 + sinh2 x) dx
= sinhx+
sinh3 x
3
+ C.
b) Put x = 2 sin θ, then
I2 =
∫ pi/6
0
4 sin2 θ.2 cos θ
2 cos θ
dθ
= 4
∫ pi/6
0
sin2 θ dθ
= 2
∫ pi/6
0
1− cos 2θ dθ
= 2
[
θ − 1
2
sin 2θ
]pi/6
0
= pi3 −
√
3
2 .
iv) The corresponding homogeneous equation is y′′ − 5y′ + 4y = 0, which has characteristic
equation λ2 − 5λ+ 4 = 0, whose roots are λ = 1, 4.
53
Hence the homogeneous solution is yh = Ae
x +Be4x.
We propose a particular solution of the form yp = Ce
2x + Dx + E (none of whose
components is a solution of the homogeneous equation).
Substituting into the differential equation we obtain:
−2Ce2x + 4Dx+ (4E − 5D) ≡ 6e2x + 12x.
Equating coefficients and solving, we have C = −3,D = 3, E = 154 .
Hence the general solution is y = Aex +Be4x − 3e2x + 3x+ 154 .
2. i) I6 =
5
6I4 =
5
6 × 34 × 12I0 = 5pi32 .
ii) a) lnn√
n
→ 0 as n→∞, since lnn grows more slowly than √n.
b) 1− 1n approaches 1 as n goes to infinity, but (−1)n oscillates, hence the given sequence
does not have a limit.
iii) a) No, since as k →∞, ak = kk+1 → 1 6= 0.
b)
∞∑
k=1
1
2k2 + 1
<
1
2
∞∑
k=1
1
k2
and this series converges by the p-series test. Hence the
given series converges by comparison.
c) With ak =
ek
k! , we have
ak+1
ak
= ek+1 → 0 as k → ∞. Since this limit is less than 1,
the series converges by the ratio test.
iv) a) Set det(A− λI) = 0, so ∣∣∣∣
(
5− λ −8
1 −1− λ
)∣∣∣∣ = 0.
Expanding out this gives λ2 − 4λ + 3 = 0 with roots λ = 3, 1. These are the
eigenvalues.
For λ = 3, the kernel of (A− 3I) = ker
(
2 −8
1 4
)
= span
{(
4
1
)}
For λ = 1, the kernel of (A− I) = ker
(
4 −8
1 −2
)
= span
{(
2
1
)}
.
Thus
{(
4
1
)}
and
{(
2
1
)}
are the corresponding eigenvectors.
b) Put P =
(
4 2
1 1
)
and D =
(
3 0
0 1
)
, then A = PDP−1.
c) Hence, A8 = PD8P−1 so A8P = PD8 =
(
4 2
1 1
)(
38 0
0 1
)
=
(
4× 38 2
38 1
)
.
v) The columns of B are the images of the standard basis vectors of R3.
T (

10
0

) = 1

13
5

 .
T (

01
0

) = 3

13
5

 .
54
T (

00
1

) = 5

13
5

 .
Hence B =

1 3 53 9 15
5 15 25

 .
3. i) Define the following events:
A - ‘Team A wins the final’, B- ‘Team B loses the final’, C- ‘Team C loses the final’, D-
‘Team D loses the final’.
Then P (A) = P (B ∩ (C ∪D)) = P (B)× P (C ∪D)
= P (B)× (P (C) + P (D)− P (C ∪D)) = 0.8 × (0.3 + 0.6 − 0.3× 0.6) = 0.567.
ii) a) E(X) = (0× 0.2) + (1× 0.3) + (2× 0.1) + (3× 0.2) + (4× 0.2) = 1.9
Var(X) = E(X2)− (E(X))2 = 5.7 − 1.92 = 2.09
b) VAr(Y ) = (−2)2Var(Y ) = 8.36.
iii) a) Binomial(15, 12 , k).
b) P (X ≥ 12) =
∑
k≥12
Binomial(15, 12 , k)
=
1
215
[(
15
12
)
+
(
15
13
)
+
(
15
14
)
+
(
15
15
)]
=
9
512
≈ 0.0176.
c) Yes, as 0.0176 is small (less than 0.05).
iv) We can write b =
(
1
1
)
= 23
(
2
1
)
− 13
(
1
−1
)
. Hence
Ab = A(
2
3
v1 − 1
3
v2) = 2v1 − 1
3
v2 =
1
3
(
11
1
)
.
4. i) Re-write the equation in standard linear form:
dy
dx
+ sec2 xy = sec2 x. (∗)
The integrating factor is φ(x) = e
∫
sec2 x dx = etan x.
Multiplying the differential equation (*) by the integrating factor, we obtain
d
dx
(
yetan x
)
= sec2 xetan x.
Integrating we have
yetanx = etan x + C ⇒ y = 1 + Ce− tanx.
Finally, we apply the initial condition, y(0) = 3 to obtain C = 2 and so the solution is
y = 1 + 2e− tan x.
55
ii) a) The formula for the tangent plane to the surface z = F (x, y) is given by z =
z0 + Fx(x − x0) + Fy(y − y0). Now Fx = 2x + y2 = 5 at P , Fy = 2xy + 4y3 = 8 at
P . Substitution gives:
z = 7 + 5(x− 2) + 8(y − 1)⇒ 5x+ 8y − z = 11.
b) The vector normal to the surface is the vector normal to the tangent plane which
we can read off as

 58
−1

.
iii)
f(x) = 1x f(1) = 1
f ′(x) = −1
x2
f ′(1) = −1
f ′′(x) = 2
x3
f ′′(1) = 2
f ′′′(x) = −6
x4
f ′′′(1) = −6
f4(x) = 24
x5
f4(1) = 24
Hence the Taylor polynomial is
P4(x) = 1−(x−1)+ 2
2!
(x−1)2− 6
3!
(x−1)3+24
4!
(x−1)4 = 1−(x−1)+(x−1)2−(x−1)3+(x−1)4.
iv) a) The velocity vector is given by v =
(
x˙(t)
y˙(t)
)
=
(
2t
3t2
)
so the speed is |v| =
√
4t2 + 9t4 = t
√
9t2 + 4.
b) The arc length is the given by
∫
|v| dt =
∫ 1
0
t
√
9t2 + 4 dt =
[
1
27
(9t2 + 4)
3
2
]1
0
=
1
27
(13
√
13− 4).
v) The volume in the tank satisfies the equation dVdt = 3 − 1 = 2 and the initial condition
V = 40 at t = 0. The volume of liquid in the tank at time t is therefore V = 40 + 2t,
hence the ratio of salt to liquid in the tank at time t is x(t)40+2t . The inflow rate of salt is
6gm/min. The outflow rate of salt is x(t)40+2t × 1.
Hence the differential equation that models this situation is given by:
dx
dt
= inflow rate of salt− outflow rate of salt,
dx
dt
= 6− x(t)
40 + 2t
.
56
MATH1231 November 2011 Solutions
1. i) a) Yes, since it contains more than 3 vectors in R3.
b) No, at least three vectors in S must be linearly independent.
c) Let A = (v1|v2|v3|v4|) then row-reducing,
A→

 1 3 1 −10 1 1 −3
0 0 0 −6


Hence {v1,v2,v4} forms a basis for R3.
ii) 0 ∈ S since A0 = 0.
If x,y ∈ S then Ax = 0 and Ay = 0. Hence A(x+ y) = Ax+Ay = 0+ 0 = 0.
Also, A(λx) = λAx = λ0 = 0. Hence S is a subspace of Rn.
iii) Since rank(T )+ nullity(T ) = m, nullity(T ) = m− k.
iv) a) Let u = sin θ, then dudθ cos θ. Hence
I1 =
∫
u3(1− u2)2 du =
∫
u3 − 2u5 + u7 du− 1
4
sin4 θ − 1
3
sin6 θ +
1
8
sin8 θ + C.
b) From the MAPLE,
I2 =
∫ −2
x2 + 1
− 5
x2 + 2
+
2
x+ 3
dx = −1
2
log(x2+2)− 5√
2
tan−1
x√
2
+2 log |x+3|+C.
v) a) lim
n→∞ an = limn→∞
2 + lnnn
3 + 1n
=
2
3
so the sequence converges.
b) Since lim
n→∞ an 6= 0 the series
∞∑
(−1)nan does not converge.
vi) Using the ratio test with bn =
5n
n!
, we consider
bn+1
bn
=
5
n+ 1
→ 0
as n→∞. Since this limit is less than 1, the series converges by the ratio test.
Alternatively, one might observe that the series equals e5 − 1.
2. i) a) Try yH = e
λx. The characteristic equation is λ2 + 4 = 0 ⇒ λ = ±2ı.
Therefore
yH = A cos 2x+B sin 2x .
To find a particluar solution, try yP = C cos x + D sinx (OR yP = C cos x since
there is no y′ term in the DE). Therefore yP ′′ = −C cosx−D sinx.
Thus
yP
′′ + 4yP = 3C cos x+ 3D sinx = cos x .
Hence C = 13 and D = 0 (as expected).
The general solution is this y(x) = A cos 2x+B sin 2x+ 13 cos x .
57
b) The forcing function cos 2x is contained in the homogeneous solution yH = A cos 2x+
B sin 2x. Hence we try the particular solution
yP = Cx cos 2x+Dx sin 2x .
ii) a) x = 6Ω, y = 12Ω. Therefore 1z =
1
6 +
1
12 ⇒ z = 4Ω.
b)
∂
∂x
(
1
z
)
= − 1
z2
∂z
∂x
=
∂
∂x
(
1
x
+
1
y
)
= − 1
x2
.
Thus
− 1
z2
∂z
∂x
=
1
x2
⇒ ∂z
∂x
=
z2
x2
.
∂z
∂y
=
z2
y2
since the expression for z is symmetric in x and y.
c)
|∆z| ≤
∣∣∣∣∂z∂x
∣∣∣∣ |∆x|+
∣∣∣∣∂z∂y
∣∣∣∣ |∆y|
=
42
62
0.1 +
42
122
0.1
=
(
4
9
+
1
9
)
0.1 =
5
90
=
1
18
Ω ≈ 0.0556Ω
iii) If a transformation is linear it should satisfy T (0) = 0, T (x1 + x2) = T (x1)+T (x2) and
T (λx) = λ (x). If it fails any of these conditions then it is NOT a linear transformation.
Consider T
((
1
1
))
=

 21
−1

 .
Now T
((
2
2
))
=

 44
−2

 6= 2× T ((1
1
))
=

 42
−2


Therefore the transformation is NOT linear.
iv) a) T (x) = Ax− 2x = Ax− 2I3x = (A− 2I3)x = Bx. Therefore
B = A− 2I3 =

1 2 31 2 3
1 2 3

 .
b) For a vector x to be an element of ker (T ) it must satisfy T (x) = 0. Thus
T



