MATH1131 Mathematics 1A
and
MATH1141 Higher Mathematics 1A
PAST EXAM PAPERS
AND
SOLUTIONS
Copyright 2020 School of Mathematics and Statistics, UNSW

Contents
PAST EXAM PAPERS 4
NOVEMBER 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
JUNE 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
JUNE 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
JUNE 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
JUNE 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
JUNE 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
PAST HIGHER EXAM PAPERS 33
JUNE 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
JUNE 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
JUNE 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
JUNE 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
JUNE 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
PAST EXAM SOLUTIONS 49
NOVEMBER 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
JUNE 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
JUNE 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
JUNE 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
JUNE 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
JUNE 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
PAST HIGHER EXAM SOLUTIONS 89
JUNE 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
JUNE 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
JUNE 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
JUNE 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
JUNE 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
Table of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
The 2017 to 2019 exam papers will be provided on Moodle.
3
4PAST EXAM PAPERS
Please note that the emphasis of the exam may change from year to year and that it is not possible
to test all aspects of the course in a 2-hour examination. Hence, these papers are only a guide as
to the style and level of difficulty of our first year examinations.
The exam format may be different in 2020 — check Moodle for details.
The solutions to the examination papers contained here have been written by many members of
staff of the School of Mathematics and Statistics. While every care is taken to excluded errors, we
cannot guarantee that the solutions are error-free. Please report any serious errors to the Director
of First Year Mathematics.
Exam papers from 2019 will be provided on Moodle for practice before the exam. Students are
encouraged to produce their own solutions to the 2019 exam papers and discuss these on Moodle.
c©2020 School of Mathematics and Statistics, UNSW Sydney
5MATH1131 November 2010
1. i) Let z = 1 + i.
a) Find |z|.
b) Find Arg(z).
c) Use the polar form of z to evaluate z28 and then express your answer in Cartesian
form.
ii) Suppose that (x+ iy)(3 + 2i) = (4 + 7i) where x, y ∈ R.
Find the value of x and the value of y.
iii) a) Find all solutions to the equation z6 = 1, where z ∈ C.
b) Hence, or otherwise, express z6−1 as a product of three quadratic polynomials with
real coefficients.
iv) Let the region S be defined as
S = {z ∈ C : |z + i| = 1}.
Sketch the region S on a carefully labelled Argand diagram.
v) Consider the following MAPLE session.
> with(LinearAlgebra):
> A:=<<1,0,-sqrt(3)>|<0,2,0>|>;
A :=


1 0
√
3
0 2 0
−√3 0 1


> A^2; 

−2 0 2√3
0 4 0
−2√3 0 −2


> A^4; 

−8 0 −8√3
0 16 0
8
√
3 0 −8


> A^6; 

64 0 0
0 64 0
0 0 64


Use the above Maple session to find the inverse of the matrix A2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
6vi) A car dealer sells three brands of cars: Audis, BMWs and Chevrolets. Her costs consist
of registration, GST and the manufacturer’s price of the car. She has to pay registration
costs of $200 per Audi, $600 per BMW and $400 per Chevrolet. The GST charges are
$1, 800 per Audi, $2, 400 per BMW and $2, 000 per Chevrolet. To the manufacturer she
pays $20, 000 per Audi, $30, 000 per BMW and $26, 000 per Chevrolet. Last year her
total registration bill was $12, 000, the total GST tax bill was $65, 600 and she paid a
total of $784,000 to manufacturers.
Let a, b and c denote the number of Audis, BMWs and Chevrolets she sold last year.
a) Write down a system of linear equations in a, b and c determined by the above
information.
b) Convert the system of equations to an augmented matrix form and hence find a, b
and c by performing Gaussian elimination on the augmented matrix.
2. i) Let P =
(
1 0 1
2 3 0
)
.
a) Evaluate the matrix product PP T .
b) State the size of the matrix P TP .
ii) Let A,B and C be points in R3 with position vectors
a =

12
3

 , b =

10
1

 and c =

34
7

 respectively.
a) Find
−−→
AB and
−→
AC.
b) Find a parametric vector equation of the plane that passes through the points A,B
and C.
iii) Determine the coordinates of the point of intersection of the line

xy
z

 =

12
5

+ t

 1−1
1

 for t ∈ R
and the plane x− 3y + z = 15.
iv) Let v =


1
3
−1
5

 and w =


−2
−6
β
−10

 be two vectors in R4.
Find the value of β so that the vectors v and w are perpendicular.
v) Let u =

21
7

 be a vector in R3.
a) Find the magnitude of u.
b) Write down a vector with a magnitude of 10 which is parallel to u.
c©2020 School of Mathematics and Statistics, UNSW Sydney
7vi) Find the determinant of the matrix
C =

3 1 04 1 7
1 2 0

 .
vii) Suppose that u, v and w are distinct non-zero vectors in Rn with the property that
projw(u) = projw(v).
Prove that (u− v) is perpendicular to w.
3. i) Evaluate the limit
lim
x→0
x2ex
1− cos(πx) .
ii) Determine all real values of a and b such that the function f given by
f(x) =
{
x2 for x ≤ 1,
−x2 + ax+ b for x > 1,
is differentiable at x = 1.
iii) Let
L = lim
x→∞
x2 − 2
x2 + 1
.
a) Evaluate L.
b) Given any ε > 0, find a constant M (which may depend on ε) such that we have∣∣∣∣x2 − 2x2 + 1 − L
∣∣∣∣ < ε
whenever x > M .
iv) Let f(x) = x3 +
√
3 x− 5 for all real x.
a) Use the Intermediate Value Theorem to prove that f has at least one positive real
root.
b) By considering f ′, or otherwise, show that f has only one real root.
v) Use logarithmic differentiation to calculate
dy
dx
for y = (sinx)x.
vi) A curve in R2 is given in polar coordinates as
r = 6 sin θ, where 0 ≤ θ ≤ π/2.
a) Express the equation of the curve using Cartesian coordinates x and y and state the
range of x and the range of y.
b) Hence, or otherwise, sketch the curve in the xy-plane.
4. i) Evaluate the following integrals:
c©2020 School of Mathematics and Statistics, UNSW Sydney
8a) I1 =
∫
1− x
(1 + x)3
dx;
b) I2 =
∫ π
0
x cos 2x dx.
ii) Determine whether the improper integral
K =
∫ ∞
1
1 + sinx
3x2
dx
converges or diverges.
iii) a) State the definitions of sinhx and coshx in terms of the exponential function.
b) Hence find an expression for
d
dx
cosh(ax) in terms of sinh(ax), where a is a constant.
c) Simplify the expression cosh−1
(
cosh(−4726)).
iv) A continuous function f satisfies the equation
∫ x
0
f(t) dt =
∫ 1
x
t2f(t) dt+
x16
8
+
x18
9
− 1
9
.
By differentiating this equation with respect to x, and using the First Fundamental
Theorem of Calculus where needed, find f(x).
v) Let g be a function defined by
g(x) = tan−1(x) + tan−1(1/x).
a) What is the (maximal) domain of g? Give reasons for your answer.
b) By examining g′(x), show that g is piecewise constant on its domain.
c) Hence, or otherwise, determine the exact value of
tan−1(−eα) + tan−1(−e−α), for all α ∈ R.
c©2020 School of Mathematics and Statistics, UNSW Sydney
9MATH1131 JUNE 2011
1. i) Let z = −1− i.
a) Find |z|.
b) Find Arg(z).
c) Use the polar form of z to evaluate z102 and then express your answer in Cartesian
form.
ii) a) Simplify (2 + 4i)2.
b) Hence, or otherwise, solve the quadratic equation z2 − 4z + (7− 4i) = 0.
iii) Sketch the following region on the Argand diagram
S = {z ∈ C : 0 ≤ Arg(z − i) ≤ π
4
}.
iv) Evaluate the limit
lim
x→∞
1
x−√x2 − 6x− 4 .
v) Evaluate the improper integral ∫ ∞
1
x−5/4 dx .
vi) A curve in the plane is defined implicitly by the equation
x2 − 3xy2 + 11 = 0 .
a) Show that the curve has slope at the point (x, y) given by
dy
dx
=
2x− 3y2
6xy
.
b) Find the equation of the tangent to the curve at the point (1, 2).
c) Write a Maple command to plot the curve in the region
1 ≤ x ≤ 4 and −5 ≤ y ≤ 5.
2. i) Use De Moivre’s Theorem to prove that
cos(4θ) = 8 cos4 θ − 8 cos2 θ + 1.
ii) Consider the line ℓ with parametric vector equation
xy
z

 =

12
4

+

32
5

 t, for t ∈ R.
a) Give two points on the line.
b) Give a vector parallel to the line.
c) Explain why the line ℓ is perpendicular to the plane P with Cartesian equation
9x+ 6y + 15z = 24.
c©2020 School of Mathematics and Statistics, UNSW Sydney
10
d) Find a point on the line whose y-coordinate is 0.
iii) Consider the following Maple session:
> with(LinearAlgebra):
> A:=<<1,1,0>|<-1,1,0>|<0,0,sqrt(2)>>;
A :=


1 −1 0
1 1 0
0 0
√
2


> A^2; 

0 −2 0
2 0 0
0 0 2


> A^4; 

−4 0 0
0 −4 0
0 0 4


> A^8; 

16 0 0
0 16 0
0 0 16


Use the above MAPLE session to find the inverse of A7.
iv) Evaluate the limit
lim
x→1
(x− 1)2
1 + cos(πx)
.
v) Evaluate the indefinite integral ∫
x sin(2x) dx.
vi) The function f has domain [0, 1] and is defined by f(x) = ex + ax, where a is a positive
constant.
a) Prove that 2 is in the range of f .
b) Prove that f has an inverse function f−1.
c) Find the domain of f−1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
11
3. i) Let A =
(
1 4 7
2 1 3
)
and B =

1 30 5
1 0

.
a) Find AB.
b) What is the size of the matrix BA?
ii) A pet shop has x hamsters, y rabbits and z guinea pigs.
Each hamster eats 50g of dry food and 40g of fresh vegetables,
and needs 1m2 of space.
Each rabbit eats 300g dry food and 320g of fresh vegetables,
and needs 5m2 of space.
Each guinea pig eats 100g of dry food and 200g of fresh vegetables,
and needs 3m2 of space.
Altogether they eat 2900g of dry food and 3920g of fresh vegetables, and need 63m2 of
space.
a) Explain why 5x+ 30y + 10z = 290.
b) Write down a system of linear equations that determine x, y and z.
c) Reduce your system to echelon form and solve to find the number of hamsters,
rabbits and guinea pigs.
iii) The points A, B and C in R3 have position vectors
a =

10
1

 , b =

23
4

 and c =

41
5


respectively.
Write down a parametric vector equation of the plane passing through A, B and C.
iv) Consider the following system of linear equations.
x+ y − z = 2
2x+ 3y + z = 6
a) Using Gaussian Elimination find the general solution to the system of equations.
b) Hence, or otherwise, find a solution to the system with the property that the sum
of the x, y and z coordinates of the solution is 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
12
v) Suppose A,B are two points in R3 with position vectors