−30
1



 = B

−30
1

 =

1 2 31 2 3
1 2 3



−30
1

 =

00
0

 = 0 .
Hence

−30
1

 ∈ ker (T ).
58
c) Note the matrix B is row equivalent to the matrix

1 2 30 0 0
0 0 0

.
# of non-leading columns in row equivalent matrix for B = nullity(T ) = 2
# of leading columns in row equivalent matrix for B = rank(T ) = 1
≡ dim(R3)− nullity(T )
= 3− 2 = 1
v) a) Use Av = λv. Thus (
3 2
0 −1
)(
1
0
)
=
(
3
0
)
= 3
(
1
0
)
.
Hence the eigenvector
(
1
0
)
has associated eigenvalue 3. Similarly
(
3 2
0 −1
)(−1
2
)
=
(
1
−2
)
= −1
(−1
2
)
.
Hence the eigenvector
(−1
2
)
has associated eigenvalue −1.
b) The general form of the solution is given by
x(t) = α
(
1
0
)
e3t + β
(−1
2
)
e−t , α, β ∈ R .
The initial condition x(0) =
(
2
1
)
together with the general solution satisfy
x(0) =
(
2
1
)
= α
(
1
0
)
+ β
(−1
2
)
.
Solving for α and β yields β = −12 ⇒ α = 52 . Overall
x(t) =
5
2
(
1
0
)
e3t − 1
2
(−1
2
)
e−t .
3. i) a)
A
(
2
2
)
=
(
5 −10
2 −4
)(
2
2
)
=
(−10
−4
)
.
b) We solve
A
(
x
y
)
=
(
0
0
)
.
Row reducing A gives the matrix (
5 −10
0 0
)
,
so x = 2y, where y ∈ R. Thus a basis for kerA is
{(
2
1
)}
.
59
c) The image of A is the column space of A, and the two columns of A are multiples
of each other. Thus a basis for the kernel of A is
{(
5
2
)}
.
d) If b belongs to the image of A, then b = Ax for some x ∈ R2. Hence
Ab = A2x = Ax = b.
Alternatively,
A
(
5
2
)
=
(
5 −10
2 −4
)(
5
2
)
=
(
5
2
)
,
and so
Aλ
(
5
2
)
= λ
(
5 −10
2 −4
)(
5
2
)
= λ
(
5
2
)
,
that is,
Ab = b
for any vector in the image of A, by part (c).
Alternatively, from part (c), if b is in the image of A, then 2b1 = 5b2, and so
A
(
b1
b2
)
=
(
5 −10
2 −4
)(
b1
b2
)
=
(
5b1 − 10b2
2b1 − 4b2
)
=
(
5b1 − 4b1
5b2 − 4b2
)
=
(
b1
b2
)
,
as required.
e) If b ∈ kerA, then Ab = 0 = 0b, and so b is an eigenvector with eigenvalue 0. From
(b), we may take b =
(
5
2
)
.
From (d), if b ∈ image A, then Ab = 1b, and so b is an eigenvector with eigenvalue
1. From (c), we may take b =
(
2
1
)
.
ii) We may present the information using a tree diagram or a table. We write L for Liverpool
supporters, and L¯ for nonsupporters; we also write R for red shirt wearers and R¯ for the
others.
L L¯
R 0.05 × 0.80 0.95 × 0.30
R¯ 0.05 × 0.20 0.95 × 0.70
With our notation
P (L|R) = P (L ∩R)
P (R)
=
0.05 × 0.80
0.05 × 0.80 + 0.95× 0.30 =
0.04
0.325
=
8
65
.
iii) a) A binomial distribution could be used, with p = 12 , n = 11.
b) There are 8 months in which the temperature increased, 3 months in which the
temperature decreased, and one month with no change. We ignore the month in
which there is no change.
Let the random variable X be the number of months in which the temperature
increases. Then
P (X ≥ 8) =
11∑
k=8
(
11
k
)
1
2k
1
211−k
=
1
211
11∑
k=8
(
11
k
)
= 0.113 . . . .
60
iv) a) Of the 16 possible outcomes of four letters, each of which is H or T , two lead to
X = 4, namely, HHHH and TTTT . Thus
P (X = 4) =
2
16
=
1
8
.
Alternate solution: if X = 4, then the possibilities were HHHH and TTTT . The
first toss can give any outcome, and the subsequent tosses then have to match the
first toss; since the match occurs with probability 12 and there are 3 subsequent
tosses,
P (X = 4) =
1
2
× 1
2
× 1
2
=
1
8
.
b) Since P (X = 0) + P (X = 2) + P (X = 4) = 1,
P (X = 0) = 1− 1
2
− 1
8
=
3
8
.
Alternative solution: there are
(4
2
)
(that is, 6) outcomes that lead to X = 0, hence
P (X = 0) =
6
16
=
3
8
.
c)
E(X) =
∑
x
xP (X = x) = 0 + 2× 1
2
+ 4× 1
8
=
3
2
.
d) If X > E(X), then either X = 2 or X = 4. Hence
P (X > E(X)) =
1
2
+
1
8
=
5
8
.
4. i) Application of the reduction formula for n = 1 yields
I2 =
1
4
+
1
2
I1.
On the other hand,
I1 =
∫ 1
0
dx
1 + x2
= arctan x
∣∣∣1
0
=
π
4
so that
I2 =
1
4
+
π
8
.
ii) a) The total force acting on the ship is given by
F = Fthrust + Fresistance = L− kv.
Newton’s second law states that
m
dv
dt
= F,
leading to the equation of motion
m
dv
dt
= L− kv.
61
b) The differential equation in a) has standard form so that the integrating factor is
given by
IF = exp
(∫
k
m
dt
)
= exp
(
k
m
t
)
without loss of generality. By definition of the integrating factor, multiplication by
IF produces
d
dt
(
e
k
m
tv
)
=
L
m
e
k
m
t.
Intergration of both sides yields
e
k
m
tv =
L
m
m
k
e
k
m
t + C
so that
v =
L
k
+Ce−
k
m
t.
Since the ship is initially at rest, we are led to the condition
0 = v(0) =
L
k
+ C ⇒ C = −L
k
and, hence,
v =
L
k
(
1− e− km t
)
.
c) Since the exponential function is strictly positiv, we conclude that
1− e− km t < 1
so that v < L/k.
d) If T denotes the time at which the speed of the ship reaches L/(2k) then
1
2
L
k
= v(T ) =
L
k
(
1− e− kmT
)
.
Division by L/k and rearrangement of terms yield
e−
k
m
T =
1
2
so that application of the logarithmic function produces
− k
m
T = ln
1
2
.
Hence,
T = −m
k
ln
1
2
=
m
k
ln 2.
iii) a) At x = 1, we obtain the alternating series
∞∑
n=0
(−1)nan, an = 1
2n + 1
.
62
It is evident that the sequence (an) is non-negativ, non-decreasing and approaches
0 for large n, that is,
an ≥ 0
an+1 ≤ an
an → 0 as n→∞.
The alternating series test (Leibniz test) then implies that the alternating series
converges.
b) The Maclaurin series for the function tan−1 converges at 1, Thus,
∞∑
n=0
(−1)n 1
2n+ 1
= tan−1 1 =
π
4
.
(Note: Technically Abel’s theorem is required here which implies that the limit
coincides with tan−1 x at x = 1. This is not in the course, but may be tacitly
assumed here.)
iv) a) The Fundamental Theorem of Calculus implies that
dx
dt
=
d
dt
(∫ t
1
cosu
u
dt
)
=
cos t
t
.
Similarly,
dy
dt
=
sin t
t
.
b) At t = t1 = π/2, we see that
dx
dt
=
cos(π/2)
π/2
= 0,
dy
dt
=
sin(π/2)
π/2
=
2
π
6= 0.
Accordingly, the curve has a vertical tangent at
(
x(π/2), y(π/2)
)
.
c) The next value of t for which the reasoning of b) may be repeated is t = t2 = 3π/2.
d) By definition of arc length,
L =
∫ t2
t1
√(
dx
dt
)2
+
(
dy
dt
)2
dt
=
∫ t2
t1
√(
cos t
t
)2
+
(
sin t
t
)2
dt
=
∫ t2
t1
1
t
dt
= ln t
∣∣∣t2
t1
(t > 0)
= ln t2 − ln t1
= ln
t2
t1
= ln 3.
63
MATH1231 February 2012 Solutions
1. i) a) Suppose x =
(
x1
x2
)
∈ S and λ ∈ R, then λx =
(
λx1
λx2
)
.
Now λx1.λx2 = λ
2x1x2 ≥ 0, since x1x2 ≥ 0. Hence λx ∈ S.
b) For example,
( −1
−3
)
∈ S, and
(
2
2
)
∈ S but x+ y =
(
1
−1
)
6∈ S.
ii) a) Consider the system of equationa λ1v1 + λ2v2 + λ3v3 + λ4v4 = b.
Putting this into matrix form and reducing to echelon form we have:

1 2 1 3 −2
0 0 2 2 −3
0 0 0 1 −2
0 0 0 0 0

 .
Since the (A|b) system has solution, we conclude that b belong to span(S).
b) No, since the row echelon form of A, has a non-leading column, the system (A|b),
for arbitrary b will not always have a solution.
c) Using the row-echelon form above, we see that a basis for span(S) is:



1
−2
1
2

 ,


1
0
5
0

 ,


3
−4
8
7



 = {v1,v3,v4}.
iii) a) For x,y ∈ R3, we have
T (x+y) =
(
(x1 + y1)− (x3 + y3)
2(x1 + y1) + (x2 + y2)
)
=
(
x1 − x3
2x1 + x2
)
+
(
y1 − y3
2y1 + y2
)
= T (x)+T (y).
Also
T (λx =
(
λx1 − λx3)
2λx1 + λx2
)
= λ
(
x1 − x3
2x1 + x2
)
= λT (x).
b)
A =
(
1 0 −1
2 1 0
)
.
iv) a) rank(T ) = 3 since there are 3 leading columns.
b) nullity(T ) = 2 since 5− 3 = 2.
c) Putting x5 = λ gives x4 = λ and putting x3 = µ we have x2 = 2µ − λ and
x1 = −3µ− 2λ. Hence a basis for the kernel of T is



−2
−1
0
1
1

 ,


−3
2
1
0
0




.
64
d) b1 + b4 = 0, b2 = b4 − b5 = 0.
e) One possible basis is:



1
−1
2
−1
0

 ,


3
3
−4
−3
6

 ,


1
2
3
−1
3

 ,


1
0
0
0
0

 ,


0
1
0
0
0




.
2. i) a) Set det(A− λI) = 0, then∣∣∣∣ 4− λ −27 −5− λ
∣∣∣∣ = (λ− 2)(λ + 3) = 0.
Hence λ = 2 or λ = −3.
For λ = 2, ker(A− 2I) = ker
(
2 −2
7 −7
)
= span
(
1
1
)
.
For λ = −3, ker(A + 3I) = ker
(
7 −2
7 −2
)
= span
(
2
7
)
.
So we take
(
1
1
)
,
(
2
7
)
as the eigenvectors.
b) Since the system can be written as ddty = Ay, the general solution is
y =
(
y1
y2
)
= c1e
2t
(
1
1
)
+ c2e
−3t
(
2
7
)
.
ii) a) Let P (A) = x then we can draw up the following diagram.
diagram
Hence
P (B) = x× 0.95 + (1− x)× 0.15 = 0.65.
Solving we obtain x = 58 = 0.625.
b)
P (A|B) = P (A ∩B)
P (B)
=
0.95 × 58
0.65
= 0.913.
iii) a) Since f is a density function,
1 =
∫ 1
0
α(1− x2) dx = 2
3
α,
hence α = 32 .
b)
E(X) =
∫ ∞
−∞
xf(x) dx =
3
2
∫ 1
0
x(1− x2) dx = 3
8
.
iv) a) Let X be the number of sixes in 720 rolls of a fair die, the X follows a binomial
distribution, B(720, 16) and so
P (X ≤ 100) =
100∑
k=0
(
720
k
)(
1
6
)k (5
6
)720−k
.
65
b) X can be approximated by the normal random variable Y ∼ N(µ, σ2), where µ =
np = 120 and σ2 = npq = 100.
Then, using the standard normal distribution,
P (X ≤ 100) ≈ P (Y ≤ 100.5) = P (Z ≤ 100.5 − 120
10
= P (Z ≤ −1.95) ≈ 0.026,
using the Normal table.
c) Since 0.026 < 0.05 the tail probability is significantly low. Hence there is good
evidence that the die is biased.
v) Let v be an eigenvalue for B with eigenvalue λ, then since Bv = λv, we have B2 =
Bλv = λ2v. Thus, multiplying both sides of B2 = 3B − 2I by v we have
B2v = 3Bv− 2Iv ⇒ λ2v = 3λv − 2v.
Since v 6= 0, we conclude that λ2 − 3λ+ 2 = 0 and so λ = 1 or λ = 2.
3. i) a) We have the surface z =
√
x2 + y2. Using the formula for the normal we find (at
x0)
n =

 zxzy
−1

 =


x√
x2+y2
y√
x2+y2
−1

 =

 8106
10
−1

 = 1
5

 43
−5

 .
b) The generic plane perpedicular to n has the form
4x+ 3y − 5z = d
substituting x0 gives
32 + 18− 50 = 0 = d.
So the equation is 4x+ 3y − 5z = 0.
c) x = 8, y = 6,∆x = 0.02,∆y = −0.03.
z =
√
8.022 + 5.872
≈ z0 + 4
5
0.02 +
3
5
(−0.03)
= 0.998
ii) z = xf(xy)
∂z
∂x
= f(xy) + xyf ′(xy)
∂z
∂y
= x2f ′(xy)
therefore
x
∂z
∂x
− y ∂z
∂x
= xf + x2yf ′ − x2yf ′(xy)
= xf
= z
66
iii) a)
I1 =
∫
cos3(θ) sin4(θ)dθ
=
∫
cos(θ)(1− sin2(θ) sin4(θ)dθ
=
∫
cos(θ) sin4(θ)− cos(θ) sin6(θ)dθ
=
sin5(θ)
5
− sin
7(θ)
7
+ C.
b) We begin by considering a partial fraction expansion of the form
x2 + x
(x− 2)(x2 + 2) =
A
x− 2 +
Bx+ C
x2 + 2
which implies
x2 + x = A(x2 + 2) + (Bx+ C)(x− 2).
Considering x = 2 gives A = 1. Comparing x2 terms, yields B = 0. And comparing
the constant term gives 2− 2C = 0 =⇒ C = 1.
So
I2 =
∫
1
x− 2 +
1
x2 + 2
dx
= ln |x− 2|+ 1√
2
tan−1
(
x√
2
)
+ C.
iv) We can calculate and integrating factor
F (x) = exp
(
−3
∫
1
x
dx
)
= exp(−3 ln(x)) = 1
x3
.
So therefore
d
dx
(
y
1
x3
)
=
1
x7
,
y
1
x3
=
−1
6x6
+ C,
y =
−1
6x3
+ Cx3.
And y(1) = 4 =⇒ C = 256 .
v) a) Calculating the appropriate partial derivatives gives
M = 2xy
N = x2 + y3
∂M
∂y
= 2x =
∂N
∂x
.
So therefore the equation is exact.
67
b) Solution is F (x, y) = C where
∂F
∂x
= 2xy,
∂F
∂y
= x2 + y3.
This implies
F = x2y + φ(y)
F = x2y +
y4
4
+ ψ(y)
comparing the equations above gives the solution
x2y +
y4
4
= C.
4. i) Let M(t) denote the mass of salt in the tank after t minutes for 0 ≤ t ≤ 150. *Intially
tank contains 250L of water.
* 5 g/L and Brine is run into the tank at 6L/m
* The mixed liquid in the tank is constantly run off at 1L/m
Note that dMdt = rate of change of M(t) = rate of salt in – rate of salt out. Now
Rate of salt in = 5× 6 = 30.
Hence, after any time t, the amount of salt in the tank is given by
M(t)
250 + 6t− t .
Multiply by the rate of outflow yields that
rate of salt out =
M(t)
250 + 5t
× 1.
Thus, we have
dM
dt
= 30− M(t)
250 + 5t
.
ii) Assume that y = eλx is a solution then λ2 − 4λ+ 4 = 0, so λ = 2.
Thus, y1(x) = e
2x and y2(x) = xe
2x.
So the solution to the homogeneous equation is yH(x) = Ay1(x)+By2(x). Now we guess
the particular solution
yp(x) = αx+ β.
Substituting we have −4α+ 4αx + 4β = x. This implies
α = 1/4 = β.
Therefore,
yG(x) = yH(x) + yp(x) = Ae
2x +Bxe2x +
1
4
(x+ 1)
is the general solution.
68
iii) a) Since an := (−1)n cos(π/n2) 6→ 0 then the series diverges by the kth term test.
b) Let an :=
sin2(3n)
n2
and see that an ≤ 1/n2 for all n ≥ 1. If we consider the series∑∞
n=1 1/n
2 then since
∑∞
n=1 1/n
2 converges by the p − series test then ∑ an con-
verges.
iv) Let an :=
(x−1)n
lnn . If we consider
lim
n→∞ |an+1|/|an|,
then we have that
lim
n→∞
|an+1|
|an| = |x− 1| limn→∞
lnn
ln(n+ 1)
= |x− 1|.
Thus for convergence of the series via the ratio test, we require |x− 1| < 1. This implies
the interval of convergence is (0, 2)
v) a)
1
1 + x
= 1− x+ x2 − x3 + · · · =
∞∑
n=1
(−1)n+1xn−1, x ∈ (−1, 1].
b) Note that from i) that
1
1 + x2
= 1− x2 + x4 − x6 + · · · =
∞∑
n=1
(−1)n+1x2n, x ∈ (−1, 1].
Thus, P6(x) = 1− x2 + x4 − x6.
c) By the previous part, it follows that∫ 1
0
1
1 + x2
dx =
∫ 1
0
1− x2 + x4 − x6 dx = 1− 1
3
+
1
5
− 1
7
= 76/105.
vi) a) By substituting, θ = t, we have that
x′(θ) = r′(θ) cos θ − r(θ) sin θ,
and
y′(θ) = r′(θ) sin θ + r(θ) cos θ.
It now follows that
[x′(θ)]2 + [y′(θ)]2 = [r′(θ) cos θ − r(θ) sin θ]2 + [r′(θ) sin θ + r(θ) cos θ]2
= [r′(θ)]2 + [r(θ)]2 (cos2 θ + sin2 θ = 1)
By changing variables, the boundary transforms to θ1 ≤ θ ≤ θ2 and the desired
formula is obtained.
69
b) Here θ1 = −pi2 , θ2 = pi2 and r = 1 + sin θ. By using the above formula and cos2 θ +
sin2 θ = 1, we have∫ θ2
θ1
2πr(θ) cos θ
√
r2 + (
dr
dθ
)2 dθ =
∫ pi
2
−pi
2
2π(1 + sin θ) cos θ
√
(1 + sin θ)2 + (cos θ)2 dθ,
=
∫ pi
2
−pi
2
23/2π(1 + sin θ) cos θ
√
(1 + sin θ dθ
= 23/2π
∫ pi
2
−pi
2
cos θ(1 + sin θ)3/2 dθ
= 23/2π
[
2
5
(1 + sin θ)5/2
]pi
2
−pi
2
=
25
5
π =
32
5
π.
70
MATH1231 November 2012 Solutions
1. i) Putting the vectors into a matrix, we have
 1 3 0 32 4 −2 2
3 6 3 1