12
4

 and

31
5

 respectively.
We let O denote the origin.
a) Find |
−→
OB |.
b) Find the area of triangle AOB.
c) Hence, or otherwise, find the perpendicular distance from A to the line through O
and B.
vi) Suppose that u and v are non-zero, non-parallel vectors in R3 of the same magnitude.
Prove that u− v is perpendicular to u+ v.
4. i) a) Give the definition of cosh x.
b) Use the definition to prove that
4 cosh3 x = cosh 3x+ 3cosh x .
ii) Find
a)
d
dx
∫ x
0
cos t√
1 + t2
dt ;
b)
d
dx
∫ sinhx
0
cos t√
1 + t2
dt . Give your answer in simplest form.
iii) a) Sketch the curve whose equation in polar coordinates is r = 6 sin 2θ.
b) Find the gradient, dydx , of this curve at the point where θ =
1
6π.
c©2020 School of Mathematics and Statistics, UNSW Sydney
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iv) A statue 2 metres high stands on a pillar 2.5 metres high. A person, whose eye is 1.5m
above the ground, stands at a distance x metres from the base of the pillar.
θ
2
1
1.5
A
B
C✷
φ
x
The diagram shows the above information, with the person’s eye being at C.
a) Prove that
d
dt
(
cot−1 t
)
=
−1
1 + t2
.
b) Show that
θ = cot−1
(x
3
)
− cot−1 x
c) Hence find the distance x that maximises the angle θ.
c©2020 School of Mathematics and Statistics, UNSW Sydney
14
MATH1131 June 2012
1. i) Let u = 3 + 2i and w = 1− 5i.
a) Find u− 2w in Cartesian form.
b) Find u/w in Cartesian form.
ii) Let z =
√
3− i.
a) Calculate |z| and Arg(z).
b) Express z in polar form.
c) Hence, or otherwise, express z10 + (z)10 in Cartesian form.
iii) Evaluate the determinant ∣∣∣∣∣∣
1 −1 4
0 2 7
0 3 1
∣∣∣∣∣∣ .
iv) a) Evaluate lim
x→∞
3x2 + sin(2x2)
x2
.
b) Evaluate lim
x→0
3x2 + sin(2x2)
x2
.
v) Consider the curve in the plane defined by
x2 − 5x sin y + y2 = 4.
Find the equation of the tangent line to this curve at the point (2, 0).
vi) Let p(x) = x5 + 5x+ 7.
a) Explain why p has at least one real root.
b) Prove that p has exactly one real root.
2. i) Use De Moivre’s Theorem to show that
cos(3θ) = 4 cos3 θ − 3 cos θ.
ii) Find the intersection of the line x =

 10
2

 + λ

 11
2

 , λ ∈ R, with the plane 5x −
2y + z = 17.
iii) Consider the following MAPLE session:
> with(LinearAlgebra):
> A:=<<0,0,1>|<0,1,0>|<-1,0,0>>;
A :=


0 0 −1
0 1 0
1 0 0


> A^2; 

−1 0 0
0 1 0
0 0 −1


c©2020 School of Mathematics and Statistics, UNSW Sydney
15
> A^3; 

0 0 1
0 1 0
−1 0 0


> A^4; 

1 0 0
0 1 0
0 0 1


Using the Maple session above, find the inverse of A2001.
iv) The points C and D have position vectors
c =

12
3

 and d =

41
5

 .
a) Find the cross product c× d.
b) Hence, or otherwise, find the area of the parallelogram with adjacent sides OC and
OD, where O is the origin.
v) Evaluate the indefinite integral ∫
x4 lnx dx.
vi) Consider the three functions:
f : R→ R, f(x) = x
2
1 + x2
,
g : (0, 3) → R, g(x) = (x− 1)2,
h : [1, 5]→ R, h(x) =
√
1 + lnx+ sinx cos x.
Only one of these functions has a maximum value (on its given domain). Which one is
it? Give reasons for your answer.
vii) Sketch the polar curve r = 2 − 2 cos θ. You should show any lines of symmetry, and
clearly identify where the curve intersects the x and y axes.
viii) Suppose that y = xsinx. Find
dy
dx
.
3. i) The points A and B in R3 have position vectors
a =

10
1

 and b =

12
4

 .
a) Find a parametric vector equation of the line l passing through A and B.
c©2020 School of Mathematics and Statistics, UNSW Sydney
16
b) By evaluating an appropriate dot product, show that the line l from part (a) is
perpendicular to the line

xy
z

 =

14
6

+

 1−3
2

 t; t ∈ R.
ii) Let P =
(
1 2 1
3 −1 4
)
and Q =
(
1 −1 1
2 5 0
)
.
a) Evaluate PQT .
b) What is the size of P TQ?
c) Does the matrix product PQ exist? Explain your answer.
iii) A system of three equations in three unknowns x, y and z has been reduced to the
following echelon form

 1 2 3 50 2 4 8
0 0 α2 − 9 α− 3

 .
a) For which value of α will the system have no solution?
b) For which value of α will the system have infinitely many solutions?
c) For the value of α determined in part (b), find the general solution.
iv) The number of $10, $20, and $50 notes in the cash register at Bill’s Burger Barn is x, y
and z respectively. The total value of all the notes in the register is $1020. There are
44 notes in total. Also, the number of $10 notes is equal to the sum of the number of
$20 notes and the number of $50 notes.
a) Explain why x+ 2y + 5z = 102.
b) By setting up two further equations and solving the system of three equations in
three unknowns x, y and z, determine how many of each type of note is in the cash
register.
v) The (non-zero) point Q has position vector

 ab
c

. The vector −→OQ makes angles α, β
and γ respectively with the X,Y and Z axes.
a) By considering the vector e1 =

 10
0

 show that
a =
√
a2 + b2 + c2 cosα.
b) Deduce that
cos2 α+ cos2 β + cos2 γ = 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
17
c) If the angles α and β are complementary, that is, their sum is 90◦, what can be said
about the vector
−→
OQ?
4. i) Find a quadratic function q(x) = x2 + bx+ c such that the function
h(x) =
{
e3x, if x ≤ 0
q(x), if x > 0
is differentiable at x = 0.
ii) Each of the following calculations is expressed in MAPLE. Write each in normal math-
ematical notation and evaluate.
a) arcsin(sin(7*Pi/3));
b) diff(int(exp(t^2),t=0..x^2),x);
iii) Prove that lim
x→∞
x2 − 2
x2 + 3
= 1 as follows:
Given any real number ǫ > 0, find a real number M (expressed in terms of ǫ), such that
if x > M then
∣∣∣∣x2 − 2x2 + 3 − 1
∣∣∣∣ < ǫ.
iv) Let f : R→ R be defined by
f(x) = x3 + sinhx+ 1.
a) Explain why f has a differentiable inverse g.
b) What is the domain of g?
c) Evaluate g′(1).
v) A chemical process produces Factor X, which flows into a 50 litre tank which is initially
empty.
At time t ≥ 0, Factor X flows into the tank at the rate of 100
10 + t2
litres per hour.
Will the tank eventually overflow? Explain your answer.
c©2020 School of Mathematics and Statistics, UNSW Sydney
18
MATH1131 June 2013
1. i) Evaluate the following limits:
a)
lim
x→∞
10x2 + 3x+ sinx
5x2 + 3x− 2 ,
b)
lim
x→0
e3x − 1
sin (7x)
.
ii) A function f : [0, 5]→ R has the following properties:
• lim
x→2+
f(x) = 3,
• lim
x→2−
f(x) = 1,
• f(2) = 4.
Draw a possible sketch of the graph of f . (You do not need to give a formula for
your function.)
iii) a) State the definitions of coshx and sinhx in terms of the exponential function.
b) Prove that cosh2 x− sinh2 x = 1.
iv) Let z = 5 + 5i and w = 2 + i.
a) Find 2z + 3w.
b) Find z(w − 1).
c) Find z/w.
v) Suppose that (x+ iy)(3 + 4i) = 13 + 9i, where x, y ∈ R.
Find the value of x and the value of y.
vi) Let the set S in the complex plane be defined by
S =
{
z ∈ C : |z| ≤ 3 and 0 ≤ Im(z) ≤ 3
}
.
a) Sketch the set S on a labelled Argand diagram.
b) By considering your sketch, or otherwise, find the area of the region defined by S.
c©2020 School of Mathematics and Statistics, UNSW Sydney
19
vii) Consider the following MAPLE session.
> with(LinearAlgebra):
> A:=<<0,1,-1>|<1,0,1>|<-1,1,0>>;
A :=


0 1 −1
1 0 1
−1 1 0


> B:=A^2;
B :=


2 −1 1
−1 2 −1
1 −1 2


> C:=A^3;
C :=


−2 3 −3
3 −2 3
−3 3 −2


> F:=3*A-C;
F :=


2 0 0
0 2 0
0 0 2


Without carrying out any row reduction, use the above Maple session to find the inverse
of the matrix A.
2. i) A function h is defined by
h(x) =
{
ax2 + 3x, if x ≥ 1
2x+ d, if x < 1.
Given that h is differentiable at x = 1, find the values of a and d.
ii) Evaluate ∫ ln 2
0
9xe3x dx.
iii) Find the equation of the tangent at the origin to the curve implicitly defined by
ex + sin y = xy + 1.
iv) Sketch the polar curve whose equation in polar coordinates is given by
r = 1 + cos 2θ.
v) Let z =
√
2−√2i.
a) Find |z|.
b) Find Arg(z).
c©2020 School of Mathematics and Statistics, UNSW Sydney
20
c) Use the polar form of z to evaluate z6. Express your answer in Cartesian form.
vi) Let A =
(
2 0
1 −1
)
and B =
(
3 0 1
1 2 0
)
.
a) Evaluate AB or explain why this product does not exist.
b) Evaluate ABT or explain why this product does not exist.
vii) a) Find, in polar form, all solutions to the equation z5 = −1, where z ∈ C.
b) Hence, or otherwise, express z5 + 1 as a product of real linear and real quadratic
factors.
3. i) Determine the point of intersection of the line
xy
z

 =

 1−1
−1

+ t

12
1

 for t ∈ R .
and the plane 4x− 5y + 3z = 0.
ii) Let M =

1 2 02 5 1
0 2 α

.
a) Evaluate the determinant of M .
b) Determine the value(s) of α for which M does not have an inverse.
c) Find the inverse of M when α = 1.
iii) Let u =

21
0

 and v =

31
1

.
a) Find the cross product u× v.
b) Hence find the Cartesian equation of the plane parallel to u and v and passing
through the point

14
2

.
iv) Let u =


2
1
4
3

 and v =


0
3
−3
β

 be two vectors in R4.
a) Find the value of β so that the vectors u and v are orthogonal.
b) For the value β = 1, find the projection, proju(v), of v onto u.
v) Let A, B and D be three points on some circle with centre C in R2 with position vectors
a =
(
2
1
)
, b =
(
2
2
)
, and d =
(
4
2
)
.
a) Let M be the midpoint of the line joining A and B. Find the position vector m of
the point M .
c©2020 School of Mathematics and Statistics, UNSW Sydney
21
b) Find a non-zero vector u that is perpendicular to
−→
AB.
c) Hence or otherwise, find the parametric vector equation for the line whose points
are equidistant from A and B.
d) Given that the parametric vector equation for the line whose points are equidistant
from B and D is x =
(
3
2
)
+ µ
(
0
1
)
, µ ∈ R, find the centre C of the circle.
4. i) Find
∫
cos(ln(x))
x
dx.
ii) Use the Pinching theorem to evaluate
lim
x→∞ e
−x sin(x).
iii) Show that the improper integral
∫ ∞
0
dx
x2 + ex
converges.
iv) The following calculation is expressed in MAPLE
> F:=diff(int(sin(sqrt(t)),t=0..x^2),x);
a) Write the calculation using standard mathematical notation.
b) Evaluate F .
v) Let p(x) = x3 + 4x− 7.
a) Use the Intermediate Value theorem to show that p has at least one real root in the
interval [1, 2].
b) Show that p has exactly one real root in the interval [1, 2].
c) Let g be the inverse of p and α be the unique root of p, whose existence is guaranteed
in part b). Express g′(0) in terms of α.
vi) Use the Mean Value Theorem to prove that, for x > 0,
ln(1 + x) >
x
1 + x
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
22
MATH1131 June 2014
1. i) Let z = 7 + i and w = 4 + 3i.
a) Find 2z − w in a+ ib form.
b) Find 5(w − i)/z in a+ ib form.
c) Find |zw|.
d) Find Arg(zw).
e) Hence, or otherwise, show that Arg(z) + Arg(w) =
π
4
.
f) Use the polar form of zw to evaluate (zw)40.
ii) Consider the following system of equations:
x − y − z = 1
x − 3y + z = 1
2x − 3y − z = 2.
a) Write the system in augmented matrix form and reduce it to row echelon form.
b) Solve the system.
iii) Evaluate the limits
a)
lim
x→∞
6x2 + sinx
4x2 + cos x
;
b)
lim
x→0
e2x − 2x− 1
4x2
.
iv) A function g is defined by
g(x) =


|x2 − 16|
x− 4 if x 6= 4
α if x = 4.
By considering the left and right hand limits at x = 4, show that no value of α can make
g continuous at the point x = 4.
v) Let f(x) = x5 + x3 + x− 2.
a) Prove that f has at least one real root in the interval [0, 2], naming any theorems
you use.
b) State, with reasons, the number of real roots of f .
2. i) Use a substitution to find the integral∫
dx
x(1 + (log x)2)
.
ii) a) Give the definitions of sinhx and cosh x in terms of the exponential function.
b) Use your definitions to prove that sinh(2x) = 2 sinhx coshx.
iii) Evaluate the integral ∫ π/3
0
x sin(2x) dx.
c©2020 School of Mathematics and Statistics, UNSW Sydney
23
iv) Simplify the matrix expression (ATA)−1(ATA)T , where A is an invertible matrix.
v) Consider the three points A,B,C in R3 with position vectors

 01
3

,

 31
4

, and

 32
4

 respectively.
a) Find a parametric vector form for the plane Π that passes through points A, B,
and C.
b) Calculate the cross product n =
−−→
AB ×−→AC, showing your working.
c) Hence, or otherwise, find a Cartesian equation for the plane Π.
d) Find the area of the triangle ABC.
e) Find the minimal distance from the point P

 53
0

 to the plane Π.
vi) Consider the following MAPLE session.
> with(LinearAlgebra):
> A := <<0,1,1>|<1,-1,-1>|<-2,1,0>>;
A :=


0 1 −2
1 −1 1
1 −1 0


> B := A^2;
B :=


−1 1 1
0 1 −3
−1 2 −3


> C:=A^3;
C :=


2 −3 3
−2 2 1
−1 0 4


> F := 2A + B + C;
F :=


1 0 0
0 1 0
0 0 1


Using the above MAPLE session, or otherwise, find the 3×3 matrix which is the inverse
of the matrix A.
3. i) A block of wood is subject to 3 vector forces:
F1 = 1, in the direction West, F2 = 1 in the direction South, F3 = 2 in the direction
North-East, each measured in Newtons.
Let F = F1 + F2 + F3 be the resultant force on the block.
a) On a scale diagram draw the 3 forces F1, F2, F3 and the resultant force F.
b) Find the exact value of |F| and the direction of F.
c©2020 School of Mathematics and Statistics, UNSW Sydney
24
ii) Let M =

 1 −1 12 1 3
−1 2 1

 and N = (3 1
4 2
)
.
a) Evaluate the determinant of M .
b) Write down the inverse of N .
iii) Let p(z) = z4 + 2z2 − 3.
a) Show that p(1) = p(−1) = 0.
b) Factor p(z) into two real quadratic polynomials q(z) and r(z).
c) Find the roots of p(z).
d) Factor p(z) into four complex linear polynomials.
iv) Consider the line ℓ and the plane Π given by the following equations:
ℓ : x =

xy
z

 =

32
1

+ λ

12
3

 , λ ∈ R,
Π : 6x+ 8y − 9z = 0.
Determine the point of intersection of the line ℓ and the plane Π.
v) a) Use De Moivre’s Theorem to prove that 4 cos3 θ = cos 3θ + 3cos θ.
b) Deduce that 2 cos
π
9
is a root of the polynomial q(z) = z3 − 3z − 1.
vi) Let u =

 01
−1

 and v =

21
β

 be two vectors in R3.
a) Find the value of β so that the vectors u and v are orthogonal.
b) For the value β = 0, find the projection, projv u, of u onto v.
c) Find the value of β so that the angle between u and v is
π
4
.
4. i) Use the Fundamental Theorem of Calculus to find
d
dx
∫ x3
x2
cos
(
1
t
)
dt.
ii) Let f(x) =
175x2 − 350x + 10
x2 − 2x+ 2 .
Consider the following MAPLE session.
> f:=(175*x^2-350*x+10)/(x^2-2*x+2);
175x2 − 350x + 10
x2 − 2x+ 2
> subs(x=0.0,f);
5.0
> subs(x=4.0,f);
141.0
> solve(f=0,x);
c©2020 School of Mathematics and Statistics, UNSW Sydney
25
1 +
√
1155/35, 1−
√
1155/35
> evalf(%);
1.971008312, 0.0289916875
> fdash:=diff(f,x);
350x − 350
x2 − 2x+ 2 −
(
175x2 − 350x + 10) (2x− 2)
(x2 − 2x+ 2)2
> solve(fdash=0,x);
1
> subs(x=1.0,f);
−165.0
a) Use the information in the MAPLE output to give a rough sketch of f(x) =
175x2 − 350x + 10
x2 − 2x+ 2 , for 0 ≤ x ≤ 4.
b) Hence, or otherwise, find the maximum and minimum values of
f(x) =
∣∣∣∣175x2 − 350x+ 10x2 − 2x+ 2
∣∣∣∣ over the closed interval [0, 4].
iii) Determine, with reasons, whether the improper integral
K =
∫ ∞
0
dx
e2x + cos2 x
converges or diverges.
iv) Sketch in the xy-plane, the graph of the polar curve given by r = 1 − cos θ. (You are
NOT required to find the slope of the curve.)
v) a) State carefully the Mean Value Theorem.
b) Suppose −1 < x < y < 1. By applying the Mean Value Theorem to the function
f(t) = sin−1 t on the interval [x, y], prove that
sin−1 y − sin−1 x ≥ y − x.
vi) Two poles A and B, of heights a metres and b metres respectively, are c metres apart
on the horizontal ground. A single tight rope runs from the top of pole A to the point
P on the ground between A and B and then to the top of pole B.
Assume that the distance from A to P is x, and that the angles θ and φ are as shown
in the diagram.
c©2020 School of Mathematics and Statistics, UNSW Sydney
26
A BP
a
b
q f
x
c
E
F
a) Explain why the length L of the rope is given by
L =
√
a2 + x2 +
√
b2 + (c− x)2.
b) Prove that cos θ = cosφ when
dL
dx
= 0.
c) Assuming that cos θ = cosφ minimizes L, using similar triangles, or otherwise, find
the value of x that minimizes L.
c©2020 School of Mathematics and Statistics, UNSW Sydney
27
MATH1131 June 2015
1. i) Find lim
x→∞
x+ sinx
2x
. (Give brief reasons for your answer.)
ii) The function f is defined by
f(x) =
{
3− x, 0 ≤ x < 1,
(x− 2)2 + 1, 1 ≤ x ≤ 3,
and f(−x) = f(x) for all x.
a) Find f(32).
b) Sketch the graph of f(x) over the interval −3 ≤ x ≤ 3.
c) Which, if any, of the following exists? If it exists, state its value. (Give brief reasons
for your answer.)
lim
x→1
f(x), f ′(0).
iii) Find, for x > 0,
I1 =
∫
dx
1 +
√
x
.
iv) Evaluate
lim
x→0
1− cos (x2 )
x2
.
v) Let z = 5 + i and w = 3 + 2i.
a) Find z − w in a+ ib form.
b) Find 10w/(z − 2) in a+ ib form.
c) Find |(z/w)8|.
d) Find Arg(zw).
e) Use the polar form of zw to evaluate (zw)8.
vi) Consider the three points A,B,C in R3 with position vectors
 11
4

 ,

 21
1

 and

 −14
3

 ,
respectively.
Find a parametric vector form for the plane Π that passes through points A, B, and C.
c©2020 School of Mathematics and Statistics, UNSW Sydney
28
vii) Consider the following MAPLE session.
> with(LinearAlgebra):
> A := <<1,1,3>|<1,0,1>|<4,1,5>>;
A :=


1 1 3
1 0 1
4 1 5


> B := A^2;
B :=


14 4 19
5 2 8
25 9 38


> C := A^3;
C :=


94 33 141
39 13 57
186 63 274


> F := C - 6B - 9A;
F :=


1 0 0
0 1 0
0 0 1


a) Using the above MAPLE session, or otherwise, find the 3 × 3 matrix which is the
inverse of the matrix A.
b) State (AT )2.
2. i) By writing z = a+ ib, or otherwise, solve z2 = 40+42i, giving your answers in Cartesian
form.
ii) Find condition(s) on b1, b2, b3 to ensure that the following system has a solution.
x + 2y = b1
x + y − z = b2
2x + y − 3z = b3
iii) Sketch the following region on the Argand diagram:
S =
{
z ∈ C : |z − i− 1| ≤ 1 or |Im(z)| ≥ 1
}
.
iv) Let
u =

 21
−1

 and v =

 1−1
−2

 .
a) Calculate the acute angle θ between the vectors u and v.
b) Find u× v.
v) Consider the function
f(x) =
{
1 + ax2, x ≤ 1,
bx+ 2x3, x > 1,
where a, b ∈ R.
c©2020 School of Mathematics and Statistics, UNSW Sydney
29
a) Find all values of a and b such that f is continuous at x = 1.
b) Find all values of a and b such that f is differentiable at x = 1.
vi) Evaluate
I2 =
∫
x3 lnx dx.
vii) Consider the polar curve r = 1 + cos θ, for 0 ≤ θ ≤ 2π.
a) Find the slope of the tangent to the curve at the point with Cartesian coordinates
(0, 1).
b) Find the polar coordinates of the points at which the tangent to the curve is hori-
zontal.
c) Sketch the polar curve.
c©2020 School of Mathematics and Statistics, UNSW Sydney
30
3. i) Show that the set of points z in the complex plane that satisfy the
equation
2|z − 3i| = |z + 3i|
lie on a circle. State the radius and centre of the circle.
ii) Consider the complex polynomial p(z) = z4 − 3z3 + 6z2 − 12z + 8.
a) Given that p(2i) = 0, factorise p into linear and quadratic factors with real coeffi-
cients.
b) Find all roots of p.
iii) Consider the plane Π in R3 with Cartesian equation
x− 3y + 2z = 1 .
a) Find a point-normal form for the plane Π.
b) Show that the line ℓ with parametric vector form
x =

 31
4

+ λ

 20
−1

 , λ ∈ R
is parallel to the plane Π.
c) Find the shortest distance between the point P (4, 2, 2) and the plane Π.
iv) a) Find the determinant of
A =

 1 −1 33 a 2
2 1 1

 .
b) For what value of a does the matrix A not have an inverse.
c) Determine the value of a and the value of b for which A−1 = B, where
B =

 −1 4 −51 b 7
1 −3 4

 .
v) Let X be an n× n matrix with determinant |X| = 1.
Prove that the inverse of X also has determinant equal to 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
31
4. i) a) Define the functions sinhx and coshx in terms of the exponential function.
b) From the definitions in (a), show that
d
dx
(cosh 6x) = 6 sinh 6x.
c) Show that, for x > 0,
ln (sinhx) < x− ln 2.
ii) Prove carefully that the improper integral
∫ ∞
1
lnx
x3
dx converges.
iii) Find
d
dx
(∫ x3
0
cos
(
t2
)
dt
)
.
iv) Consider the polynomial p(x) = x3 + 3x+ 1 defined on R.
a) Use the Intermediate Value Theorem to show that the equation
p(x) = 0 has at least one real root.
b) Show that the function p has an inverse function g.
c) Find the value of g′(1).
v) a) State the Mean Value Theorem.
b) Apply the Mean Value Theorem to f(t) = cos t, on a suitably chosen interval, to
show that | cos y − cos x| ≤ |y − x|, for all x, y ∈ R.
c©2020 School of Mathematics and Statistics, UNSW Sydney
32
c©2020 School of Mathematics and Statistics, UNSW Sydney
33
PAST HIGHER EXAM PAPERS
c©2020 School of Mathematics and Statistics, UNSW Sydney
34
MATH1141 JUNE 2011
2. i) You may use the following Maple session to assist you in answering the question below.
> with(LinearAlgebra):
> A:=<<1,2,1>|<1,3,a>|<-1,a,3>>;
A :=