 ∼

 1 3 0 30 1 1 2
0 0 6 2


after row reducing. Hence {v1,v2,v3} is a basis for span(S).
ii) a) We solve Ax = 0. So (A|0) =
∼
(
3 1 1 0 0
0 0 0 1 0
)
.
Hence x4 = 0 and putting x2 = 3α, x3 = 3β, we have x1 = −α−β and so the kernel
is α


−1
3
0
0

+ β


−1
0
3
0

. Hence a basis for the kernel is




−1
3
0
0

 ,


−1
0
3
0



 .
b) nullity(A) = 2.
iii) a) Suppose p ∈ W , then p′(0)p′′(0) = 0. Now (λp)′(0)(λp)′′(0) = λ2p′(0)p′′(0) = 0,
hence λp ∈W .
b) The polynomials p, q where p(x) = x, q(x) = x2 belong to W but it is easy to check
that their sum p+ q, where (p + q)(x) = x+ x2 does not.
iv) a) b = −1, c = 0.
b) We can hence write
J =
∫
1
x− 1 dx−
∫
x+ 1
x2 + 2x+ 2
+
∫
1
(x+ 1)2 + 1
= log |x− 1| − 1
2
log(x2 + 2x+ 1) + tan−1(x+ 1) + C.
v) The equation is linear. The integrating factor is e− log(x−1) = 1x−1 . Multiplying by the
integrating factor we have,
d
dx
(
y
x− 1
)
=
2
x− 1
and so y = 2(x− 1) log |x− 1|+ C(x− 1).
Now y(2) = 3⇒ C = 3 and so y = y = 2(x− 1) log |x− 1|+ 3(x− 1).
vi) a)
S =
∫ t1
t0
2πy
√(
dx
dt
)2
+
(
dy
dt
)2
dt.
71
b)
S = 2π
∫ 2pi
0
(1− cos t)√2− 2 cos t dt = 2π32
3
=
64π
3
.
2. i) a)
∂P
∂V
= −k T
V 2
,
∂P
∂T
= k
1
V
.
b) Given in the problem the information
∆V
V
= +4% and
∆T
T
= −3%. The change in
the pressure is approximately given by
∆P ≈ ∂P
∂V
∆V +
∂P
∂T
∆T
= −k T
V 2
∆V + k
1
V
∆T
= −k T
V
∆V
V
+ k
T
V
∆T
T
= −P∆V
V
+ P
∆T
T
.
Thus
∆P
P
≈ −∆V
V
+
∆T
T
= −(+4%) + (−3%) = −7%
Hence the pressure decreases by 7%.
ii) a) Consider the function f(x) =
x√
4x2 − 1 which is continuous on the interval (
1
2 ,∞).
We now investigate the limit as x tends to infinity for function f .
lim
x→∞
x√
4x2 − 1 = limx→∞
x
2x
√
1− 1
4x2
=
1
2
lim
x→∞
1√
1− 1
4x2
=
1
2
× 1√
1− 0 =
1
2
.
Since the limit of the continuous function exists and is
1
2
then the limit of the
sequence exists and is also equal to
1
2
, i.e.,
lim
n→∞
n√
4n2 − 1 =
1
2
.
b) Note this is an example from the notes. Consider
bn =
n!
nn
=
1
n
2
n
· · · n
n
≤ 1
n
n
n
· · · n
n
=
1
n
whenever n ≥ 1. On the other hand, bn is always positive and thus 0 ≤ bn ≤ 1
n
.
Hence by the pinching (squeeze, sandwich) theorem bn → 0 as n→∞, i.e.,
lim
n→∞
n!
nn
= 0 .
72
iii) a) Set F = 2x+ 3y,G = 3x+ 2y then ∂F∂y =
∂G
∂x = 3. Hence the equation is exact.
b) Let the solution be u(x, y) = c then,
∂u
∂x
= 2x+ 3y ⇒ u(x, y) = x2 + 3xy + f(y)
⇓
∂u
∂y
= 3x+ 2y ⇔ ∂u
∂y
= 3x+ f ′(y)
Therefore f ′(y) = 2y and thus f(y) = y2 − constant. Overall
u(x, y) = x2 + 3xy + y2 − constant = 0
and therefore x2 + 3xy + y2 = constant is the general solution.
iv) The matrix A is constructed by first calculating the transformation on each of the
standard basis vectors e1, e2, e3 and e4 of R
4. The resulting vectors form the columns
of the matrix A, i.e.,
A =
(
T (e1) T (e2) T (e3) T (e4)
)
=

T




1
0
0
0



 T




0
1
0
0



 T




0
0
1
0



 T




0
0
0
1






=


0 1 0 0
−1 0 2 0
1 1 0 5
0 6 0 −3


v) a) The eigenvalues are calculated by solving |A− λI| = 0, i.e.,
|A− λI| = 0 =
∣∣∣∣6− λ −12 3− λ
∣∣∣∣ = (6− λ)(3− λ) + 2
= λ2 − 9λ+ 20
= (λ− 4)(λ− 5)⇒ λ = 4, 5 .
Check: Trace(A) = 6 + 3 = 9 = 4 + 5 = sum of eigenvalues.
• λ = 4 (
2 −1 | 0
2 −1 | 0
)
⇒ vλ=4 = t
(
1
2
)
, t ∈ R\{0}
• λ = 5 (
1 −1 | 0
1 −1 | 0
)
⇒ vλ=5 = t
(
1
1
)
, t ∈ R\{0}
Check: Av = λv
73
b)
P =
(
1 1
2 1
)
and D =
(
4 0
0 5
)
OR
P =
(
1 1
1 2
)
and D =
(
5 0
0 4
)
vi) a) For a transformation to be linear it must satisfy
T (λx+ µy) = λT (x) + µT (y)
where x,y ∈ R3 and λ, µ ∈ R.
Consider
T (λx+ µy) =
(λx+ µy) · b
|b|2 b
=
λx · b
|b|2 b+
µy · b
|b|2 b
= λ
x · b
|b|2 b+ µ
y · b
|b|2 b
= λT (x) + µT (y)
Hence the transformation T is linear .
b) In addition to the zero vector, the ker(T ) in this case is the set of vectors perpen-
dicular (normal, orthogonal) to the vector b since x · b will be equal to zero and
hence T (x) = 0. Thus geometrically the kernel of T is a plane through the origin
with a normal in the direction of the vector b.
3. i) The vector
(
1
1
)
lies on the line y = x, so it does not change under T , that is,
T
(
1
1
)
=
(
1
1
)
.
Hence this vector is an eigenvector and the eigenvalue is 1.
The vector
(
1
−1
)
is perpendicular to the line y = x, so it reflects across the line, that is
T
(
1
−1
)
=
(−1
1
)
= (−1)
(
1
−1
)
.
Hence this vector is an eigenvector and the eigenvalue is −1.
ii) Let B, T and L be the events that a worker used the bus, the train or the light rail
respectively. By the inclusion-exclusion formula,
P (B ∪ T ∪ L)
= P (B ∩ T ∩ L)− P (B ∪ T )− P (B ∪ L)− P (T ∪ L) + P (B) + P (T ) + P (L),
74
L
T
B
a
bc d
e f
g
Figure 1: Three sets intersecting
and hence
P (B ∩ T ∩ L) = 0.85 + 0.20 + 0.10 + 0.08− 0.50 − 0.40 − 0.30 = 0.03.
Alternative Solution
This problem can be solved by setting up a Venn diagram, as in Figure 1. Consider the
regions with probabilities a, b, c, . . . , as shown. Then we are told that
a+ b+ c+ d+ e+ f + g = 0.85 (8)
c+ d+ e+ g = 0.50 (9)
a+ e+ f + g = 0.40 (10)
b+ d+ f + g = 0.30 (11)
e+ g = 0.20 (12)
d+ g = 0.10 (13)
f + g = 0.08. (14)
Solving these equations, by subtracting equations (2) to (4) from (1) and then adding
(5) to (7), we find that g = 0.03.
Essentially the same solution was also presented just working with the Venn diagram,
and putting numbers instead of the letters a, b, and so on.
iii) a) Since f is a pdf,
1 =
∫ +∞
−∞
f(x) dx =
∫ a
0
(1 + 4x) dx = (x+ 2x2)
∣∣a
0
= a+ 2a2.
Hence (2a − 1)(a + 1) = 0, but since a > 0, the only relevant solution is a = 1/2.
75
b)
E(X) =
∫ +∞
−∞
f(x) dx =
∫ 1/2
0
x(1 + 4x) dx =
(x2
2
+
4x3
3
)∣∣∣∣1/2
0
=
1
8
+
1
6
=
7
24
.
c)
E(X2) =
∫ +∞
−∞
x2 f(x) dx =
∫ 1/2
0
x2(1 + 4x) dx =
(x3
3
+ x4
)∣∣∣∣1/2
0
=
1
24
+
1
16
=
5
48
.
Hence
Var[(]X) = E(X2)− E(X)2 = 11
576
.
iv) a) Let X be the number of the rolls that yielded a 6.
Then
P (X ≥ 68) =
300∑
k=68
(
300
k
)(1
6
)k(5
6
)300−k
.
b) Assuming that the die is fair, we see that µ = 50 and σ2 = 300 × 16 × 56 = 1253 =
41.666 . . . , so σ ≈ 6.455. Now, set
Z =
X − 50
6.455
.
Then P (X ≥ 68) = P (Z ≥ 2.788 . . . ), and Z is approximately normal, and hence
(from the tables)
P (X ≥ 68) ≈ 0.003.
c) Since this probability is much less than 0.05, we may conclude that it is likely that
the die is biased.
v) a) By definition,
col(AAT ) = {AAT~x : ~x ∈ R3}
= {A~y : ~y ∈ col(AT )}
⊆ {A~y : ~y ∈ R3}
= col(A).
Now rank(AAT ) ≤ rank(A), since rank(A) = dim(col(A).
b) First, col(A) = span
{
A
(
1
0
)
, A
(
0
1
)}
, and so rank(A) ≤ 2. It follows that rank(AAT ) ≤
2.
By the rank-nullity theorem,
nullity(AAT ) = 3− rank(AAT ) ≥ 3− 2 = 1.
The nullity of the identity matrix is 0 and hence AAT cannot be the identity matrix.
76
4. i) The series may be formulated as
∞∑
k=2
(x− 3)k
2k log k
=
∞∑
k=2
ak(x) =
∞∑
k=2
bk(x− 3)k.
Thus,
lim
k→∞
∣∣∣∣ak+1ak
∣∣∣∣ = limk→∞
∣∣∣∣ (x− 3)k+12k+1 log(k + 1) 2
k log k
(x− 3)k
∣∣∣∣
= lim
k→∞
|x− 3|
2
log k
log(k + 1)
=
|x− 3|
2
.
Therefore, according to the ratio test, the series
∞∑
k=2
|ak(x)| and, hence, the series
∞∑
k=2
ak(x)
converges if
|x− 3|
2
< 1.
Accordingly, the open interval of convergence is I = (1, 5).
Alternatively, the radius of convergence may be obtained from
R = lim
k→∞
∣∣∣∣ bkbk+1
∣∣∣∣ = limk→∞ 2log(k + 1)log k = 2
since the above limit exists. The open interval of convergence is then given by I =
(3−R, 3 +R) which is in agreement with the previous result.
ii) a) Since ∣∣∣∣∣ cos 2n(n3 + 1) 12
∣∣∣∣∣ ≤ 1(n3 + 1) 12 < 1n 32
and the p-series
∞∑
n=1
1
n
3
2
converges, the comparison test implies that the series
∞∑
n=1
cos 2n
(n3 + 1)
1
2
converges absolutely and therefore in the normal sense.
77
b) Application of the ratio test yields
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)4
(n+ 1)!
n!
n4
= lim
n→∞
(
n+ 1
n
)4 1
n+ 1
= 0 < 1
and, hence, the series
∞∑
n=1
an =
∞∑
n=1
n4
n!
converges.
iii) a) Evaluation of ∫
1
1 + x
dx =
∫
(1− x+ x2 − x3 + · · · ) dx
produces
log(1 + x) +D = x− x
2
2
+
x3
3
− x
4
4
+ · · ·
so that, at x = 0,
log(1 + 0) +D = 0 ⇒ D = 0.
b) Integration of the above result, that is,∫
log(1 + x) dx =
∫ (
x− x
2
2
+
x3
3
− x
4
4
+ · · ·
)
dx,
leads to
(x+ 1) log(x+ 1)− (x+ 1) + C = x
2
2
− x
3
2× 3 +
x4
3× 4 −
x5
4× 5 + · · · .
Evaluation at x = 0 then produces
(0 + 1) log(0 + 1)− (0 + 1) + C = 0 ⇒ C = 1.
iv) a) Differentation of
db
dt
= k1a− k2b
produces
d2b
dt2
= k1
da
dt
− k2 db
dt
= −k21a− k2(k1a− k2b)
by virtue of
da
dt
= −k1a.
Hence, the expression
d2b
dt2
+ (k1 + k2)
db
dt
+ k1k2b
vanishes.
78
b) The characteristic equation associated with the differential equation
d2b
dt2
+ (k1 + k2)
db
dt
+ k1k2b = 0
is given by
0 = λ2 + (k1 + k2)λ+ k1k2 = (λ+ k1)(λ+ k2).
Hence, if k1 6= k2 then the general solution is given by
b = Ce−k1t +De−k2t.
c) If k1 = k2 = k then λ = −k constitutes a double root and
b = Ce−kt +Dte−kt.
The initial condition b(0) = 0 then produces
0 = Ce0 ⇒ C = 0
so that
b = Dte−kt.
d) Differentiation of b as given above leads to
db
dt
= D(1− kt)e−kt
so that the condition
db
dt
= 0
is equivalent to
1− kt = 0 ⇒ t = 1
k
.
79
MATH1231 November 2013 Solutions
1. i) a) Suppose

 ab
c

 ∈ S then a3 + b3 + c3 = 0. Thus λ3a3 + λ3b3 + λ3c3 = 0 so

 λaλb
λc

 ∈ S.
b) For example the vectors a =

 10
−1

 and b =

 2−2
0

 lie in S, but their sum

 3−2
−1

 does not. Hence S is not closed under addition and so does not form a
subspace.
ii) a) The MAPLE output shows that A has 4 distinct eigenvalues and hence it has four
linearly independent eigenvectors.
b)
y(t) = Ae2t