1 1 −1
2 3 a
1 a 3


> t:=<1,3,2>;
t :=


1
3
2


> B:=;
B :=


1 1 −1 1
2 3 a 3
1 a 3 2


> G:=GaussianElimination(B);
G :=


1 1 −1 1
0 1 a+ 2 1
0 0 6− a2 − a 2− a


For which values of a will the system Ax = t have:
a) no solutions,
b) unique solution,
c) infinitely many solutions?
ii) Evaluate the determinant ∣∣∣∣∣∣
2 0 −1
1 3 0
5 7 3
∣∣∣∣∣∣ .
iii) Find the point of intersection, if any, of the line x = λ

12
3

, λ ∈ R, with the plane
x =

10
0

+ µ

11
0

+ ν

 10
−1

 , µ, ν ∈ R.
c©2020 School of Mathematics and Statistics, UNSW Sydney
35
iv) Consider the function f defined by
f(x) =
{
e−1/x
2
for x 6= 0
0 for x = 0.
a) Given that lim
x→∞xe
−x = 0, evaluate the limit
lim
h→0
e−1/h
2
h
.
b) Using the definition of a derivative, determine whether f is differentiable at x = 0.
v) Consider the function f defined by
f(x) =


2
x
for x < −1,
x2 − 1 for − 1 ≤ x ≤ 6
π
,(
72
π2
− 2
)
sin
1
x
for x >
6
π
.
a) Find all critical points of f and determine their nature. In particular, distinguish
between local and global extrema.
b) Find all horizontal asymptotes for f .
c) Given that f does not have a point of inflexion, sketch the graph of f , clearly
indicating all of the above information.
c©2020 School of Mathematics and Statistics, UNSW Sydney
36
3. i) Let
S = eiθ +
e3iθ
3
+
e5iθ
32
+
e7iθ
33
+ · · · .
a) Prove that
S =
3(3eiθ − e−iθ)
10− 6 cos(2θ) .
b) Hence, or otherwise, find the sum
T = sin(θ) +
sin(3θ)
3
+
sin(5θ)
32
+
sin(7θ)
33
+ · · · .
ii) Suppose that u and v are non-zero, non-parallel vectors in R3 of the same magnitude.
Prove that u− v is perpendicular to u+ v.
iii) Let A = (aij) be a real n×n matrix and let e1, . . . ,en be the standard basis vectors for
R
n.
a) Prove that eTi Aej = aij for all 1 ≤ i, j ≤ n.
b) Prove that if A is symmetric then xTAy = (Ax)Ty for all x,y ∈ Rn.
c) Conversely, prove that if xTAy = (Ax)Ty for all x,y ∈ Rn, then A is symmetric.
iv) The matrix
M =
1
9

 1 4 −84 7 4
−8 4 1


has the following property.
If v is the position vector of any point P in R3, then w = Mv is the vector
obtained by reflecting v in a fixed plane Π, which passes through the origin.
Hence, in the diagram, |v| = |w| and |RP | = |RQ|, where R is the foot of the perpen-
dicular from P (or Q) to the plane.
v
P
Q
0
R
w
Π
a) Show that M reflects the vector a =

 2−1
2

 to the vector −a.
b) Hence write down the Cartesian equation of Π.
c) Find a non-zero vector u such that Mu = u.
c©2020 School of Mathematics and Statistics, UNSW Sydney
37
d) Find the shortest distance from the point B with position vector b =

 6−6
0

 to the
plane.
4. i) a) Using L’Hoˆpital’s rule or otherwise, indicate why
lim
x→∞ e
−xxn = 0
for any n ∈ N.
b) Show that for all x ≥ 1,
e−xxn < Cx−2,
where C is a positive constant.
c) Hence, or otherwise, show that the improper integral∫ ∞
1
e−xxn dx
converges for any n ∈ N.
ii) Suppose that f : [0, 2] → [0, 12] is continuous on its domain and twice differentiable on
(0, 2). Further suppose that f(0) = 0, f(2) = 12.
a) Explain why f ′(c) = 6 for some real number c ∈ (0, 2).
b) Suppose further that f ′(0) = 0, prove that f ′′(d) > 3 for some real number d ∈ (0, c).
iii) Let f : R→ R be the function defined by
f(x) = x− a tanhx
for some constant a ∈ R.
a) Explain why sech x < 1 for x > 0.
b) By considering the derivative, or otherwise, find the values of a for which f(x) > 0
for all x > 0? Give reasons for your answer
c) For which values of a does the equation
x = a tanhx
have a positive solution?
c©2020 School of Mathematics and Statistics, UNSW Sydney
38
iv) The area A(t) of an arbitrary convex quadrilateral with given side lengths a, b, c, d de-
pends on the sum t = α+ β of either pair of opposite angles, and is given by
A(t) =
√
(s − a)(s− b)(s − c)(s − d)− 1
2
abcd(1 + cos t),
where
s =
1
2
(a+ b+ c+ d).
a b
c
d
α β
a) Explain why the area A of a convex quadrilateral, with fixed side lengths a, b, c, d,
is maximal if the sum of either pair of opposite angles is π.
b) Show that the area function A : [0, π] → R as defined above is invertible, and that
the inverse function B is differentiable on (A(0), A(π)).
c) Show that
B′(A0) =
4A0
abcd
,
where
A0 =
√
(s− a)(s − b)(s − c)(s − d)− 1
2
abcd.
c©2020 School of Mathematics and Statistics, UNSW Sydney
39
MATH1141 JUNE 2012
2. i) Consider the function f : (0, 2
√
π ]→ R defined by
f(x) = x2 + cos(x2).
a) Find all critical points of f and determine their nature.
b) Explain why f is invertible, state the domain of f−1 and find f−1(5π/2).
c) Where is f−1 differentiable?
ii) Let
f(x) =
∫ x2−9x
0
e−t
2
dt.
a) Use the Mean Value Theorem to show that f has a stationary point x0 in the interval
[0, 9].
b) Find the value of x0 and determine the nature of the stationary point.
iii) Suppose that z lies on the unit circle in the complex plane.
a) Show that z +
1
z
is real.
b) Find the maximum value of z +
1
z
.
iv) Use De Moivre’s theorem to prove that
cos 4θ = 8cos4 θ − 8 cos2 θ + 1.
v) Consider the plane P with parametric vector form
x =

12
0

+ λ1

 11
−3

+ λ2

 3−1
−1

 , λ1, λ2 ∈ R.
a) Does the point a =

 44
−7

 lie on P?
b) Is the vector b =

13
2

 parallel to P?
c) Is the vector c =

12
1

 orthogonal to P?
3. i) Suppose that A is a 4 × 4 matrix with det(A) = 5. The matrix B is the result of
performing the following three elementary row operations on A:
1. first multiply the 3rd row by 7;
2. then replace the second row with twice the first row plus the second row;
3. then swap the first and last rows.
What is the value of det(B)?
c©2020 School of Mathematics and Statistics, UNSW Sydney
40
ii) Consider the line in R3,
x− 4 = −y = z − 5.
a) Write this line in parametric vector form.
b) Find the point on the line closest to the origin

00
0

.
iii) Consider the following Maple session, which defines a matrix A and a vector b ∈ R3:
> with(LinearAlgebra):
> A:=<<1,3,2>|<1,2,a>|<-2,2*a,4>>;
A :=


1 1 −2
3 2 2a
2 a 4


> b:=<1,2,-2>; 
 12
−2


> GaussianElimination();

1 1 −2 1
0 −1 2a+ 6 −1
0 0 −4 + 2a2 + 2a −2− a


For which values of a will the system Ax = b have
a) a unique solution,
b) no solutions,
c) infinitely many solutions?
iv) A matrix Q ∈Mnn(R) is said to be nilpotent (of degree 2) if Q2 = 0, the zero matrix.
a) Give an example of a non-zero 2× 2 nilpotent matrix.
b) Explain why a nilpotent matrix cannot be invertible.
Suppose now that S,Q ∈ Mnn(R) commute, that S is invertible and that Q is
nilpotent (of degree 2).
c) Prove that S−1Q = QS−1.
d) Show that S +Q is invertible by finding an integer k such that
(S +Q)(S−1 − S−kQ) = I.
v) Consider the non-zero vector x =

ab
c

 in R3 which makes angles α, β, γ with the three
coordinate axes respectively.
By considering dot products with the standard basis vectors e1, e2, e3, (or otherwise),
prove that
cos2 α+ cos2 β + cos2 γ = 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
41
vi) Suppose that A is an n × n matrix with the property that every vector b ∈ Rn can be
written uniquely as a linear combination of the columns of A. Prove that every vector
b ∈ Rn can also be written uniquely as a linear combination of the rows of A.
4. i) Consider the polar curve r = 1 + cos 2θ.
a) Prove that the curve is symmetric about the x axis and also about the y axis.
b) Sketch the curve.
(You are NOT required to find the derivative.)
ii) Let f(x) = tanhx (the hyperbolic tangent).
a) Express tanhx in terms of exponentials.
b) Sketch the graph y = f(x).
c) Show that
lim
x→∞
1− tanhx
e−2x
= 2 .
d) Explain why the improper integral∫ ∞
0
(1− tanhx) dx
converges.
e) Compute ∫ ∞
0
(1− tanhx) dx,
justifying your calculations.
iii) Suppose that f is a function whose derivative is continuous and hence bounded on [a, b],
with |f ′(x)| ≤ L for all x ∈ [a, b].
a) Show that for any n > 0,
∫ b
a
f(x) sinnx dx =
K(n)
n
+
1
n
∫ b
a
f ′(x) cosnx dx,
where K(n) = f(a) cos(na)− f(b) cos(nb).
b) Explain why ∣∣∣∣
∫ b
a
f ′(x) cos nx dx
∣∣∣∣ ≤ (b− a)L.
c) Find, with reasons,
lim
n→∞
∫ b
a
f(x) sinnx dx.
c©2020 School of Mathematics and Statistics, UNSW Sydney
42
MATH1141 JUNE 2013
2. i) Consider the function f : R→ R defined by
f(x) =
{
x3 if x < 0
x2 if x ≥ 0.
a) Explain why f is differentiable everywhere and determine f ′(x).
b) Explain why the function g defined by g(x) = f ′(x) is continuous at x = 0.
c) Use the definition of the derivative to determine whether g is differentiable at x = 0.
ii) Consider the function f(x) =
1
1 + x
defined on [0, 1] and let P be the partition {0, 1n , 2n , . . . , nn}.
a) Show that the lower Riemann sum LP (f) is given by
LP (f) =
n∑
k=1
1
n+ k
.
b) Assuming that the limits of the upper and lower Riemann sums are equal, evaluate
lim
n→∞
n∑
k=1
1
n+ k
.
iii) Let A, B and C be three points in the plane with corresponding position vectors a, b
and c.
a) Let M be the midpoint of the line joining A and B. What is the position vector m
of M?
b) Write a parametric vector equation for the line through C and M .
c) Suppose that
(b− a) · (c− a) = 1
2
|b− a| |c− a| and (c− b) · (a− b) = 1
2
|c− b| |a− b|.
Explain why the triangle ABC is equilateral.
iv) Consider the system of equations
x+ y − z = 2 (1)
x− y + 3z = 6 (2)
x2 + y2 + z2 = 10 (3)
[Note that equation (3) is NOT linear.]
a) Give, in parametric vector form, the set of points which satisfy the first two equations
(that is, (1) and (2)).
b) Describe this solution set geometrically.
c) Using the answer to (a), or otherwise, find all the points which satisfy all three
equations.
c©2020 School of Mathematics and Statistics, UNSW Sydney
43
3. i) Find the shortest distance from the plane
x =

12
4

+ λ1

 02
−1

+ λ2

−11
3

 , λ1, λ2 ∈ R
to the point p =

32
3

.
ii) Let p(z) = z4 − z3 − z2 − z + 2. Denote the roots of p by α1, α2, α3, α4, where α1 is an
integer.
a) Find the value of α1.
b) Given that at least one of the roots of p is not real, deduce how many of the roots
are real.
c) By considering the sum of the roots, or otherwise, prove that at least one of the
roots has negative real part.
d) Prove that |αj | > 12 for j = 1, 2, 3, 4.
iii) Which of the following statements are true for all non-zero 2 × 2 matrices A,B,C ∈
M2,2(R)? For those statements which are not always true, give a counterexample.
a) AB = BA.
b) det(AB) = det(BA).
c) If det(AB) = det(AC) then det(B) = det(C).
d) If AB = AC then B = C.
iv) a) Define what it means for a set of vectors {u1,u2, . . . ,uk} to be an orthonormal
set in Rn.
b) LetM be the matrix whose columns consist of the n orthonormal vectors, v1,v2, . . . ,vn
in Rn. By considering MTM or otherwise, find, with reasons, all possible values for
det(M).
4. i) Consider the function f : R→ R defined by
f(x) =
∫ x3
0
e−t
2
dt.
a) Determine, with reasons,
lim
t→∞ t
2e−t
2
.
b) Does the improper integral
I =
∫ ∞
0
e−t
2
dt
converge? Give reasons for your answer.
c) Find all critical points and asymptotes of f .
d) Carefully sketch the graph of f , clearly indicating the above information and any
other relevant features.
c©2020 School of Mathematics and Statistics, UNSW Sydney
44
ii) Let f be a differentiable function on (a, b), and take c ∈ (a, b). Define
q(x) =
f(x)− f(c)− f ′(c)(x − c)
(x− c)2 ,
where a < x < b and x 6= c.
Show that if f ′′(c) exists, then
lim
x→c q(x) =
f ′′(c)
2
.
iii) An oval of Cassini is a curve on the (x, y)-plane defined implicitly by
(
x2 + y2
)2 − 2 (x2 − y2)+ 1 = b.
The shape of the curve depends on the value of the positive constant b.
The plot below shows ovals of Cassini for several different values of b.
a) Show that the points on an oval of Cassini where the tangent is horizontal either lie
on the unit circle x2 + y2 = 1 or lie on the y-axis.
b) Determine all such points (x, y) for which the corresponding tangent is horizontal
and state carefully for which values of b > 0 these exist.
iv) Suppose f : [0, 2]→ [0, 8] is continuous and differentiable on its domain.
a) By considering the function g(x) = f(x) − x3, prove that there is a real number
ξ ∈ [0, 2] such that f(ξ) = ξ3, stating any theorems you use.
b) Now suppose that f(0) = 0 and f(2) = 8. Explain why f ′(η) = 4 for some real
η ∈ (0, 2), stating any theorems you use.
c©2020 School of Mathematics and Statistics, UNSW Sydney
45
MATH1141 JUNE 2014
3. i) Let g(x) = 3x − cos 2x − 1, x ∈ R. Explain why g has a differentiable inverse function
h = g−1 and calculate h′(−2).
ii) a) State carefully the Mean Value Theorem.
b) Use the Mean Value Theorem to prove that if a < b then
0 < tan−1 b− tan−1 a ≤ b− a.
c) Using (b) or otherwise, prove that the improper integral
I =
∫ ∞
1
tan−1
(
t+
1
t2
)
− tan−1 t dt
converges.
iii) Use the ǫ-M definition of the limit to prove that
lim
x→∞
ex
coshx
= 2.
iv) Consider the polar curve r = 1 + cos 4θ.
a) Determine the values of θ ∈ [0, 2π] for which r has the smallest and largest values.
b) Hence, or otherwise, sketch this polar curve. (You are not required to find the
slope.)
v) For x > 0, let f(x) = xx lnx.
a) Evaluate f ′(x).
b) Determine the values of x for which f ′(x) > 0 and the values of x for which f ′(x) < 0.
c) Given that lim
x→0+
f(x) = 1, sketch the graph y = f(x) for 0 ≤ x ≤ 2.
4. i) Find the conditions on b1, b2, b3 which ensure that the following system has a solution.
2x − 4z = b1
3x + y − 2z = b2
−2x − y = b3
ii) Let I, J and K be the points in R3 whose position vectors are the three standard basis
vectors i, j, and k respectively.
By considering vectors of the form

 xx
x

, find the position vector of a point A, not
the origin, such that the distances from A to I, J and K are all 1.
iii) Consider the complex matrix A =
(
2 i
1 + i α
)
.
a) Find A−1 in the case when α ∈ R.
b) Find all values of α ∈ C for which det(A2) = −1.
iv) You may assume that (z9 − 1) = (z3 − 1)(z6 + z3 + 1).
a) Explain why the roots of z6 + z3 + 1 = 0 are e±
2pii
9 , e±
4pii
9 , e±
8pii
9 .
c©2020 School of Mathematics and Statistics, UNSW Sydney
46
b) Divide z6 + z3 + 1 by z3 and let x = z +
1
z
.
Find a cubic equation satisfied by x.
c) Deduce that cos 2π9 + cos
4π
9 + cos
8π
9 = 0.
v) The norm ‖M‖ of an n×n matrix M is the maximum value that |Mu| takes for all unit
vectors u ∈ Rn.
a) Show that for any vector x ∈ Rn,
|Mx| ≤ ‖M‖|x|.
b) Suppose that M and N are any two n × n matrices. By considering MNu, or
otherwise, show that
‖MN‖ ≤ ‖M‖‖N‖.
c) What is the norm of the matrix
(
0 1
−2 0
)
c©2020 School of Mathematics and Statistics, UNSW Sydney
47
MATH1141 JUNE 2015
3. i) a) Carefully state the Mean Value Theorem.
b) Assume that a differentiable function f on R is such that f ′(x) ≤ 1 for all x ∈ R.
Given that f(2) = 2, show that f(x) ≥ x for all x ≤ 2.
ii) Let f be a continuous function on R and
g(x) =
∫ x
0 f(t) dt− xf(0)
x2
.
Use L’Hoˆpital’s rule to show that if f ′(0) exists then
lim
x→0
g(x) =
f ′(0)
2
.
iii) Let
f(x) =
∫ x3
0
(t2 − 1)et2dt.
a) Show that f is an odd function, that is, f(−x) = −f(x).
b) Find the stationary points of f .
c) By examining the sign of f ′(x) in a neighbourhood of each stationary point, or
otherwise, determine the nature of each stationary point.
iv) Consider the curve in the (x, y)-plane defined by the relation
xy(x2 + y2) = 1. (1)
a) Wherever y is an implicit function of x, determine dydx in terms of x and y.
b) For any point (x, y) on the curve with x, y ≥ 0, consider the area A = xy of a
rectangle of edge lengths x and y. Find the point (x, y) on the curve for which A is
stationary.
c) Find the polar form of the curve (1) and express A in terms of the radial polar
coordinate r.
d) Explain why the stationary point of A corresponds to the rectangle which has the
largest area.
4. i) Find the shortest distance from the point P (1, 2, 0) to the line with parametric vector
equation
x =

 10
1

+ λ

 01
1

 , λ ∈ R.
ii) a) By using the dot product, or otherwise, show that for any vectors a and b, the
following identity holds:
|a+ b|2 + |a− b|2 = 2|a|2 + 2|b|2.
b) Consider a parallelogram ABCD, as pictured below.
c©2020 School of Mathematics and Statistics, UNSW Sydney
48
A
B
C
D
If
−→
AB = a and
−→
AD = b, describe clearly in words the geometric interpretation of
the result in (a).
iii) a) Use De Moivre’s Theorem to express sin 5θ as a polynomial in
x = sin θ.
b) Consider the polynomial p(x) = 16x5 − 20x3 + 5x− 1.
Show that sin π10 is a root of p(x).
c) Using the fact that
16x5 − 20x3 + 5x− 1 = (x− 1)(4x2 + 2x− 1)2
find the distinct roots of p(x).
d) Evaluate sin π10 in surd form.
iv) Consider the matrix A =

 a b cd e f
g h i

. Suppose that det(A) = 7.
Find the value of the determinant of each of the following matrices
B =

 d e fa b c
d− 3a e− 3b f − 3c

 and C =

 g − 3a 2a a− dh− 3b 2b b− e
i− 3c 2c c− f

 .
v) Recall that the dot product of two vectors a and b can be written as aTb and that a
square matrix Q is said to be orthogonal if QTQ = I.
Let Q be a real orthogonal n× n matrix and suppose v is a non-zero vector in Rn.
a) Prove that the vectors Qv and v have the same length.
b) If v is a non-zero vector in Rn, such that Qv = λv, for some λ ∈ R, show that λ = 1
or λ = −1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
49
PAST EXAM SOLUTIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
50
MATH1131 NOVEMBER 2010 Solutions
1. i) a) |z| = √12 + 12 = √2
b) Arg z = π4
c)
z =
√
2ei
pi
4
z28 =
(√
2ei
pi
4
)28
= (2)
28
2 ei
28pi
4
= 214 (cos 7π + isin 7π )
= −214(= −16384)
ii)
(x+ iy) (3 + 2i) = 4 + 7i
x+ iy =
4 + 7i
3 + 2i
=
(4 + 7i)
(3 + 2i)
(3− 2i)
(3− 2i)
=
26 + 13i
13
= 2 + i
Equating real and imaginary parts, x = 2 and y = 1.
iii) a)
z6 = 1 = e2kπi, k ∈ Z
Hence z = e2kiπ/6. Substituting appropriate values for k, the solutions are
1, eiπ/3, e−iπ/3, e2iπ/3, e−2iπ/3,−1.
b) (
z6 − 1) = (z − 1)(z − e ipi3 )(z − e−ipi3 )(z − ei 2pi3 )(z − e−i 2pi3 ) (z + 1)
= (z − 1) (z + 1) (z2 − 2z cos(π/3) + 1) (z2 − 2z cos(2π/3) + 1)
=
(
z2 − z + 1) (z2 + z + 1) .
iv) |z + i| = 1 represents a circle, centred at z = −i, radius 1.
−i
Rez
Imz
b
c©2020 School of Mathematics and Statistics, UNSW Sydney
51
v) From MAPLE, A6 = 64I. Multiplying by (A2)−1, we have
A4 = 64(A2)−1
so
(
A2
)−1
=
1
64
A4 =
1
64

 −8 0 −8
√
3
0 −16 0
8
√
3 0 −8


vi) a)
200a + 600b+ 400c = 12000
1800a + 2400b + 2000c = 65600
20000a + 30000b + 26000c = 784000
b) 
 200 600 4001800 2400 2000
20000 30000 26000
∣∣∣∣∣∣
12000
65600
784000


−→

 2 6 418 24 20
20 30 26
∣∣∣∣∣∣
120
656
784


R2 − 9R1−−−−−−−−−−−→
R3 − 10R1

 2 6 40 −30 −16
0 −30 −14
∣∣∣∣∣∣
120
−424
−416


R3 +R2−−−−−−−−→

 2 6 40 30 16
0 0 2
∣∣∣∣∣∣
120
424
8


Hence,
c = 4
30b+ 16c = 424
b = 12
2a+ 6b+ 4c = 120
a = 16
i.e. (a, b, c) = (16, 12, 4) .
2. i) a)
PP T =
(
1 0 1
2 3 0
)1 20 3
1 0

 = (2 2
2 13
)
.
b) P TP is a 3× 3 matrix.
ii) a)
−−→
AB = b− a =

 0−2
−2

 ,−→AC = c− a =

22
4

.
c©2020 School of Mathematics and Statistics, UNSW Sydney
52
b) The plane has equation x =

12
3

+ λ

 0−2
−2

+ µ

22
4

 , λ, µ ∈ R.
iii) Substituting the x = 1 + t, y = 2− y, z = 5 + t into the equation of the plane, we have
(1 + t)− 3(2− t) + (5 + t) = 15⇒ t = 3.
Hence, the point of intersection is x =