1
1
3
1

+Be3t


2
5
0
1

+Ce5t


1
2
3
1

+D


1
2
4
1

 .
iii) a) Pr( disc is a movie)= 34 × 35 + 14 × 910 = 2740 . (A tree diagram can also be used).
b) For sets A and B, PR(A|B) = Pr(A∩B)
Pr(B) and so Pr(BR|movie) = 940/2740 = 13 .
iv) a) Using partial fractions,
∫
5x2 − 6x+ 4
x(x− 1)(x− 2) dx =
∫
2
x
− 3
x− 1+
6
x− 2 dx = 2 log |x|−3 log |x−1|+6 log |x−2|+C.
b) I2 = ∫ pi/2
0
sin3 θ cos2 θ dθ =
∫ pi/2
0
sin θ × (1− cos2 θ) cos2 θ dθ
=
∫ pi/2
0
sin θ cos2 θ − sin θ cos4 θ dθ = −cos
3 θ
3
+
cos5 θ
5
∣∣∣pi20 = 215 .
v) Separating and integrating∫
y2 dy =
∫
sinx dx⇒ y3 = C − 3 cos x.
Apppying the initial condition, we find C = 4.
vi) f ′′(0) = 8 and f ′′′(0) = 0.
80
2. i) a) Let u(x, y) = 2x− 3y2 and hence F (x, y) = f (u(x, y)). Thus
∂F
∂x
=
df
du
∂u
∂x
= f ′ × (2) = 2f ′
∂F
∂y
=
df
du
∂u
∂y
= f ′ × (−6y) = −6yf ′
such that f ′ ≡ df
du
. Therefore
3y
∂F
∂x
+
∂F
∂y
= 3y × 2f ′ − 6yf ′ = 6yf ′ − 6yf ′ = 0
as required.
b) Given F (x, 0) = cos(x) for all x. Note
F (x, 0) = cos(x) = cos
(
1
2
(2x)
)
= f(2x)
and therefore f(u) = cos
(
1
2u
)
. Hence
F (x, y) = f(2x− 3y2) = cos
(
1
2
(2x− 3y2)
)
.
ii) The arc length l of the curve C, given parametrically by x = t3 and y = 2t2 between
t = 0 and t = 1, is given by the integral
l =
t=1∫
t=0
√(
dx
dt
)2
+
(
dy
dt
)2
dt =
t=1∫
t=0
√
(3t2)2 + (4t)2 dt
=
t=1∫
t=0
√
16t2 + 9t4 dt
=
t=1∫
t=0
t
√
16 + 9t2 dt
=
1
18
2
3
(16 + 9t2)
3
2
∣∣∣1
0
=
1
27
(125 − 64)
=
61
27
.
The arc length of C is 6127 units of length.
iii) a) Try yH = e
λx. Characteristic equation λ2− 5λ+6 = (λ− 3)(λ− 2) = 0 ⇒ λ = 2, 3.
Therefore
yH(x) = Ae
2x +Be3x .
81
b) Try yP = Cxe
2x since the forcing function e2x, i.e., RHS of the DE, is proportional
to the homogeneous solution yH(x) (calculated in the previous part). Thus
y
′
P = Ce
2x + 2Cxe2x and y
′′
P = 4Ce
2x + 4Cxe2x .
Substituting into the DE with yP , y
′
P and y
′′
P yields
y
′′
P − 5y
′
P + 6yP = 4Ce
2x + 4Cxe2x − 5 (Ce2x + 2Cxe2x)+ 6Cxe2x = −Ce2x
= e2x .
Therefore C = −1 and yP (x) = −xe2x. The general solution yG(x) is the addition of
the homogeneous solution yH(x) (from the previous part) and the particular solution
yP (x), i.e.,
yG(x) = yH(x) + yP (x)
= Ae2x +Be3x − xe2x .
iv) a) The vector e1 has length 1 and thus when it is rotated the transformed vector will
have length 2. Thus
T (e1) = T



1 cos(0)
1 sin(0)




=


2 cos
(
0 +
π
6
)
2 sin
(
0 +
π
6
)

 =


2 cos
π
6
2 sin
π
6

 =


2
√
3
2
2
1
2

 =


√
3
1


as required.
b) The matrix A is constructed by first calculating the transformation on each of the
standard basis vectors e1 and e2 of R
2. Note
T (e2) = T




1 cos
(π
2
)
1 sin
(π
2
)




=


2 cos
(π
2
+
π
6
)
2 sin
(π
2
+
π
6
)

 =


2 cos
2π
3
2 sin
2π
3

 =


2
(
−1
2
)
2
√
3
2

 =

−1√
3

 .
The resulting vectors form the columns of the matrix A, i.e.,
A =
(
T (e1) T (e2)
)
=
(
T
((
1
0
))
T
((
0
1
)))
=


√
3 −1
1
√
3

 .
82
v) a)
E(X) =
∞∫
−∞
xf(x) dx =
2∫
0
x
2
(2− x) dx = 1
2
2∫
0
(
2x− x2) dx
=
1
2
(
x2 − 1
3
x3
∣∣∣∣2
0
)
=
2
3
.
Similarly
E(X2) =
∞∫
−∞
x2f(x) dx =
2∫
0
x2
2
(2− x) dx = 1
2
2∫
0
(
2x2 − x3) dx
=
1
2
(
2
3
x3 − 1
4
x4
∣∣∣∣2
0
)
=
2
3
.
Therefore
Var(X) = E(X2)− (E(X))2 = 2
3
−
(
2
3
)2
=
2
3
− 4
9
=
2
9
.
b) Note P (X ≤ m) =
m∫
0
f(x) dx.
1
2
=
1
2
m∫
0
(2− x) dx ⇒ 1 = 2m− 1
2
m2 ⇒ m2 − 4m+ 2 = 0
⇒ m = 2±
√
2 .
Since the median must lie in the interval [0, 2] the median value is m = 2−√2.
vi) a) For a set S of n vectors {v1,v2, . . . ,vn} to be linearly dependent there exists scalars
λ1, λ2, . . . , λn, not all of which are zero, such that
λ1v1 + λ2v2 + · · ·+ λnvn = 0 .
Note another way to say this is at least one of the vectors in S can be written as a
linear combination of the others.
b) The vectors v1,v2 and 0 are linearly dependent since the equation
0v1 + 0v2 + 10 = 0
is true with not all the λis equal to zero.
83
3. i) Every basis for P3(R) has 4 elements and the set S has five elements. Hence S is not a
basis for P3(R).
a)
b) The matrix A consists of the coordinate vectors in R4 for the polynomials in S and
the standard basis elements of P3(R). Taking the elements corresponding to leading
columns in the part of the row-echelon form matrix corresponding to polynomials
in S, the basis is B = {p1, p2, p4}.
c) The next leading column in the full row echelon form matrix is the one corresponding
to the coordinate vector b2, that this the polynomial e(x) = x. Thus a basis for R
4
can be formed by B′ = {p1, p2, p4, e}.
ii) a) x is in ker(A) if Ax = 0. This is solved using Gaussian elemination on A:
1 2 3 42 3 5 3
1 3 4 9

←→

1 2 3 40 −1 −1 −5
0 1 1 5

←→

1 2 3 40 −1 −1 −5
0 0 0 0


Let x = (x1, x2, x3, x4)
T . As the third and fourth columns are non-leading, let
x3 = λ1 and x4 = λ2. Then −x2 = x3 + 5x4 = λ1 + 5λ2 so x2 = −λ1 − 5λ2. From
the first row
x1 = −2x2 − 3x3 − 4x3 = 2λ1 + 10λ2 − 3λ1 − 4λ2 = −λ1 + 6λ2.
Thus 

x1
x2
x3
x4

 = λ1


−1
−1
1
0

+ λ2


6
−5
0
1

 .
Thus a basis for the kernel is
B =




−1
−1
1
0

 ,


6
−5
0
1



 .
b) As the basis has two elements, the nullity of A, which is the dimension of the kernel,
is 2. (Alternatively, the row-echelon form for A has two non-leading columns.)
iii) a) The characteristic equation for C is
p(λ) = det(C − λI) = (1− λ)(5− λ) + 4 = λ2 − 6λ+ 9 = (λ− 3)2.
Thus the only eigenvalue is λ = 3.
C − 3I =
(−2 2
−2 2
)
, so (C − 3I)b = 0 has solution b = t
(
1
1
)
. In particular,
(
1
1
)
is an eigenvector for the eigenvalue 3.
b) If C = MDM−1 for some invertible matrix M and diagonal matrix D then the
columns of M consist of linearly independent eigenvectors of C. Since C does not
have two linearly independent eigenvectors, no such M can exist and hence C is not
diagonalizable.
84
iv) a) n = 100 and p = 0.2
b) The probability of getting greater than or equal to 30 bottles containing less than
1.05 litres is
P (X ≤ 30) =
100∑
k=30
(
100
k
)
0.2k 0.8100−k .
c) X is approximately normally distributed with mean µ = np = 20 and variance
σ2 = np(1− p) = 16. Thus, if Z ∼ N(0, 1),
P (X ≥ 30) = P
(
X − µ
σ
≥ 30− 20
4
)
≈ P (Z ≥ 2.5)
= 1− P (Z ≤ 2.5)
= 1− 0.9938 = 0.0062.
d) As P (X ≥ 30) ≈ 0.0062 ≪ 0.05, there is strong evidence that the Manufacturer’s
claim is false.
4. i) a) Let ak =
2k + 1
3k
. Then
ak+1
ak
=
2k+1 + 1
3k+1
× 3
k
2k + 1
=
1
3
× 2 + 2
−k
1 + 2−k
→ 2
3
as k →∞.
By the ratio test, the series
∑∞
k=0 ak converges.
b) Let ak =
1√
k(k + 1)
. Then
• {ak} is decreasing,
• ak → 0 as k →∞.
By Leibniz’ test, the alternating series
∑∞
k=1(−1)kak converges.
ii) a) By replacing x in the given formula by x2 we obtain
sin(x2) =
∞∑
k=0
(−1)k x
4k+2
(2k + 1)!
.
b) By differentiating both sides of the given formula with respect to x, noting that
we can differentiate term by term of the power series on the right hand side in its
interval of convergnence, we obtain
cos x =
∞∑
k=0
(−1)k(2k + 1) x
2k
(2k + 1)!
=
∞∑
k=0
(−1)k x
2k
(2k)!
.
c) Letting x = π/4 in the above formula we deduce
1√
2
= cos
π
4
=
∞∑
k=0
(−1)k π
2k
42k(2k)!
=
∞∑
k=0
(−1)k π
2k
24k(2k)!
.
85
iii) a) By using L’Hoˆpital’s rule we have
lim
x→∞
lnx
x
= lim
x→∞
1
x
= 0.
Hence limn→∞ an = 0.
b) By using the above result and the continuity of the exponential function we have
bn = exp(ln(n
1/n)) = exp
(
lnn
n
)
→ exp(0) = 1.
iv) a) Equation (1) is separable, thus∫
dr
r(1− r) =
∫
k dt.
Partial fractions give
1
r(1− r) =
1
r
+
1
1− r ,
so that ∫ (
1
r
+
1
1− r
)
dr =
∫
k dt.
Hence
ln r − ln(1− r) = kt+ c,
or
ln
r
1− r = kt+ c,
so that
r
1− r = e
kt+c = βekt,
where β = ec > 0.
b) A) When t = 0 we have
β =
0.01
1− 0.01 =
1
99
.
When t = 5 (days) we have
1
99
e5k =
0.2
1− 0.2 =
1
4
.
Hence
k =
1
5
ln(99/4) = 0.6418.
B) At time t when r = 80 % there holds
ekt =
1
β
× 0.8
1− 0.8 = 4×
1
β
,
so that
kt = ln
4
β
.
Hence
t =
1
k
ln
4
β
= 9.32 days.
86
c) When t→∞
r
1− r →∞
so r → 1, i.e., the whole population have heard the rumour.
87
MATH1231 November 2014 Solutions
1. i) sin(A+B) = sinA cosB + cosA sinB and sin(A−B) = sinA cosB − cosA sinB.
Adding these we have sin(A+B) + sin(A−B) = 2 sinA cosB.
Put A = 5x,B = x we have
I1 =
∫
sin(5x) cos(x) dx =
∫
sin 6x+ sin 4x dx = − 1
12
cos(6x)− 1
8
cos(4x) + C.
ii) Let x = 2 sin θ then dxdθ = 2cos θ. Now x = 1⇒ θ = pi6 and x = 0⇒ θ = 0. Substituting
and simplifying gives
I2 = 4
∫ pi
6
0
sin2 θ dθ =
∫ pi
6
0
2− 2 cos θ dθ = π
3
−
√
3
2
.
iii) a) Using the integral test, we consider
∫ ∞
3
dx
x(lnx)3
dx. This converges to 12(ln 3)2 which
is finite and so the series converges by the integral test.
b) For n ≥ 1, 0 ≤ 2
2n + 3n
<
2
2n + 2n
=
1
2n
. Now
∞∑
n=1
1
2n
is a geometric series which
converges to 1 and so the given series converges by comparison test.
iv) a) Suppose x ∈ S and λ is a scalar, then x2+ y2 = z2. Consider the components of λx.
We have
(λx)2 + (λy)2 = λ2(x2 + y2) = (λz)2
so λx ∈ S.
b) S is not closed under addition since for example