12
5

+ 3

 1−1
1

 =

 4−1
8

 .
iv) Since the vectors are perpendicular, their dot product is 0, hence
1×−2 + 3×−6− β + 5×−10 = 0⇒ β = −70.
v) a) The magnitude of u is
√
22 + 12 + 72 =
√
54 = 3
√
6.
b) The vector 10
3
√
6
u is parallel to u and has length 10.
vi) One possible method is to row-reduce first, but in this case it is probably easier to expand
across the first row.
det(C) = 3det
(
1 7
2 0
)
− 1× det
(
4 7
1 0
)
+ 0 = −35.
vii) If projw(u) = projw(v) then
u.w
|w|2w =
v.w
|w|2w
⇒ u.w = v.w
since w is a non-zero vector. Hence, using the properties of dot product, (u− v).w = 0
and so u− v is perpendicular to w.
3. i) Since both the numerator and denominator are 0 at x = 0, we apply L’Hoˆpital’s rule.
L = lim
x→0
x2ex
1− cos(πx) = limx→0
(2x+ x2)ex
π sin(πx)
.
Again, both the numerator and denominator are 0 at x = 0, so we apply L’Hoˆpital’s
rule a second time.
L = lim
x→0
(2 + 4x+ x2)ex
π2 cos(πx)
=
2
π2
.
ii) As a necessary condition, f must be continuous at x = 1 so
lim
x→1+
f(x) = f(1).
Hence
lim
x→1+
(−x2 + ax+ b) = 1⇒ a+ b = 2.
For differentiability, we need to show that
lim
h→0−
f(1 + h)− f(1)
h
= lim
h→0+
f(1 + h)− f(1)
h
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
53
Now
LHS = lim
h→0−
f(1 + h)− f(1)
h
= lim
h→0−
(x+ h)2 − x2
h
= 2
and
RHS = lim
h→0+
f(1 + h)− f(1)
h
= lim
h→0−
−(x+ h)2 + a(x+ h) + b− (−x2 + ax+ b)
h
= −2+a.
Hence we have a = 4 and from a+ b = 2 it follows that b = −2.
iii) a)
L = lim
x→∞
x2 − 2
x2 + 1
= lim
x→∞
1− 2
x2
1 + 1x2
= 1.
b) ∣∣∣∣x2 − 2x2 + 1 − 1
∣∣∣∣ = 3x2 + 1 < 3x2 .
Hence, if 3
x2
< ε then
∣∣∣x2−2x2+1 − 1
∣∣∣ < ε. Now 3x2 < ε ⇒ x >
√
3
ε so let M =
√
3
ε and
then if x > M , we have
∣∣∣x2−2x2+1 − 1
∣∣∣ < ε.
iv) a) The function f is continuouson the closed interval [0, 2] and f(0) = −5 < 0, while
f(2) = 3 + 2
√
3 > 0. Hence by the intermediate value theorem, f has at least one
positive real root in the interval [0, 2].
b) Since f ′(x) = 3x2 +
√
3 > 0, the function f is increasing. Hence f has exactly one
real positive root.
v) Using logarithms,
log y = log(sinx)x = x log(sin x).
Hence
1
y
dy
dx
= log(sinx) + x
cos x
sinx
⇒ dy
dx
= (sinx)x(log(sinx) + x cot x).
vi) a) x = r cos θ = 6 sin θ cos θ, y = r sin θ = 6 sin2 θ,
so x2 + y2 = 36 sin2 θ(sin2 θ + cos2 θ) = 36 sin2 θ = 6y. Completing the square,
x2 + (y − 3)2 = 9.
From the above, since 0 ≤ θ ≤ π2 , we see that x ∈ [0, 3], y ∈ [0, 6] and so we have a
semicircle centre 3 radius 3 in the right half plane.
b)
3
3 b
b
c©2020 School of Mathematics and Statistics, UNSW Sydney
54
4. i) a) Using integration by parts with u = 1− x, v′ = 1(1+x)3 , we have
I1 = (1− x).
( −1
2(1 + x)2
)
−
∫
(−1). (−1)
2(1 + x)2
dx
=
x− 1
2(1 + x)2
+
1
2(1 + x)
+ C.
Alternatively you can make the substitution x = u− 1.
b) Using integration by parts with u = x, v′ = cos 2x we have
I2 =
[
x(
1
2
sin(2x))
]π
0
−
∫ π
0
1.
1
2
sin(2x) dx =
[
x
2
sin(2x) +
1
4
cos(2x)
]π
0
=
1
4
− 1
4
= 0.
ii)
K =
∫ ∞
1
1 + sinx
3x2
dx ≤
∫ ∞
1
2
3x2
dx
= lim
N→∞
∫ N
1
2
3x2
dx = lim
N→∞
2
3
− 2
3N
=
2
3
.
Hence the improper integral converges by comparison.
iii) a)
sinhx =
ex − e−x
2
, coshx =
ex + e−x
2
.
b) Replacing x with ax, we have
d
dx
cosh(ax) =
d
dx
eax + e−ax
2
=
aeax − ae−ax
2
= a sinh(ax).
c) Since cosh is an even function,
cosh−1
(
cosh(−4726)) = cosh−1(cosh(4726)) = 4726.
iv) Using the First Fundamental Theorem of Calculus,
d
dx
∫ x
0
f(t) dt =
d
dx
∫ 1
x
t2f(t) dt+
d
dx
(
x16
8
+
x18
9
− 1
9
)
⇒ f(x) = − d
dx
(∫ x
1
t2f(t) dt
)
+ 2x15 + 2x17 = −x2f(x) + 2x15 + 2x17.
Solving for f(x), we have
f(x) =
2x15 + 2x17
1 + x2
= 2x15.
v) a) Since tan−1 is defined for all real x, g(x) = tan−1(x) + tan−1(1/x) is defined for all
real x except x = 0. Hence the maximal domain for g is (−∞, 0) ∪ (0,∞).
b) For x > 0,
g′(x) =
1
1 + x2
+
−1
x2
1 + ( 1x)
2
=
1
x2 + 1
− 1
1 + x2
= 0.
Hence on (0,∞) g is constant. The same calculations hold for x ∈ (−∞, 0). Hence
g is piecewise constant on its domain. (Note that the constant is not the same in
the two intervals. On (0,∞), g(x) = π2 , while on (−∞, 0), g(x) = −π2 .
c) Since the expression is constant, put α = 0 then tan−1(−1) + tan−1(−1) = −π2 .
c©2020 School of Mathematics and Statistics, UNSW Sydney
55
MATH1131 June 2011 Solutions
1. i) a) |z| = √2.
b) Arg(z) = −3π4 .
c) z =
√
2e−3πi/4 and hence z102 = (
√
2)102e−153πi/2 = 251e−πi/2 = −251i.
ii) a) (2 + 4i)2 = −12 + 16i.
b) Applying the quadratic formula, (or by completing the square), we have z = 4±
√−12+16i
2 .
By (a),this simplifies to z = 4±(2+4i)2 , and so the two solutions are z = 3 + 2i or
z = 1− 2i.
iii) Diagram as below
π
4
Im
Re
i
iv) Rationalising the denominator,
lim
x→∞
1
x−√x2 − 6x− 4 = limx→∞
x+
√
x2 − 6x− 4
(x−√x2 − 6x− 4)(x+√x2 − 6x− 4)
= lim
x→∞
x+
√
x2 − 6x− 4
(x2 − (x2 − 6x− 4))
= lim
x→∞
x+
√
x2 − 6x− 4
6x+ 4
.
We now divide by the highest power of x in the denominator to obtain:
lim
x→∞
1 +
√
1− 6x − 4x2
6 + 4x
=
1
3
.
v) Applying the definition,∫ ∞
1
x−5/4 dx = lim
M→∞
∫ M
1
x−5/4 dx = lim
M→∞
[
−4x− 14
]M
1
= lim
M→∞
4− 4
M
1
4
= 4.
vi) a) Differentiating implicitly,
2x− 3y2 − 6xy dy
dx
= 0
and the result follows.
b) At the point (1, 2),
dy
dx
= −5
6
, so the equation of the tangent is y − 2 = −56(x− 1)
or 5x+ 6y = 17.
c©2020 School of Mathematics and Statistics, UNSW Sydney
56
c) implicitplot(x2 − 3xy2 + 11 = 0, x = 1..4, y = −5..5).
2. i) By De Moivre’s Theorem,
cos(4θ) + i sin(4θ) = (cos θ + i sin θ)4.
Expanding and equating the real parts on both sides, we have
cos(4θ) = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ.
Replacing the sine terms,
cos(4θ) = cos4 θ − 6 cos2 θ(1− cos2 θ) + (1− cos2 θ)2
and the result follows.
ii) a) Using the values t = 0, t = 1 we obtain the two points

12
4

 and

44
9

.
b)

32
5


c) The vector

 96
15

 = 3

32
5

 is normal to the plane and parallel to the direction of
the line. Hence the line is perpendicular to the plane.
d) t = −1 gives the required point

−20
−1

.
iii) From the MAPLE, we see that A8 = 16I. Multiplying both sides by (A7)−1, we have
(A7)−1 = 116A =
1
16

−1 −1 01 1 0
0 0
√
2

 .
iv) The expression is indeterminate at x = 1 and so we apply L’Hoˆpital’s Rule (twice),
lim
x→1
(x− 1)2
1 + cos(πx)
= lim
x→1
2(x− 1)
−π sin(πx) = limx→1
2
−π2 cos(πx) =
2
π2
.
v) Using integration by parts,∫
x sin(2x) dx = −x
2
cos 2x+
1
2
∫
cos 2x dx = −x
2
cos 2x+
1
4
sin 2x+C.
vi) a) f(0) = 1 < 2 and f(1) = e + a > 2 for a > 0. Since f is continuous, by the
intermediate value theorem, there is a c ∈ (0, 1) such that f(c) = 2 and so 2 is in
the range of f .
b) f ′(x) = ex + a > 0 for a > 0 and for all x. Thus the function f is continuous and
increasing and so f has an inverse.
c) The domain of f−1 is a the range of f and since f is increasing, the range is
[f(0), f(1)] = [1, e + a].
3. i) a) AB =
(
8 23
5 11
)
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
57
b) BA is a 3× 3 matrix.
ii) a) Looking at the dry fruit eaten, we have
50x+ 300y + 100z = 2900.
Dividing this equation by 2 gives the desired result.
b)
5x+ 30y + 10z = 290
4x+ 32y + 20z = 392
x+ 5y + 3z = 63
c) 
 5 30 10 2904 32 20 392
1 5 3 63

→

 1 5 3 630 1 −1 −5
0 0 0 20


Back substitution yields, x = 8, y = 5, z = 10 so there are 8 hamsters, 5 rabbits, 10
guinea pigs.
iii) x =

10
1

+ λ

13
3

+ µ

31
4

, λ, µ ∈ R.
iv) a) (
1 1 −1 2
2 3 1 6
)
→
(
1 1 −1 2
0 1 3 2
)
Let z = λ, then by back substitution, x = 4λ, y = 2− 3λ.
So x =

02
0

+ λ

 4−3
1

 , λ ∈ R.
b) x+ y + z = 2 + 2λ = 0 when λ = −1 and then x =

−45
−1


v) a) |
−→
OB | = √35.
b) Area of triangle AOB = 12 |a× b
=
1
2
∣∣∣∣∣∣
∣∣∣∣∣∣
i j k
1 2 4
3 1 5
∣∣∣∣∣∣
∣∣∣∣∣∣ =
1
2