02
2

 ∈ S and

20
2

 ∈ S but their
sum

22
4

 6∈ S, since 22 + 22 6= 42.
v) a) Set det(A− λI) = 0 giving λ2− 3λ+2 = 0, which has solutions λ = 1, 2. These are
the eigenvalues.
For λ = 1 the eigenspace is the kernel of A− I = ker
( −1 −1
2 2
)
= α
(
1
−1
)
so
we take
(
1
−1
)
as an eigenvector.
For λ = 2 the eigenspace is the kernel of A− 2I = ker
( −2 −1
2 1
)
= α
(
1
−2
)
so
we take
(
1
−2
)
as an eigenvector.
b) Tkae M =
(
1 1
−1 −2
)
and D =
(
1 0
0 2
)
.
88
vi) To find the kernel we solve Ax = 0. Using the output from the MAPLE session, we put
x4 = λ then x3 = 7λ, x2 = −4λ and x1 = −5λ. Hence ker(A) =

λ


−5
−4
7
1

 , λ ∈ R


and so




−5
−4
7
1



 is a basis for the kernel of A.
2. i) a) The integrating factor for the first order linear ODE
dy
dx
+
(
2 +
1
x
)
y =
2
x
, x > 0 is
given by
e
∫
(2+ 1
x
) dx = e2x+ln |x| = |x|e2x = xe2x since x > 0.
b) Multiplying the ODE by the integrating factor yields
d
dx
(
xe2xy
)
= 2e2x
⇒ xe2xy = e2x + C
⇒ y = 1
x
(
1 +Ce−2x
)
.
Applying the initial condition y(1) = 0 determines the value of C:
y(1) = 0 =
1
1
+
C
1
e−2 ⇒ C = −e2 .
Thus the solution to the first order ODE is y(x) =
1
x
(
1− e−2(x−1)).
ii) Homogeneous solution y′′ + 2y′ + 2y = 0.
Try yH = e
λx. Characteristic equation λ2 + 2λ + 2 = (λ + 1)2 + 1 = 0 ⇒ λ = −1 ± ı.
Therefore
yH(x) = Ae
−x cos x+Be−x sinx .
Particluar solution y′′ + 2y′ + 2y = 20e2x.
Try yP = Ce
2x. Thus
y′P = 2Ce
2x and y′′P = 4Ce
2x .
Substituting into the DE with yP , y
′
P and y
′′
P yields
y′′P + 2y
′
P + 2yP = 4Ce
2x + 4Ce2x + 2Ce2x = 10Ce2x
= 20e2x .
Therefore C = 2 and yP (x) = 2e
2x. The general solution yG(x) is the addition of the
homogeneous solution yH(x) and the particular solution yP (x), i.e.,
yG(x) = yH(x) + yP (x)
= Ae−x cos x+Be−x sinx+ 2e2x .
89
iii) Applying the Ratio Test to the power series
∞∑
n=1
nn(x− 1)n
n!
along with the Maple output
yields
lim
n→∞
∣∣∣∣an+1an
∣∣∣∣ = limn→∞
∣∣∣∣∣∣
(n+1)n+1(x−1)n+1
(n+1)!
nn(x−1)n
n!
∣∣∣∣∣∣ = e|x− 1| .
For the power series to be convergent e|x − 1| < 1. Thus the interval of convergence is
I =
(
1− e−1, 1 + e−1).
iv) a) To find scalars λ and µ such that u = λv1+µv2 we construct an augmented matrix
and use Gaussian elimination:
 1 2 11 3 −1
2 −1 12

−−−→

 1 2 10 1 −2
0 0 0

 .
Using back substitution, µ = −2 and λ = 5 and hence

 1−1
12

 = 5

11
2

− 2

 23
−1

.
b) [α)] Since T is a linear transformation then T (u) = −2T (v1) + 5T (v2).
But T (v1) = 2v1 and T (v2) = −v2 since v1 and v2 are eigenvectors of T with
eigenvalues 2 and −1 respectively. Thus T (u) = −2T (v1) + 5T (v2) = −4v1 − 5v2.
[β)] Consider T 2(u) = T (−4v1 − 5v2) = −4T (v1)− 5T (v2) = −8v1 + 5v2 and
T 3(u) = T (−8v1 + 5v2) = −8T (v1) + 5T (v2) = −16v1 − 5v2. The pattern for
positive integer n is
T n(u) = −2n+1v1 + (−1)n+1 (5v2) .
To formally prove this pattern you would use the Principle of Mathematical Induc-
tion.
v) a) If advertising has no effect on people’s preference then P (AppleAde) = 50% = 12
and P (BananAde) ≡ P (AppleAde)c = 12 . Hence the tail probability for 60 or more
people preferring AppleAde is given by the binomial distribution, i.e.,
P (X ≥ 60) =
100∑
k=60
(
100
k
)(
1
2
)k (1
2
)100−k
.
where X is the binomial random variable for the number of people who prefer
AppleAde.
b)
P (X ≥ 60) ≈ P (Y > 59.5) = P

Z > 59.5 − 50√
100× 12 × 12


= P (Z > 1.9)
= 1− P (Z < 1.9)
= 1− 0.971
= 0.029
90
where Y is the normal random variable approximating X and Z is the standard
normal variable.
Note the answer without the correction factor is P (Z > 2) = 0.023 or 2.3%.
c) There is a 2.90% chance a group of 60 or more people would prefer AppleAde
assuming no effect of advertising. This percentage is lower than our 5% threshold
and since 60 out of 100 DID prefer AppleAde we conclude there is some evidence
that the advertising campaign increased the percentage of the population who prefer
AppleAde.
3. i) Define
T (v) := 2a+ (b− c)x for all v = (a, b, c).
Suppose v,w ∈ R3 where
v := (a, b, c), w := (d, e, f).
We see that by applying T , we have
T (v+w) = 2(a+ d) + (b+ e− (c+ f))x = 2a+(b− c)x+2d+(e− f)x = T (v) +T (w).
Also, suppose λ ∈ R then
T (λv) = 2λa+ (λb− λc)x = λT (v).
This proves that T is a linear map.
ii) a) The three vectors from the given maple output are {v1,vi,v5} where i = 2, 3, 4, for
example: 

21
12
−6
−31
−12

 ,


11
20
−14
−29
−21

 ,


16
−17
−31
28
−21


b) The dimension of the span(S) is 3.
iii) a) The tree diagram is:
0.4
0.6
Y
O
F
C
P
F
P
C
0.25
0.15
0.6
0.45
0.2
0.35
91
From the tree diagram,
P (fulltime) = 0.4 × 0.25 + 0.6 × 0.45 = 37/100.
b) Using the conditional probability formula,
P (O|F ) = P (O ∩ F )
P (F )
.
Hence,
P (O|F ) = 27
37
.
iv) a) The sum of all the probabilities is 1:
1 = P (all events) = 0.3 + 4c+ 0.35 + 6c+ 0.05.
Hence, c = 0.03.
b) P (Y = 4) = P (X2 = 4) = 0.3 + 0.18 = 0.48.
v) a) The cumulative density function is:
F (x) :=


0 for x < 0
x3 for 0 ≤ x ≤ 1
1 for x > 1
b) The expected value for the continuous function is
E(X) =
∫ ∞
0
xf(x) dx =
∫ 1
0
3x3 dx =
3
4
.
c) Using
V ar(X) = E(X2)− [E(X)]2,
the variance is
V ar(X) = 3/5− 9/16 = 3/80.
vi) a) If α, β, γ ∈ R then the set B is linearly independent if and only if the only solution
to
αv1 + βv2 + γv3 = 0
is α = β = γ = 0.
b) Suppose
αv1 + βv2 + γv3 = 0.
Taking the dot product of both sides with v1 gives α|v1|2 = 0 and since v1 is not
the zero vector, we have α = 0. Repeating the process with v2 and v3, we find
β = γ = 0 also. Hence the vectors are linearly independent.
c) By the above, the 3 vectors are linearly independent and since R3 has dimension 3,
the three vectors also span R3 and so form a basis for R3.
92
4. i) a) S =
∫ h
0
2πy(x)
√
1 + y′(x)2 dx. Now y(x) =
r
h
x so
√
1 + y′(x)2 =
√
1 +
r2
h2
and so
S = 2π
∫ h
0
r
h
x
√
1 +
r2
h2
dx = 2π
r
h
√
1 +
r2
h2
∫ h
0
x dx = 2π
r
h
√
1 +
r2
h2
h2
2
= πr
√
r2 + h2.
b)
∂S
∂r
= π
√
r2 + h2 + πr · 2r
2
√
r2 + h2
=
π(2r2 + h2)√
r2 + h2
.
∂S
∂h
= πr · 2h
2
√
r2 + h2
=
πrh√
r2 + h2
.
c) Using the total differential approximation, the maximum absolute error is
|∆S| ≈
∣∣∣∣∂S∂r∆r + ∂S∂h∆h
∣∣∣∣
=
∣∣∣∣π(2r2 + h2)√r2 + h2 ∆r + πrh√r2 + h2∆h
∣∣∣∣
=
∣∣∣∣34π5 ∆r + 12π5 ∆h
∣∣∣∣
≤ 34π
5
(0.05) +
12π
5
(0.05)
=
23π
50
≈ 1.45cm2.
ii) a) This equation is separable:
dP
P − 10 = −k dt, so∫
dP
P − 10 = ln |P − 10| = −
∫
k dt = −kt+ C.
Thus (if P > 10)
P = 10 + exp(−kt+ C) = 10 + ce−kt
for some positive constant c.
Now P (0) = 20 = 10 + c, so c = 10. Also P (1) = 18 so
18 = 10 + 10e−k
so
−k = ln 8
10
= ln
4
5
or k = ln
5
4
.
b) 6am corresponds to t = 8.
P (8) = 10 + 10e−8 ln(5/4) = 11.68◦C.
iii) a) For all n, n1/n ≥ 1 and so 1
2
≤ n
1/n
2
. Each term an is equal to either the lower value
or the upper value.
93
b) Let bn = lnn
1/n =
lnn
n
. By L’Hoˆpital’s Rule (formally applied to f(x) =
lnx
x
,
x ≥ 1)
lim
n→∞ bn = limn→∞
lnn
n
= lim
n→∞
1/n
1
= 0
and so, by the continuity of the exponential function,
lim
n→∞n
1/n = lim
n→∞ e
bn = exp( lim
n→∞ bn) = e
0 = 1.
c) From (a),
1
2
≤ an ≤ n
1/n
2
. Now, from (b)
lim
n→∞
1
2
=
1
2
and lim
n→∞
n1/n
2
=
1
2
.
The Pinching Theorem then implies that lim
n→∞ an =
1
2
too.
d) If
∑
xn converges then xn → 0 as n → ∞ (the n-th term test). Here (−1)nan
doesn’t converge at all and hence
∑
(−1)nan doesn’t converge.
iv) a) The second Taylor polynomial about 0 is
p2(x) = f(0) + f
′(0)x+
f ′′(0)
2!
x2
= x− x2
from the given data.
b) The remainder is given by
R3(x) = f(x)− p2(x) = f
′′′(c)
3!
x3
for some c between 0 and x. In particular, if x = 12 then, using the bound on f
′′′,
|R3(1/2)| ≤ 6
3!
(
1
2
)3
=
1
8
.
Now p2(1/2) =
1
4 so this implies that
−1
8
≤ f(1/2) − p2(1/2) = f(1/2) − 1
4
≤ 1
8
or
1
8
≤ f(1/2) ≤ 3
8
c) For any x > 0,
x− f(x)
x2
=
x− (p2(x) +R2(x))
x2
=
x2 +R3(x)
x2
= 1 +
f ′′′(cx)x3
6x2
= 1 +
f ′′′(cx)
6
x.
Since |f ′′′(cx)/6| ≤ 1, the Pinching Theorem tells us that x− f(x)
x2
→ 1 as x→ 0+.
94
PAST HIGHER EXAM SOLUTIONS
95
MATH1241 November 2010 Solutions
2. i) Let P = (1, 2, 6). Then
Fx = −50 x(x2+y2)2 = −2 at P
Fy = −50 y(x2+y2)2 = −4 at P
The equation of the tangent plane is given by z = 6+Fx(x− 1)+Fy(y− 2) = 6− 2(x−
1)− 4(y − 2) so the desired equation is 2x+ 4y + z = 16.
ii) Using the chain rule,
∂u
∂t
= f ′(u+ t
∂u
∂t
)⇒ ∂u
∂t
(1− tf ′) = uf ′.
Also
∂u
∂x
= f ′(1 + t
∂u
∂x
)⇒ ∂u
∂x
(1− tf ′) = f ′.
Hence ∂u∂t = u
∂u
∂x .
iii) Using the method of partial fractions,
6x2 − 11x+ 9
(x− 1)(x2 − 2x+ 2) =
A
x− 1 +
Bx+ C
x2 − 2x+ 2 .
⇒ A(x2 − 2x+ 2) + (Bx+ C)(x− 1) ≡ 6x2 − 11x+ 9.
Putting x = 1, gives A = 4; equating the coefficients of x2 gives B = 2 and finally by
considering the constant terms we have C = −1.
Hence
I =
∫
4
x− 1 +
2x− 1
x2 − 2x+ 2 dx =
∫
4
x− 1 +
2x− 2
x2 − 2x+ 2 +
1
(x− 1)2 + 1 dx
= 4 ln |x− 1|+ ln(x2 − 2x+ 2) + tan−1(x− 1) + C.
iv) The zero polynomial satifies the differential equation and so belongs to S. Now suppose
p, q ∈ S and consider the polynomial λp+ µq.
(x− 3)(λp + µq)′ − 2(λp + µq) = (x− 3)λp′ + (x− 3)µq′ − 2λp− 2µq
= λ[(x− 3)p′ − 2p] + µ[(x− 3)q′ − 2q] = 0 + 0 = 0
hence λp+ µq ∈ S and so S is a subspace of P3(R)
v) Let I be the event ‘a particular person is injured’, and S be the event ‘a particular
person survives 10 years’. We draw a weighted tree diagram as follows.
96
I
S
Sc
Sc
Ic
0.8
0.6
0.4
0.2
0.4
0.6
b
b
b
b
b
b
b
We want P (Ic|S) = P (Ic∩S)P (S) = 0.160.48+0.16 = 25%.
vi) a) since G is a basis,
x = λ1v1 + ...+ λnvn.
Now for i = 1, 2, ..., n, by orthogonality, xvi = λivi.vi so λi =
x.vi
vi.vi
. Hence
x =
x.v1
v1.v1
v1 + ...+
x.vn
vn.vn
vn.
b) Since T is linear,
T (x) =
n∑
i=1
x.vi
vi.vi
T (vi)
=
n∑
i=1
x.vi
vi.vi
γi(vi.vi)
=
n∑
i=1
γi(x.vi).
3. i) a) Set det(A− λI) = 0, so ∣∣∣∣
(
5− λ −8
1 −1− λ
)∣∣∣∣ = 0.
Expanding out this gives λ2 − 4λ + 3 = 0 with roots λ = 3, 1. These are the
eigenvalues.
For λ = 3, the kernel of (A− 3I) = ker
(
2 −8
1 4
)
= span
{(
4
1
)}
For λ = 1, the kernel of (A− I) = ker
(
4 −8
1 −2
)
= span
{(
2
1
)}
.
Thus
{(
4
1
)}
and
{(
2
1
)}
are the corresponding eigenvectors.
b) Put P =
(
4 2
1 1
)
and D =
(
3 0
0 1
)
, then A = PDP−1.
97
c) Hence, A8 = PD8P−1 so A8P = PD8 =
(
4 2
1 1
)(
38 0
0 1
)
=
(
4× 38 2
38 1
)
.
ii) a) p = 103365 .
b) binomial(10, p, 5) =
(
10
5
)
p5(1− p)5 ≈ 8.59%
c) geometric(1 − p, 5) = (1− (1− p))4(1− p) = p4(1− p) = 0.0455%
iii) a) For f, g ∈ S and λ, µ ∈ R,
T (λf + µg) = (λf + µg)′ − 2(λf + µg)
= λ(f ′ − 2f) + µ(g′ − 2g) = λT (f) + µT (g)
and so T is linear.
b) T (e−x) = −3e−x, T (sinx) = −2 sinx + cos x, and T (cos x) = − sinx − 2 cos x.
Hence the matrix representing the linear map with respect to the basis B is
C =