 67
−5

 = 1
2
√
110.
c) Since the area of a triangle is half the base time perpendicular height, the perpen-
dicular distance from A to the line through O and B is 12
√
110 ÷ 12
√
35 =
√
154/7.
vi) (u − v)(u + v) = u.u + u.v − v.u − v.v = |u|2 − |v|2 = 0 since the vectors have the
same magnitude. Hence the vectors u− v and u+ v are perpendicular.
4. i) a) coshx = 12(e
x + e−x).
b) 4 cosh3 x = 4(12 (e
x + e−x))3 = 12(e
3x + e−3x + 3ex + 3e−x) = cosh 3x+ 3cosh x.
c©2020 School of Mathematics and Statistics, UNSW Sydney
58
ii) a) By the Fundamental Theorem of Calculus,
d
dx
∫ x
0
cos t√
1 + t2
dt =
cosx√
1 + x2
.
b) By the Fundamental Theorem of Calculus, and the chain rule,
d
dx
∫ sinhx
0
cos t√
1 + t2
dt =
cosh x× cos(sinhx)√
1 + sinh2 x
= cos(sinhx).
iii) a) Diagram as below,
b) x = r cos θ = 6cos θ sin(2θ) and y = r sin θ = 6 sin θ sin(2θ).
Hence dydθ = 12 cos(2θ) sin θ+6 sin(2θ) cos θ =
15
2 at θ =
π
6 . Similarly,
dx
dθ = 12 cos(2θ) cos θ−
6 sin(2θ) sin θ = 32
√
3 at θ = π6 .
Hence, dydx =
dy
dθ/
dx
dθ = 5
√
3/3.
iv) a) Write y = cot−1 t, then t = cot y so
dt
dy
= −cosec2 y = −(1 + cot2 y) = −(1 + t2).
The result follows.
b) Write θ = (θ + φ)− φ and use the two right triangles.
c)
dθ
dx
= − 3
9 + x2
+
1
1 + x2
= 0 for a maximum. Solving this yields x =
√
3 (since
x > 0). To show it is a maximum, we note that d
2θ
dx2
is negative at x =
√
3.
c©2020 School of Mathematics and Statistics, UNSW Sydney
59
MATH1131 June 2012 Solutions
1. i) a) u− 2w = 1 + 12i.
b) u/w = − 726 + i1726 .
ii) a) |z| = 2 and Arg(z) = −π6 ..
b) z = 2e−i
pi
6 .
c) z10 = 210e−
5pi
3 and so z10 = 210e
5pi
3 . Hence z10 + (z)10 = 2× 210 cos 5π
3
= 210.
iii) Expanding down the first column,∣∣∣∣∣∣
1 −1 4
0 2 7
0 3 1
∣∣∣∣∣∣ =
∣∣∣∣2 73 1
∣∣∣∣ = −19.
iv) a) lim
x→∞
3x2 + sin(2x2)
x2
= lim
x→∞
3 + sin(2x2)/x2
1
= 3.
b) Applying L’Hopital’s Rule,
lim
x→0
3x2 + sin(2x2)
x2
= lim
x→0
6x+ 4x cos(2x2)
2x
= lim
x→0
6 + 4 cos(2x2)
2
= 5.
v) Differentiating implicitly,
2x− 5 sin y − 5x cos y dy
dx
+ 2y
dy
dx
= 0
and so
dy
dx
=
2
5
at (2, 0). Hence the equation of the tangent at this point is y = 25(x− 2)
or 2x− 5y = 4.
vi) a) p is a polynomial and so is continuous. Since p has degree 5, for large positive x,
p(x) > 0 and for large negative x, p(x) < 0, so by the Intermediate Value Theorem,
p(x) has at least one real root.
b) p′(x) = 5x4 + 5 > 0 for all x and so p(x) is strictly increasing. Hence p has at most
one real root.
2. i) We use De Moivre’s theorem and then the binomial formula to find
cos 3θ + i sin 3θ = (cos θ + i sin θ)3
= cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ.
Equating real parts then gives
cos 3θ = cos3 θ − 3 cos θ sin2 θ
= cos3 θ − 3 cos θ(1− cos2 θ)
= 4 cos3 θ − 3 cos θ.
The result is shown.
ii) We solve equations simultaneously using x = 1 + λ, y = λ, z = 2 + 2λ to find
17 = 5(1 + λ)− 2λ+ (2 + 2λ) = 5λ+ 7.
c©2020 School of Mathematics and Statistics, UNSW Sydney
60
Solving for λ we find λ = 2. The point of intersection can now be obtained from the
parametric equation for the line as
x =

10
2

+ 2

11
2

 =

32
6

 .
iii) Below we use the MAPLE output giving A4 = I and the output for A3. The inverse to
A2001 is
A−2001 = A4×(−501)+3 = (A4)−501A3 = I−501A3 = A3 =

 0 0 10 1 0
−1 0 0

 .
iv) a) We compute
c× d =

e1 e2 e31 2 3
4 1 5

 =

 77
−7

 .
b) The area of the parallelogram with sides c,d is |c× d| =
√
73 + 73 + (−7)3 = 7√3.
v) We use the integration by parts formula
∫
udv = uv − ∫ vdu with u = lnx, v = 15x5 so
du = 1xdx, dv = x
4dx.
∫
x4 lnxdx =
1
5
x5 lnx−
∫
1
5
x5
1
x
dx
=
1
5
x5 lnx−
∫
1
5
x4dx
=
1
5
x5 lnx− 1
25
x5dx+ C
where C is an arbitrary constant
vi) We first note that the functions lnx, sinx, cos x are all continuous on their domains.
Hence the same is true of h(x). Furthermore, the domain of h is the closed interval
[1, 5] which has finite length. It follows by the max-min theorem that h(x) must attain
a maximum value.
vii) First note that r is an even function of θ so the graph symmetric about the x-axis.
We now compute some values
r(0) = 2− 2 cos 0 = 0
r(π/2) = 2− 2 cos π/2 = 2− 0 = 2
r(π) = 2− 2 cos π = 2− 2(−1) = 4
Note that in general, as θ increases from 0 to π, r(θ) decreases from 0 to 4. Hence the
polar curve spirals out as you rotate anti-clockwise or clockwise from the positive x-axis
to the negative x-axis.
Note that the x-intercepts are ±2 while the y-intercepts are 0,−4. The only axis of
symmetry is the x-axis.
c©2020 School of Mathematics and Statistics, UNSW Sydney
61
viii) We differentiate implicitly with respect to x, the equation ln y = ln(xsinx) = sinx lnx.
1
y
dy
dx
= cos x lnx+
sinx
x
Hence
dy
dx
= y(cos x lnx+
sinx
x
) = xsinx(cos x lnx+
sinx
x
)
3. i) a) We have
−−→
AB = b− a =

 12
4

−

 10
1

 =

 02
3

. Therefore l is
x =

 10
1

+ λ

 02
3

 , λ ∈ R.
b)

 02
3

 ·

 1−3
2

 = 0− 6 + 6 = 0. Therefore the two lines are perpendicular.
ii) a) Note that QT =

 1 2−1 5
1 0

. Thus
PQT =
(
1 2 1
3 −1 4
) 1 2−1 5
1 0

 = ( 0 12
8 1
)
.
b) P T is a 3× 2 and Q is a 2× 3 matrix. Therefore, P TQ is a 3× 3 matrix.
c) No as the number of columns of P doesn’t equal to the number of rows of Q.
c©2020 School of Mathematics and Statistics, UNSW Sydney
62
iii) a) Note that α2 − 9 = (α− 3)(α+ 3) and so the matrix becomes:
 1 2 3 50 2 4 8
0 0 (α− 3)(α + 3) α− 3

 .
If α = −3 the last column will be leading, and so this will lead to no solutions.
b) If α = 3 both the thirst and last columns will be non-leading, implying infinitely
many solutions.
c) If α = 3 we get 
 1 2 3 50 2 4 8
0 0 0 0

 .
Let z = λ. Back substituting, we get 2y + 4λ = 8 and so y = 4 − 2λ. Also,
x+ 2(4 − 2λ) + 3λ = 5 and thus we get x = −3 + λ. Thus
x =

 −34
0

+ λ

 1−2
1

 , λ ∈ R
iv) a) Since the total value if $1020 we have
10x + 20y + 50z = 1020
and after dividing by 10 we get
x+ 2y + 5x = 102.
b) Since there are 44 notes in total, we get
x+ y + z = 44.
The last line in the question implies x = y + z or equivalently
x− y − z = 0.
Putting this in augmented matrix form and row reducing we get:
 1 −1 −1 01 1 1 44
1 2 5 102

 R2→R2−R1−−−−−−−→
R3→R3−R1

 1 −1 −1 00 2 2 44
0 3 6 102


R3→ 12R3−−−−−−→
R3→ 13R3

 1 −1 −1 00 1 1 22
0 1 2 34

 R3→R3−R2−−−−−−−→

 1 −1 −1 00 1 1 22
0 0 1 12

 .
Back substituting we get z = 12, y = 10, x = 22.
v) a) We have
a · e1 = |a||e1| cosα
and so a =
√
a2 + b2 + c2 cosα.
c©2020 School of Mathematics and Statistics, UNSW Sydney
63
b) Similarly to (a) we get b =
√
a2 + b2 + c2 cos β and c =
√
a2 + b2 + c2 cos γ. Squaring
and adding the above three equations gives a2 + b2 + c2 = (a2 + b2 + c2)(cos2 α +
cos2 β + cos2 γ) and so we get
cos2 α+ cos2 β + cos2 γ = 1.
c) If α+β = 90◦ then cos β = sinα. From this we get cos2 α+cos2 β = cos2 α+sin2 α =
1 and so from (b) we see that cos2 γ = 0. This implies γ = π/2 i.e.
−−→
OQ is parallel
to the XY -plane.
4. i) Let p(x) = e3x. For h to be differentiable at x = 0, we need
• p(0) = q(0), so that h is continuous at 0;
• p′(0) = q′(0).
As p′(x) = 3e3x and q′(x) = 2x+ b, these conditions say
• 1 = c;
• 3 = b,
and so q(x) = x2 + 3x+ 1.
ii) a) In mathematical notation this is sin−1 (sin(7π/3)). Evaluating (or just writing this
down from thinking about the quadrant)
sin−1 (sin(7π/3)) = sin−1(
√
3/2) =
π
3
.
b) In mathematical notation this is
d
dx
∫ x2
0
et
2
dt.
To evaluate this, let F (u) =
∫ u
0 e
t2 dt and let u(x) = x2. Then
d
dx
∫ x2
0
et
2
dt =
d
dx
F (u(x))
= F ′(u(x))u′(x) (by the chain rule)
= eu(x)
2
2x (by the FTC)
= 2xex
4
.
iii) Suppose that ǫ > 0 is given. Put M =
√
5
ǫ then for x ≥M ,∣∣∣∣x2 − 2x2 + 3 − 1
∣∣∣∣ =
∣∣∣∣x2 − 2− x2 − 3x2 + 3
∣∣∣∣ = 5x2 + 3 ≤ 5x2 ≤ 5M2 = ǫ.
iv) a) f is differentiable and
f ′(x) = 3x2 + cosh x ≥ coshx ≥ 1
and so f ′(x) is never zero. It follows by the inverse function theorem that f has a
differentiable inverse g.
b) As f(x)→ −∞ as f → −∞ and f(x)→∞ as x→∞
Dom(g) = Ran(f) = R.
c©2020 School of Mathematics and Statistics, UNSW Sydney
64
c) The inverse function theorem says that
g′(x) =
1
f ′(g(x))
.
We need to find y = g(1). That is, solve f(y) = y3 + sinh y + 1 = 1. By inspection,
the solution is y = 0. Plugging this into f ′(x) from (a):
g′(1) =
1
f ′(g(1))
=
1
f ′(0)
=
1
1
= 1.
v) Let V (t) denote the volume of Factor X in the tank at time t ≥ 0.
We are given that V ′(t) =
100
10 + t2
.
Thus after T hours the amount in the tank is
V (T ) =
∫ T
0
100
10 + t2
dt =
[
10
√
10 tan−1
t√
10
]T
0
= 10
√
10 tan−1
T√
10
.
As T →∞,
tan−1
T√
10
→ π
2
and so V (T ) increases towards its limit 5
√
10π ≈ 49.6729 < 50.
In particular, the tank is just big enough to never overflow.
c©2020 School of Mathematics and Statistics, UNSW Sydney
65
MATH1131 June 2013 Solutions
1. i) a) 2
b)
3
7
ii)
x
f(x)
20
1
3
4
5
iii) a) coshx = 12(e
x + e−x) sinhx = 12(e
x − e−x)
b) Proof.
cosh2 x− sinh2 x =
(1
2
(ex + e−x)
)2
−
(1
2
(ex − e−x)
)2
=
1
4
(e2x + e−2x + 2)− 1
4
(xe2x + e−2x − 2) = 1
4
× 4 = 1 .
iv) a) 16 + 7i
b) 10i
c) 3 + i
v) x = 3
y = −1
vi) a)
R
iR
3
b)
9π
2
vii) A =
1
2
(3I −A2) = 1
2