−3 0 00 −2 1
0 −1 −2


c) By inspection the rank of C is 3.
d) Since the matrix C is invertible, so T is a bijection and has an inverse. Hence
y′ − 2y = g has a unique inverse solution (y = T−1(g)).
e) The coordinate vector of e−x with respect to the basis above is

10
0

 and C−1 =
−13

1 0 00 ∗ ∗
0 ∗ ∗

 so C−1

10
0

 =

−1/30
0

. Hence the solution is y = −13e−x. (Note:
We have only found the entries in C−1 that were needed to perform the desired
calculation.)
4. i) The equation can be written as
dy
dx
=
3x2 − 5xy + 2y2
x2 − xy
and so is a homogeneous equation. We put y = xu(x). This gives
u+ xu′ =
3x2 − 5x2u+ 2u2x2
x2 − x2u =
3− 5u+ 2u2
1− u .
This equation is now separable. After a little algebra we can separate the variables and
obtain ∫
1− u
3u2 − 6u+ 3 du =
∫
dx
x
−1
3
∫
1
u− 1 du =
∫
dx
x
98
ln |u− 1| = ln |x−3|+K
Exponentiating both sides and replacing u with yx we obtain the solution, y = x +
A
x2
,
where A is a constant.
ii) a) 0 ≤ k
ln k!
<
k
k ln k − k + 1 =
1
ln k − 1 + 1k
→ 0 as k → ∞. Hence the limit is 0 by
the pinching theorem.
b) Note the k! = k(k − 1)...1 < k.k.....k︸ ︷︷ ︸
k times
= kk.
Hence
k
ln k!
>
k
ln kk
=
1
ln k
. So,
∞∑
k=2
k
ln k!
>
∞∑
k=2
1
ln k
>
∞∑
k=2
1
k
and the latter series diverges. Hence
∞∑
k=2
ak diverges by comparison.
iii) With an =
1
2n
√
n
, the radius of convergence is given by
R = lim
n→∞
∣∣∣∣ anan+1
∣∣∣∣ = limn→∞ 2n+1
√
n+ 1
2n
√
n
= 2
iv) a)
f(x) = sinhx f(0) = 0
f ′(x) = coshx f ′(0) = 1
f ′′(x) = sinhx f ′′(0) = 0
f ′′′(x) = coshx f ′′′(0) = 1
and the pattern repeats.
Hence the Taylor polynomial is
P4(x) = x+
1
3!
x3 +
1
5!
x5.
The remainder term is given by R6(x) =
sinh c
6! x
6.
b) ∫ x
0
sinh t
t
dt =
∫ x
0
t+ 13! t
3 + 15!t
5
t
dt
=
∫ x
0
1 +
1
3!
t2 +
1
5!
t4 dt = x+
x3
18
+
x5
600
.
v) The surface area formula is given by
S =
∫
2πy
√(
dx
dt
)2
+
(
dy
dt
)2
dt.
Substitution yields,
S = 2πb
∫ pi
0
sin t
√
a2 sin2 t+ b2 cos2 t dt
99
= 2πb
∫ pi
0
sin t
√
a2 + (b2 − a2) cos2 t dt.
We can make the change of variable u = cos t to give
S = 2πb
∫ 1
−1
√
a2 + (b2 − a2)u2 du
= 2πb
√
b2 − a2
∫ 1
−1
√
u2 +
a2
b2 − a2 du.
This integral is hard to do and wise students moved on to the next question at this stage.
For completeness of the solutions, I continue.
The integral J =
∫ √
u2 + k2 du can be done as follows: Put u = k sinh θ. This gives
J =
∫
k2 cosh2 θ dθ =
k2
2
∫
1 + cosh 2θ dθ
=
k2
2
(
θ +
1
2
sinh 2θ
)
=
k2
2
(
sinh−1
u
k
+
u
k
√
1 +
(u
k
)2)
using sinh 2θ = 2 sinh θ cosh θ.
Hence ∫ 1
−1
√
u2 + k2 du = k2 sinh−1
1
k
+
√
k2 + 1.
Using this result, with k2 = a
2
b2−a2 , and noting that 1 + k
2 = b
2
b2−a2 , we have
S = 2πb
√
b2 − a2
(
a2
b2 − a2 sinh
−1
√
b2 − a2
a
+
b√
b2 − a2
)
= 2πb
(
a2√
b2 − a2 sinh
−1
√
b2 − a2
a
+ b
)
.
vi) The volume of liquid in the tank at time t is V = 40+2t, hence the ratio of salt to liquid
in the tank at time t is x(t)40+2t . The inflow rate of salt is 6gm/min.
The outflow rate of salt is x(t)40+2t × 1.
Hence the differential equation that models this situation is given by:
dx
dt
= inflow rate of salt− outflow rate of salt,
dx
dt
= 6− x(t)
40 + 2t
.
100
MATH1241 November 2011 Solutions
2. i) a)
fy(x0, y0) = lim
h→0
f(x0, y0 + h)− f(x0, y0)
h
whenever the limit exists
b) Using the above formula, we see that
fy(0, 0) = lim
h→0
f(0, 0 + h)− f(0, 0)
h
= lim
h→0
0
h
= 0.
Alternatively, noticing that f(0, y) = 0 then it follows that fy(0, 0) = 0.
ii) a) Total resistance, 1z0 =
1
6 +
1
12 =
1
4 , so z0 = 4.
b) Differentiating both sides of
1
z
=
1
x
+
1
y
with respect to x yields,
−1
z2
dz
dx
=
−1
x2
and from here by rearranging, the desired result follows.
c) Let x0 = 6, y0 = 12. To compute the maximum absolute error in the calculated
value of the total resistance z; |∆z|, consider
∆z ≈ ∂z
∂x
∆x+
∂z
∂y
∆y.
By taking absolute values and using the triangle inequality, we have
|∆z| ≤ |∂z
∂x
∆x|+ |∂z
∂y
∆y|.
Since |∆x| ≤ 0.1 and |∆y| ≤ 0.1; we have
|∆z| ≤ | z
2
0
x20
|0.1 + |z
2
0
y20
|0.1 = 1
10
(
16
36
+
16
144
) = 0.0556.
iii) a) Since there are 3 vectors (same dimension as R3), it suffices to determine if the
vectors are linearly independent. To save time, consider
α1v1 + α2v2 + α3v3 = w.
This can be written as the augmented matrix;
 1 2 7 a−1 1 2 b
−3 2 3 c

 .
By row-reducing this matrix, we have
 1 2 7 a0 3 9 a+ b
0 0 0 a− 8b+ 3c

 .
101
If w = 0 then this implies that the vectors; v1, v2, v3 are linearly dependent and thus
{v1, v2, v3} does not span R3. Alternatively, at this stage; we see that there are only
2 ”leading” columns resulting in the same conclusion.
The condition such that w belongs to the span(v1, v2, v3) is a− 8b+ 3c = 0.
b) No, since 5− 48 + 0 = −43 6= 0.
iv) a) Let CR be the event that a student gets a CR or higher,
P (CR) = 0.65 × 0.45 + 0.35 × 0.30 = 0.3975.
Thus,
P (CRc) = 0.6025.
b)
P (A|CR) = P (A ∩CR)
P (CR)
=
0.65 × 0.45
0.3975
= 0.736.
Thus as a percentage, we have the final answer to be 73.6
3. i) a) The differential equation is homogeneous since
dy
dx
=
xy − y2
x2
=
y
x
−
(y
x
)2
meaning we can use the substitution
y = xu,
dy
dx
= u+ x
du
dx
.
The resulting differential equation is separable:
u+ x
du
dx
= u− u2
∫
− 1
u2
du =
∫
1
x
dx
and has solution
1
u
= ln |x|+ c. Hence
x
y
= ln |x|+ c
y =
x
ln |x|+ c .
b) If (x, y) = (1, 1), then 1 =
1
0 + c
and c = 1. Hence the particular solution is
y =
x
ln |x|+ 1 .
102
ii) a)
b) We use the comparison test to show convergence. First, we use a comparison to
eliminate the factorial. Since k! < kk,
ln k!
k3
<
ln kk
k3
=
k ln k
k3
=
ln k
k2
.
Now we use a comparison to elimate the logarithm. For any α > 0, ln k < kα when
k is sufficiently large. For our purposes, choose α = 1/2, so for k sufficiently large,
ln k
k2
<
k1/2
k2
=
1
k3/2
.
Finally, since
∑ 1
k3/2
converges,
∞∑
k=2
ln k!
k3
converges via the comparison test.
c) First, we show the alternating series converges using the Leibniz Test. To see that
the sequence converges to zero, notice that for k sufficiently large,
ak =
ln k√
k2 + 1
<
k1/2√
k2 + 1
<
k1/2√
k2
=
1
k1/2
so via the Pinching Theorem, limk→∞ ak = 0. Next, we need the sequence to be
(eventually) decreasing. Let
f(x) =
ln k√
k2 + 1
f ′(x) =
1
k
√
k2 + 1− ln k k√
k2+1
k2 + 1
=
1− k2(ln k − 1)
k(k2 + 1)3/2
which is negative for k sufficiently large (k ≥ 4), which shows the sequence is
(eventually) decreasing. Hence,
∞∑
k=2
(−1)k ln k√
k2 + 1
converges by the Leibniz Test. [Note,
the fact that the sequence might not be initially decreasing does not prohibit us from
using the Leibniz Test since convergence depends only on the tail of the series and
not a finite number of initial terms.]
We now show the convergence is not absolute using the comparison test. For k ≥ 3,
ln k√
k2 + 1
>
1√
k2 + 1
>
1√
k2 + 3k2
=
1
2k
.
Since
∑ 1
2k
diverges, the series
∞∑
k=2
ln k√
k2 + 1
also diverges. We conlude that
∞∑
k=2
(−1)k ln k√
k2 + 1
converges conditionally.
iii) a) We use the Maclaurin series for ex:
ex =
∞∑
k=0
xk
k!
= 1 + x+
x2
2!
+
x3
3!
+
x4
4!
· · ·
ex
2
=
∞∑
k=0
(x2)k
k!
=
∞∑
k=0
x2k
k!
= 1 + x2 +
x4
2!
+
x6
3!
+
x8
4!
· · ·
103
b) The ODE is separable:
∫
1
y
dy =
∫
ex
2
dx =
∫ ( ∞∑
k=0
x2k
k!
)
dx.
The Maclaurin series can be integrated term by term:
∫ ( ∞∑
k=0
x2k
k!
)
dx =
∞∑
k=0
(∫
x2k
k!
dx
)
=
∞∑
k=0
x2k+1
(2k + 1)k!
+ c.
Hence, the general solution is ln |y| =
∞∑
k=0
x2k+1
(2k + 1)k!
+ c. By letting x = 0 and
y = 1, we get 0 = 0 + c so the particular solution is ln |y| =
∞∑
k=0
x2k+1
(2k + 1)k!
.
c) We use the ratio test to see when the series is convergent:
lim
k→∞
[
|x|2(k+1)+1
(2(k + 1) + 1)(k + 1)!
× (2k + 1)k!|x|2k+1
]
= lim
k→∞
(
|x|2 × 2k + 1
2k + 3
× 1
k
)
= |x|2 ×
(
lim
k→∞
2k + 1
2k + 3
)
×
(
lim
k→∞
1
k
)
= |x|2 × 1× 0
= 0
which is less than 1 for every x. Hence the solution is defined for all real x.
iv) a) We proceed by induction to show ak+1 > ak. The base case is k = 1:
a2 =
√
1 + a1 =
√
2 > 1 = a1.
Now assume the hypothesis for k = n: i.e. an+1 > an. We will verify it for k = n+1:
an+2 =
√
1 + an+1 >
√
1 + an = an+1.
Hence, ak+1 > ak is true for all k ≥ 1.
b) We show ak < 3 by induction. The base case is k = 1, and a1 = 1 < 3. Now assume
an < 3. Then an+1 =
√
1 + an <
√
1 + 3 = 2 < 3. Hence {ak}∞k=1 is bounded above
by 3.
c) Since the sequence is increasing and bounded above, it must converge: i.e. there is
a number L such that limk→∞ ak = L.
d) Since the sequence converges, ak+1 =
√
1 + ak ≈ ak when k is large. Thus we
solve for L in the equation L =
√
1 + L. Equivalently, it is the positive solution of
L2 − L− 1 = 0. Hence L = 1+
√
5
2 , which is the Golden Ratio.
104
4. i) a) The eigenvalues of A are the roots of the characteristic polynomial
pA(x) = det(A− xI) =
∣∣∣22− x −100
5 −23− x
∣∣∣
= (22− x)(−23 − x)− (−100)5
= x2 + x− 6
= (x− 2)(x + 3) ,
that is, λ = −3, 2.
Let us now find the eigenvectors corresponding to λ = −3;
these are the non-zero vectors in ker(A+ 3I):(
25 −100
5 −20
)
←→
(
1 −4
0 0
)
Thus, the eigenvectors corresponding to λ = −3 are t
(
4
1
)
for all t 6= 0.
Let us now find the eigenvectors corresponding to λ = 2;
these are the non-zero vectors in ker(A− 2I):(
20 −100
5 −25
)
↔
(
1 −5
0 0
)
Thus, the eigenvectors corresponding to λ = 2 are t
(
5
1
)
for all t 6= 0.
b) D =
(−3 0
0 2
)
and P =
(
4 5
1 1
)
, or D =
(
2 0
0 −3
)
and P =
(
5 4
1 1
)
,
or these matrices with columns scaled by non-zero constants.
c)
AnP = PDn =
(
4 5
1 1
)(−3 0
0 2
)n
=
(
4 5
1 1
)(
(−3)n 0
0 2n
)
=
(
4(−3)n 5× 2n
(−3)n 2n
)
ii) a) Since
1 =
4∑
k=0
pk =
4∑
k=0
c
2k
=
(
1 +
1
2
+
1
4
+
1
8
+
1
16
)
c =
31
16
c ,
we see that c =
16
31
.
b) P (X = 2) =
16
31× 22 =
4
31
.
c)
P ((X − 2)2 < 4) = P (X = 1) + P (X = 2) + P (X = 3)
=
16
31 × 2 +
16
31× 4 +
16
31 × 8
=
14
31
.
105
iii) a) Suppose that non-zero matrix C ∈M22 satisfies CT = λC.
Then CT = λC = λ(CT )T = λ(λC)T = λ2CT .
Since CT is non-zero, λ2 = 1, so λ = ±1.
b) Since
T (v1) = v1 T (v2) = v3 T (v3) = v2 T (v4) = v4
we see that
B =


1 0 0 0
0 0 1 0
0 1 0 0
0 0 0 1


c) rankT = rankB = 4.
d) Let λ be an eigenvalue of T .
Then CT = T (C) = λC for some matrix C ∈M22, so by part a), λ = ±1.
Conversely, λ = −1 and λ = 1 are both eigenvalues of T :
T (I) = I = 1× I and T
(
0 1
−1 0
)
=
(
0 −1
1 0
)
= (−1)×
(
0 1
−1 0
)
.
[Part b) could also have been used here.]
e) A basis for the eigenspace of T corresponding to λ = −1 is
{(
0 1
−1 0
)}
.
A basis for the eigenspace of T corresponding to λ = 1 is{(
1 0
0 0
)
,
(
0 0
0 1
)
,
(
0 1
1 0
)}
.
106
MATH1241 November 2012 Solutions
2. i)
∂u
∂x
= f ′(s)
1
y
,
∂u
∂y
= −f ′(s) x
y2
.
ii)
∂u
∂x
=
2x
x2 + y2
,
∂u
∂y
=
2y
x2 + y2
,
∂2u
∂x2
=
2(y2 − x2)
x2 + y2
,
∂2u
∂y2
=
2(x2 − y2)
x2 + y2
,
∂2u
∂x2
=
∂2u
∂y2
.
iii) a) area A =
=
∫ a/2
−a/2
2b
a
√
a2 − x2dx (15)
=
2b
a
∫ 2pi/3
pi/2
a2 sin2(θ)dθ, x = a cos(θ) (16)
= 2ab
∫ 2pi/3
pi/3
1− cos(2θ)
2
dθ = ab
(
π
3
+
√
3
2
)
(17)
b) ∆A = ∂A∂a∆a+
∂A
∂b ∆b.
∆A
A =
∆a
a +
∆b
b = 0.04 − 0.01 = 3%.
iv) a) 

1 2 0 5 a
2 3 1 7 b
1 4 −2 11 c
3 1 5 0 d

 ∼


1 2 0 5 a
0 1 −1 3 2a− b
0 0 0 0 5a− 2b− c
0 0 0 0 7a− 5b+ d

 (18)
b) 5a− 2b− c = 0 and 7a− 5b+ d = 0
c) a = (−3a+ 2b+ γ)v1 + (2a− b− 3γ)v2 + γv3 for some parameter γ.
d) A hyper plane in R4
v) a)
P (4− 2T ≥ 0) = P (T ≤ 2) =
∫ 2
−∞
f(x, λ)dx = [− exp(−λx)]x=2x=0 = 1− exp(−2λ)
(19)
b)
E(T 2) =
∫ 0
−∞
x2λ exp(−λx)dx = lim
R→∞
[−x2 exp(λx)]x=R
x=0
−
∫ ∞
0
(−2x) exp(−λx)dx
(20)
=
2
λ2
(21)
107
3. i) a) Let X be the number of times a 6 is rolled out of 300 rolls. Assuming the die is
unbiased, X ∼ Bin(300, 16) and the tail probability that measures the chance of
getting a 6 at least 68 times is P (X ≥ 68).
b) We have E(X) = 600 · 16 = 50 and Var(X) = 600 · 16 · 56 = 1253 and so X
approx.∼
N
(
50, 1253
)
. Thus
P (X > 67.5) = P

Z > 67.5− 50√
125
3


= P (Z > 2.71)
= 1− 0.9966
= 0.0026
c) Yes, since 0.0026 < 0.05.
ii) a) P (stop after 13-th roll)= P (don’t stop after 6 rolls)× P (not roll a six on 7-th
roll)×P (roll six 6’s in a row) =
(
1− (16)6) · (56) (16)6.
b) P (roll at least ten times) = 1− P (stop after 6,7,8 or 9 times). Now
P (stop after 6 rolls) =
(
1
6
)6
P (stop after 7(or 8 or 9) rolls) = P (don’t roll 6 on 1st (2nd or 3rd) roll )
× P ( roll six 6’s in a row)
=
(
5
6
)
·
(
1
6
)6
P (roll at least ten times) = 1−
((
1
6
)6
+
(
5
6
) · (16)6 · 3)
iii) a) Note: column space and image mean the same thing, when speaking about matrices.
v ∈ col(AAT ). That is, there exists w ∈ R2 such that v = AATw. Let u = ATw.
Then v = Au and so we see that v ∈ col(A), which implies col(AAT) ⊆ col(A).
Since rank(A) = dim col(A) we get rank(AAT ) ≤ rank(A).
b) Note that AAT is a 3 × 3 matrix. Suppose AAT = I then rank(AAT ) = 3. How-
ever, since A is 3 × 2, rank(A) ≤ 2 and so (a) implies rank(AAT ) ≤ 2. This is a
contradiction, and so AAT 6= I.
iv) a) If MMT = I then det(MMT ) = 1. Since detMT = detM and det(MMT ) =
detM · detMT we get (det M)2 = 1 and so det M = ±1.
b) Note that mi = Mei and recall that if u,v ∈ Rn then v.u = vTu. So, mi.mj =
Mei.Mej = (Mei)
TMej = e
T
i M
TMej = e
T
i Iej = e
T
i ej = ei.ej which is zero if
i 6= j and one, otherwise.
c) (Av).w = (Av)Tw = vTATw = v.(ATw).
d) |Mv|2 =Mv.Mv = v.MTMv = v.Iv = v.v = |v|2. Thus |Mv| = |v|.
e) Since M is 3 × 3 the characteristic polynomial of M is real and of degree 3. Any
real polynomial of degree 3 must have at least one real root and thus M must have
at least one real eigenvalue. Furthermore a real eigenvalue corresponds to a real
eigenvector and vice versa.
108
f) If u is a real eigenvector then it corresponds to a real eigenvalue λ. We have
Mu = λu and so |Mu| = |λ||u|. However, part (d) implies |Mu| = |u| and so
|λ||u| = |u|. Since λ is real, we conclude that λ = 1 or −1.
4. i) a) The characteristic equation is
λ2 + 6λ+ 9 = 0 ⇐⇒ (λ+ 3)2 = 0 ⇐⇒ λ = −3,−3.
So the general solution takes the form y = Ae−3x +Bxe−3x. Substituting y(0) = 0
yields A = 0. Hence the general solution is
y = Bxe−3x.
b)
dy
dx
= Be−3x − 3Bxe−3x = Be−3x(1− 3x) = 0 ⇐⇒ x = 1
3
ii) a)
dy
dx
∣∣∣∣
(2a,a)
=
2a · a
(2a)2 − a2 =
2
3
Thus the gradient of normal at (2a, a) is −3/2. The equation of normal is therefore
y − a = −3
2
(x− 2a) ⇐⇒ y = −3
2
x+ 4a.
b) This is a separable equation.∫
1
y
dy =
∫
x
x2 − a2 dx ⇐⇒ ln |y| =
1
2
ln |x2 − a2|+ c
⇐⇒ y2 = A|x2 − a2|, A ≥ 0.
When |x| > a, we obtain y2 = A(x2 − a2), A ≥ 0, which is a hyperbola. When
|x| < a, we obtain y2 = −A(x2 − a2), A ≥ 0, which is an ellipse.
iii) a) Sketch: f(0) = 0, f(9) = −1/2, f(10) = 10. f(x)→ −∞ as x→ (√99)−, f(x)→∞
as x→ (√99)+, and f(x)→ 0+ as x→∞.
b)
∞∑
n=0
(−1)nn
n2 − 99 =
9∑
n=0
(−1)nn
n2 − 99 +
∞∑
n=10
(−1)nn
n2 − 99
For n ≥ 10, we see from the previous sketch that an = n/(n2−99) is decreasing and
bounded below by 0. By Leibniz’s alternating series test,
∑∞
n=10
(−1)nn
n2−99 converges,
and so does
∑∞
n=0
(−1)nn
n2−99 .
iv) a)
f(t) =
∞∑
k=1
(−1)k+1t2k
(2k)!
109
b)
Cin(x) =
∞∑
k=1
(−1)k+1 ∫ x0 t2k−1 dt
(2k)!
=
∞∑
k=1
(−1)k+1x2k
(2k)! (2k)
The series converges for all x.
v) a) Note that
x(1− x) = −
(
x− 1
2
)2
+
1
4
<
1
4
for x 6= 1
2
.
We have a1 = 1/4 < 1/2. Suppose that ak < 1/2 for some k ≥ 1. Then
ak+1 = 2ak(1− ak) < 2 · 1
4
=
1
2
.
b) Using an < 1/2, we obtain
an+1 − an = 2an(1− an)− an = an − 2a2n = an(1− 2an) > an · 0 = 0.
Hence the sequence is strictly increasing
c) Since the sequence is strictly increasing and bounded above, by the monotone con-
vergence theorem it converges.
d) Suppose ℓ = limn→∞ an. Solving ℓ = 2ℓ(1 − ℓ) gives ℓ = 0, 1/2. Since the limit
cannot be 0, we conclude that limn→∞ an = 1/2.
110
MATH1241 November 2013 Solutions
3. i) a) The variance of the exponential distribution is 1λ2 . Hence Var(8−2T ) = 22Var(T ) =
4
λ2 .
b) P (λT ≥ 1) = P (T ≥ 1λ) = 1− P (T < 1λ ) = 1− (1− e−1) = 1e .
ii) Let X = the number of defective bulbs in a box. Then E(X) = 2 and Var(X) =
2 × 0.98 = 1.96. We can approximate the binomial by X ∼ B(100, 0.02) by the normal
X ∼ N(2, 1.96). So
P (X > 3.5) = P (Z >
3.5− 2√
1.96
) = P (Z > 1.07) = 1− P (Z ≤ 1.07) ≈ 0.1423.
iii) a) There are a red balls out of a total of a+ b balls.
b) P (R2) =
a
a+b × a+c−1a+b+c−1 + ba+b × aa+b+c−1 = aa+b = P (R1).
iv) a) One can either give an inductive proof or perhaps write
Ak = A....A︸ ︷︷ ︸
k times
=M−1DM ×M−1DM...M−1DM︸ ︷︷ ︸
k times
=M−1DkMk.
b) Best shown by induction.
c) By (ii) the general entry in the matrix sum Dn + an−1Dn−1 + ... + a0I has the
form λni + anλ
n−1
i + ... + a0, which is zero, since λi is a root of the characteristic
polynomial. The result follows.
d) P (A) =M−1P (D)M = 0).
e) The characteristic polynomial is x2−x− 6 and so by (iii), A2−A+6I = 0 and the
result follows.
4. i) Setting y = xv, we get
x2
d
dx
(xv) = 4x2 + x(xv) + (xv)2.
By the product rule,
x2(v + x
dv
dx
) = 4x2 + x2v + x2v2,
or
x
dv
dx
= 4 + v2
(assuming x 6= 0). Then separating variables, we have∫
1
x
d =
∫
1
4 + v2
dv.
Integrating, we have
1
2
tan−1(
v
2
) = ln|x|+ C,
or
y = 2xtan(2ln|x|+ C).
111
ii) By the ratio test, the radius of convergence is
lim
n→∞
2nn2
n3+1
2n+1(n+1)2
(n+1)3+1
=
1
2
,
so the power series converges converges on the open interval
{|x| : |x− 1| < 1
2
} = (1
2
,
3
2
).
At the endpoints, the series becomes
∞∑
n=1
(∓1)n n
2
n3 + 1
,
which converges in the “−” case by the alternating series test, and diverges in the “+”
case by the integral test. Therefore the interval of convergence is [12 ,
3
2).
iii) a) (We use “X” units in g and “t” units in hours.)
(a)
Given k1 = k2 = 2, we have after separating variables∫
(2−X)−2dx =
∫
kdt.
Integrating,
(2−X)−1 = kt+ C.
The initial condition X(0) = 0 gives c = 12 , and then X(1) = (1) gives k =
1
2 .
Finally, setting X = 1.8 gives t = 9 as desired.
b) Using k = 12 from part (a) and k1 = 20, k2 = 2, we have after separating variables∫
(2−X)−1(20−X)−1dx =
∫
1
2
dt.
Integrating, we get
1
18
ln(
20 −X
2−X ) =
1
2
t+ C.
The initial condition X(0) = 0 gives C = ln(10)18 , and then setting X = 1.8 gives
t = 19 ln(
91
10) ≈ .25.
iv) a) Since f is nondecreasing, we have ck+1 = f(ck) ≥ ck for all k; therefore the sequence
is monotone. Since f([a, b]) ⊆ [a, b], we have a ≤ ck ≤ b for all k (by induction),
so the sequence is bounded. Therefore the sequence converges by the monotone
convergence theorem.
b) We have
f(L) = f( lim
k→∞
ck) = lim
k→∞
f(ck) = lim
k→∞
ck+1 = L,
where interchanging the function and the limit is allowed because f is continuous.
112
newpage
MATH1241 November 2014 Solutions
3. i) For q, h ∈ P(R) and λ, µ ∈ R,
T (λq + µh) =
(
λq(0) + µh(0)
λq(1) + µh(1)
)
= λ
(
q(0)
q(1)
)
+ µ
(
h(0)
h(1)
)
= λT (q) + µT (h).
ii) a) N2 is the 2× 2 zero matrix.
b) By (a), eN = I +N =
(
1 1
0 1
)
.
c) If P 2 = P , then Pn = P for all n ≥ 1. Hence
eP = I + P +
1
2!
P +
1
3!
P + ... = I − P + P (1 + 1
1!
+
1
2!
+
1
3!
+ ...)
= I − P + eP = I + (e− 1)P.
iii) a) T (cos x) = cos x+2 sinx and T (sinx) = sinx−2 cos x, These have coordinates (1, 2)
and (−2, 1) respectively with respect to the given basis, so C =
(
1 −2
2 1
)
.
b) Since the determinant of C is not 0, the columns are linearly independent, and so
the rank of T equals the rank of C which is 2.
c) By the rank-nullity theorem, the nullity of T is 0 and hence the map T is invertible
and so the d.e. has a unique solution in S.
d) The solution is y = T−1(cos x). Now the coordinates of the y w.r.t. the basis above
are given by
C−1
(
1
0
)
=
(
1 −2
2 1
)−1(
1
0
)
=
1
5
(
1 2
−2 1
)(
1
0
)
=
1
5
(
1
−2
)
.
Hence y = 15 cos x− 25 sinx.
iv) Let p = 0.05 and let X be the number of warranty periods when the first claim occurs.
Then from λ = ln 11−p we have λ ≈ 0.5129. Hence P (X ≤ 1) =
∫ 1
0
λe−λt dt = 1− e−λ =
p = 0.05.
v) a) The sample space is S×S which has 81 elements. The event A = {(v1,v2) ∈ S×S :
v1,v2 are linearly independent }. Since there are 8 choices for v1 and 6 choices for
v2 it follows that |A| = 48. Hence P (A) = 4881 = 1627 .
b)
k 0 1 2
pk
1
81
32
81
16
27
c) E(X) = 12881 .
d) Since the event b ∈ span{v1,v2} is statistically independent of the selection of
{(v1,v2)} ∈ A, we have
P (B|A) = P (B)P (A)
P (A)
= P (B) = 1.
113
4. i) The equation for tangent plane at a point P (x0, y0, z0) on a surface S ⊂ R3 given by an
equation of the form f(x, y) = z is
fx(P )(x− x0) + fy(P )(y − y0) = (z − z0).
In this question
f(x, y) = c(
x2
a2
+
y2
b2
),
and therefore
fx =
2cx
a2
fy =
2cy
b2
.
From the above the equation of the tangent plane at P is
2cx0(x− x0)
a2
+
2cy0(y − y0)
b2
= (z − z0).
The left hand side can be re-arranged as
−2c(x
2
0
a2
+
y20
b2
) + 2c(
xx0
a2
+
yy0
b2
)
= −2z0 + 2c(xx0
a2
+
yy0
b2
).
Therefore the equation of the tangent is
−2z0 + 2c(xx0
a2
+
yy0
b2
) = (z − z0)
which after adding 2z0 to both sides and then dividing both sides by 2c becomes
(
xx0
a2
+
yy0
b2
) =
z + z0
2c
.
ii) a) Since dudt > 0 it is not equal to zero and therefore
dt
du
= (
du
dt
)−1.
So
t =
1
k
∫
1√
2u+ ǫu2
du.
If ǫ = 1, then we need to evaluate∫
1√
2u+ u2
du.
We make the substitution u+ 1 = cosh(θ) to get
du = sinh(θ)dθ.
114
Then completing the square gives∫
du√
2u+ ǫu2
=
∫
du√
(u+ 1)2 − 1
=
∫
sinh(θ)dθ√
cosh2(θ)− 1
=
∫
sinh(θ)dθ√
sinh2(θ)
=
∫
dθ
= cosh−1(u+ 1) + C.
So
kt = cosh−1(u+ 1) + C
kt− C = cosh−1(u+ 1)
u+ 1 = cosh(kt− C)
u = cosh(kt− C)− 1.
Since u(0) = 1 we have 2 = cosh(−C) i.e. 4 = e−C + eC . Re-arranging we get
e2C − 4eC + 1 = 0 and so
eC =
4±√12
2
i.e.
C = ln(2±
√
3).
To determine which value C takes we observe that dudt > 0 and
du
dt
(0) = sinh(−C)
=
1
2
(
1
2±√3 − 2±
√
3)
is greater than zero only if C = ln(2−√3).
Now if u(t) = 0 then
cosh(kt− C) = 1
so
ekt−C + e−kt+C = 2
and if x = ekt−C we have
x2 − 2x+ 1 = 0
i.e.
x = 1
or
1 = ekt−C
115
which implies that
kt = C = ln(2−
√
3) < 0
which is impossible as k and t are ≥ 0.
b) We need to evaluate ∫
1√
2u− u2du.
We make the substitution u − 1 = sin(θ) to get (following the same reasoning as
above) ∫
1√
2u− u2du = sin
−1(u− 1) + C.
Therefore
u = sin(kt− C) + 1.
Now u(0) = 1 so
sin(−C) = 0
i.e.
C = mπ
where m ∈ Z.
Note that since
sin(x− 2π) = sin(x)
we havem = 0, 1, i.e. C = 0, π. Moreover since u is assumed to be positive (initially,
once u = 0 it becomes negative, at which point the mathematical model is no longer
modeling the physical universe) we must have m = 0. That is C = 0. If u(t) = 0
then
sin(kt) = −1
which happens whenever kt is a multiple of 3π/2, and which will happen first when
t = 3pi2k .
Note: (this was not needed to answer the question, but it is worth stating) The
question states that current observations indicate that dudt > 0; however if ǫ = −1,
then once t > 3pi2k we have
du
dt < 0, in which case present observations imply that
0 < t < 3pi2k .
iii) a)
dx
dt
= −8t2 + 24t
dy
dt
= 2(t+ 1)(t− 3)2 + 2(t+ 1)2(t− 3)
= 2(t+ 1)(t− 3)(t− 3 + t+ 1)
= 2(t+ 1)(t− 3)(2t − 2).
Now
dx
dt
= 0
116
means
8t(t− 3) = 0
so t = 0 or t = 3, so there are at most 2 vertical tangents at t = 0 or t = 3. If t = 0
then
dy
dt
= 2× 1× (−3) × (−2) 6= 0
and therefore there is a vertical tangent at t = 0. If t = 3 then
dy
dt
= 0
and so there is no vertical tangent at t = 3.
b) Recall the arc-length formula for a parametric curve from t = 0 to t = 3 is
∫ 3
0
√
(
dx
dt
)2 + (
dy
dt
)2dt.
Now the maple output tells us that
(
dx
dt
)2 + (
dy
dt
)2 = 16(t − 3)2(t2 + 1)2
and the positive squareroot of this last expression for values of t ∈ [0, 3) is
4(3− t)(t2 + 1)
so the arc-length is given by
4
∫ 3
0
(3− t)(t2 + 1)dt
= 4
∫ 3
0
(−t3 + 3t2 − t+ 3)dt
= 4[−t4/4 + t3 − t2/2 + 3t]30
= 4(−81/4 + 27− 9/2 + 9)
= (−81 + 108− 18 + 36)
= 45.
iv) The Maclaurin series for y has the form
y = a0 + a1x+ a2x
2 + a3x
3 + · · ·
and so
dy
dx
= a1 + 2a2x+ 3a3x
2 + · · · .
We are told that
y(0) = 0
117
so
a0 = 0.
Now the Maclaurin series for cos(x) is
cos(x) = 1− x
2
2!
+
x4
4!
− · · · .
Expanding y2 we get
y2 = a21x
2 + · · ·
Substituting the two expressions for y and dydx into the equation
dy
dx
+ y2 = cos(x)
we therefore obtain
(a1 + 2a2x+ 3a3x
2 + · · · ) + (a21x2 + · · · ) = 1−
x2
2!
+ · · · .
Comparing co-efficients of powers of x we therefore get
a1 = 1
a2 = 0
3a3 + 1 = −1/2.
Therefore the first two terms of the Maclaurin series for y are given by
y = x− 1
2
x3 + · · · .
The above argument assumes that we may differentiate the Maclaurin series of y to
obtain that for dydx and that the Maclaurin series of y
2 is just the square of the Macclaurin
series of y. We can reason alternatively as follows:
The Macclaurin series is given by
y = y(0) + y′(0)x+ y′′(0)x2/2! + y′′′(0)x3/3! + · · ·
so we in fact want to find the first two non-zero derivatives of y (since y(0) = 0).
From the equation
dy
dx
+ y2 = cos(x)
we see that
y′(0) = 1.
Now implicit differentiation, gives
d2y
dx2
+ 2
dy
dx
y = − sin(x)
and therefore
y′′(0) = 0.
118
Implicit differentiating one more time gives
d3y
dx3
+ 2
d2y
dx2
y + 2(
dy
dx
)2 = − cos(x)
and so
y′′′(0) = −3
which yields the same result.
v) We use that fact that
L = lim
n→∞n(c
1/n − 1) = lim
x→∞x(c
1/x − 1).
Define a continuous function
f(x) = (c1/x − 1)
then
f ′(x) =
− ln(c)
x2
c1/x.
Then L is given by
lim
x→∞xf(x)
and can be re-written as
L = lim
x→∞
f(x)
1
x
and as the numerator and denominator both approach zero as x approaches infinity, we
can apply l’Hoˆpital’s rule to get
L = lim
x→∞
− ln(c)
x2 c
1/x
−1/x2
= lim
x→∞ ln(c)c
1/x
= ln(c).
119
Standard normal probabilities P (Z ≤ z)
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
−2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014
−2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019
−2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026
−2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036
−2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048
−2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064
−2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084
−2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110
−2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143
−2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183
−1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233
−1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294
−1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367
−1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455
−1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559
−1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681
−1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823
−1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985
−1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170
−1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379
−0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611
−0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867
−0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148
−0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451
−0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776
−0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121
−0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483
−0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859
−0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247
−0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549
0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852
0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177
1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319
1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441
1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545
1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817
2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857
2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890
2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916
2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936
2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952
2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964
2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974
2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981
2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
120
BASIC INTEGRALS∫
1
x
dx = ln |x|+ C = ln |kx|, C = ln k∫
eax dx =
1
a
eax + C∫
ax dx =
1
ln a
ax + C, a 6= 1∫
sin ax dx = −1
a
cos ax+ C∫
cos ax dx =
1
a
sin ax+ C∫
sec2 ax dx =
1
a
tan ax+ C∫
cosec2ax dx = −1
a
cot ax+ C∫
tan ax dx =
1
a
ln | sec ax|+ C∫
cot ax dx =
1
a
ln | sin ax|+ C∫
sec ax dx =
1
a
ln | sec ax+ tan ax|+ C∫
sinh ax dx =
1
a
cosh ax+ C∫
cosh ax dx =
1
a
sinh ax+ C∫
sech2ax dx =
1
a
tanh ax+ C∫
cosech2ax dx = −1
a
coth ax+ C∫
dx
a2 + x2
=
1
a
tan−1
x
a
+ C∫
dx
a2 − x2 =
1
a
tanh−1
x
a
+ C, |x| < a
= 1a coth
−1 x
a + C, |x| > a > 0
=
1
2a
ln
∣∣∣∣a+ xa− x
∣∣∣∣+ C, x2 6= a2∫
dx√
a2 − x2 = sin
−1 x
a
+ C∫
dx√
x2 + a2
= sinh−1
x
a
+ C∫
dx√
x2 − a2 = cosh
−1 x
a
+ C, x > a > 0

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