 1 1 −11 1 1
−1 1 1


2. i)
h(x) =
{
ax2 + 3x, if x ≥ 1
2x+ d, if x < 1.
Both branches are polynomials and are therefore elementary functions. This means they
are both continuous and differentiable x 6= 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
66
Differentiating for x 6= 1:
h′(x) =
{
2ax+ 3, if x > 1
2, if x < 1.
If h(x) is differentiable at x = 1 then h(x) is continuous at x = 1. i.e.
h(1) = lim
x→1+
h(x) = lim
x→1−
h(x)
lim
x→1+
ax2 + 3x = lim
x→1−
2x+ d
a+ 3 = 2 + d
d = a+ 1.
Also, h′(x) has to be continuous at x = 1, i.e. we require
lim
x→1+
h′(x) = lim
x→1−
h′(x)
lim
x→1+
2ax+ 3 = lim
x→1−
2
2a+ 3 = 2
a = −1
2
.
But d = a+ 1, so
a = −1
2
and d =
1
2
.
ii)
∫ ln 2
0
9xe3xdx =
[
9x
e3x
3
]ln 2
0
−
∫ ln 2
0
9
e3x
3
dx
= 3(ln 2)e3 ln 2 − 0− 3
[
e3x
3
]ln 2
0
= 3(ln 2)23 − e3 ln 2 + e0
= 24 ln 2− 7.
iii)
ex + sin y = xy + 1.
Differentiating implicitly with respect to x throughout, using the chain rule and the
product rule:
ex + cos y
dy
dx
= x
dy
dx
+ 1y + 0
Evaluate this at (x, y) = (0, 0):
e0 + cos 0
dy
dx
= 0
dy
dx
+ 0 + 0
1 +
dy
dx
= 0
dy
dx
= −1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
67
The tangent is a line passing through (x, y) = (0, 0) with gradient 1. The equation is
therefore
y − 0 = −1(x− 0)
y = −x.
iv) Diagram as below,
bb
b
x
y
2−2
v) a)
|z| = √2 + 2 = 2.
b)
Arg(z) = tan−1
(
−√2√
2
)
= −π
4
.
c)
z = 2e−
pi
4
i
z6 = 26e−
6pi
4
i
= 26e−
3pi
2
i
= 26i
= 64i.
vi) a)
AB =
(
2 0
1 −1
)(
3 0 1
1 2 0
)
=
(
6 0 2
2 −2 1
)
.
b) ABT does not exist as A is order (2×2) and BT is order (3×2), and so the number
of columns in A is not equal to the number of rows in BT .
vii) a)
z5 = −1 = ei(π+2kπ), k ∈ Z.
c©2020 School of Mathematics and Statistics, UNSW Sydney
68
So,
z = ei
pi
5
(2k+1)
= ei
pi
5 , e−i
pi
5 , ei
3pi
5 , e−i
3pi
5 ,−1, listing the principal set.
b) Using these solutions,
z5 + 1 =
(
z − eipi5
)(
z − e−ipi5
)(
z − ei 3pi5
)(
z − e−i 3pi5
)
(z + 1)
=
(
z2 − 2 cos
(π
5
)
z + 1
)(
z2 − 2 cos
(
3π
5
)
z + 1
)
(z + 1) .
3. i) The line

xy
z

 =

 1−1
−1

+ t

12
1

 can be written as


x = 1 + t
y = −1 + 2t
z = −1 + t
.
For the point of intersection, we substitute the parametric equations for the line into the
Cartesian equation of the plane.
4(1 + t)− 5(−1 + 2t) + 3(−1 + t) = 0
−3t+ 6 = 0
t = 2
Hence the point of intersection is

 1 + 2−1 + 2(2)
−1 + 2

 =

33
1

.
ii) a) Expand det(M) along the first row.
∣∣∣∣∣∣
1 2 0
2 5 1
0 2 α
∣∣∣∣∣∣ =
∣∣∣∣5 12 α
∣∣∣∣− 2
∣∣∣∣2 10 α
∣∣∣∣ = (5α − 2)− 2(2α) = α− 2.
b) M does not have an inverse if and only if det(M) = 0. That is, α = 2.
c) When α = 1, the matrix M is invertible. We can find the inverse of M by reducing
c©2020 School of Mathematics and Statistics, UNSW Sydney
69
the augmented matrix (M |I) to reduced row-echelon form.
 1 2 0 1 0 02 5 1 0 1 0
0 2 1 0 0 1

 R2 = R2 − 2R1−−−−−−−−−−−−−−−→

 1 2 0 1 0 00 1 1 −2 1 0
0 2 1 0 0 1


R3 = R3 − 2R2−−−−−−−−−−−−−−−→

 1 2 0 1 0 00 1 1 −2 1 0
0 0 −1 4 −2 1


R3 = −R3−−−−−−−−−−−−−−−→

 1 2 0 1 0 00 1 1 −2 1 0
0 0 1 −4 2 −1


R2 = R2 −R3−−−−−−−−−−−−−−→

 1 2 0 1 0 00 1 0 2 −1 1
0 0 1 −4 2 −1


R1 = R1 − 2R2−−−−−−−−−−−−−−−→

 1 0 0 −3 2 −20 1 0 2 −1 1
0 0 1 −4 2 −1


Hence M−1 =

 −3 2 −22 −1 1
−4 2 −1

.
iii) a) The cross product
u× v =
∣∣∣∣∣∣
e1 e2 e3
2 1 0
3 1 1
∣∣∣∣∣∣ = e1
∣∣∣∣1 01 1
∣∣∣∣− e2
∣∣∣∣2 03 1
∣∣∣∣+ e3
∣∣∣∣2 13 1
∣∣∣∣ =

 1−2
−1

 .
b) The vector u× v is perpendicular to the plane. Hence, a point-normal form of the
plane is 


xy
z

−

14
2



 ·

 1−2
−1

 =

00
0


Therefore, the Cartesian equation of the plane is x− 2y − z = −9.
iv) a) The vectors u and v are orthogonal if and only if u · v = 0. That is,
0 + 3− 12 + 3β = 0, i.e. β = 3.
b) For the value β = 1,
projuv =
(
u · v
|u2|
)
u =
0 + 3− 12 + 3
22 + 12 + 42 + 32


2
1
4
3

 = −15


2
1
4
3

 .
v) a) m =
1
2
(a+ b) =
(
2
3
2
)
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
70
b) Since
−−→
AB =
(
0
1
)
, we may take u =
(
1
0
)
which is perpendicular to
−−→
AB.
c) The perpendicular bisector of AB is the line whose points are equidistant from A
and B. A parametric vector equation for this line is
x =
(
2
3
2
)
+ λ
(
1
0
)
, λ ∈ R.
d) Since the centre is equidistant from A, B and D, the centre is the intersection of
the two perpendicular bisectors. At the point of intersection, we have(
2
3
2
)
+ λ
(
1
0
)
=
(
3
2
)
+ µ
(
0
1
)
λ
(
1
0
)
− µ
(
0
1
)
=
(
1
1
2
)
At the intersection, the value of λ is 1. Hence the position vector of the point of
intersection is
(
2
3
2
)
+
(
1
0
)
=
(
3
3
2
)
.
4. i) Let u = ln(x), and so du =
dx
x
. Substituting:
∫
cos(ln(x))
x
dx =
∫
cos u du = sinu+ C = sin(ln(x)) +C.
ii) Let f(x) = e−x and g(x) = −e−x. Then for all x
g(x) ≤ e−x sin(x) ≤ f(x).
Since
lim
x→∞ g(x) = limx→∞ f(x) = 0
the Pinching theorem implies that lim
x→∞ e
−x sin(x) exists and also equals 0.
iii) For all x ≥ 0, we have 0 ≤ 1
x2 + ex
≤ e−x.
Now
∫ ∞
0
e−x dx = lim
R→∞
[−e−x]R0 = limr→∞
(−e−R + 1) = 1 converges.
Thus, by the comparison test
∫ ∞
0
dx
x2 + ex
converges too.
iv) a) F =
d
dx
∫ x2
0
sin(
√
t) dt.
b) Let u(x) = x2 and let g(u) =
∫ u
0 sin(
√
t) dt. Then, using the chain rule, F =
d
dx
g(u(x)) = g′(u(x))u′(x).
By the Fundamental Theorem of Calculus g′(u) = sin(
√
u), and so
F = sin(
√
x2) · 2x = 2x sin(|x|).
c©2020 School of Mathematics and Statistics, UNSW Sydney
71
v) a) Since p is a polynomial it is continuous on [1, 2]. Now p(1) = −2 < 0 and p(2) = 9 >
0, so by the Intermediate Value theorem there is a value c ∈ (1, 2) so that p(c) = 0.
b) p′(x) = 3x2 + 4 > 0 for all x ∈ [1, 2]. Therefore p is increasing and hence can only
take on any value at most once. Combined with (a) this implies that p has exactly
one root in [1, 2].
c) The Inverse Function theorem says that g is differentiable with
g′(0) =
1
p′(α)
=
1
3α2 + 4
.
vi) Let
f(x) = ln(1 + x)− x
1 + x
.
We need to show that f(x) > 0 for all x > 0.
Suppose then that x > 0. Since f is continuous on [0, x] and differentiable on (0, x), the
Mean Value Theorem says that there exists c ∈ (0, x) such that
f ′(c) =
f(x)− f(0)
x− 0 =
f(x)
x
.
Now, differentiating f at c,
f ′(c) =
1
1 + c
− (1 + c)− c
(1 + c)2
=
(1 + c)− 1
(1 + c)2
=
c
(1 + c)2
and so for any c ∈ (0, x), f ′(c) > 0. It follows then that
f(x) = xf ′(c) > 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
72
MATH1131 June 2014 Solutions
1. i) a) 2z − w = 10 + 5i
b) 5(w − i)/z = 3 + i.
c) |zw| = |z||w| = √50× 5 = 25√2.
d) zw = 25 + 25i and so Arg(zw) = π4 .
e) Since both z and w lie in the first quadrant, the sum of their arguments lies between
0 and π. Hence Arg(z)+ Arg(w) =Arg(zw) = π/4.
f) zw = 25
√
2e
ipi
4 so (zw)40 = (25
√
2)e10πi = 580220.
ii) a)
[A|b] =

 1 −1 −1 11 −3 1 1
2 −3 −1 2

→

 1 −1 1 10 1 −1 0
0 0 0 0


b) Let z = λ then y = λ and x = 1 + 2λ. Hence the general solution is
 xy
z

 =

 10
0

+ λ

 21
1

 , λ ∈ R.
iii) a)
lim
x→∞
6x2 + sinx
4x2 + cos x
= lim
x→∞
6 + sinxx2
4 + cos x
x2
=
3
2
since cos x and sinx are bounded.
b) Using L’Hoˆpital’s Rule twice, and checking the necessary conditions,
lim
x→0
e2x − 2x− 1
4x2
= lim
x→0
2e2x − 2
8x
= lim
x→0
4e2x
8
=
1
2
.
iv) For x > 4, g(x) = x
2−16
x−4 = x+ 4 and so lim
x→4+
g(x) = 8.
For 3 < x < 4, g(x) = −(x
2−16)
x−4 = −(x+ 4) and so lim
x→4−
g(x) = −8.
Since these limits have different values, so value of g(4) may be given to make g contin-
uous at x = 4.
v) a) f is a polynomial and so is continuous on [0, 2].
Now f(0) = −2 < 0 and f(2) = 40 > 0 and so by the intermediate value theorem,
f has at least one zero in the interval (0, 2).
b) f ′(x) = 5x4 + 3x2 + 1 > 1 for all real x and so f is an increasing function. Hence f
has only one real root.
2. i) Let u = log x, dudx =
1
x ∫
dx
x
(
1 + (log x)2
) = ∫ du
1 + u2
= tan−1 u+ C
= tan−1(log x) + C
c©2020 School of Mathematics and Statistics, UNSW Sydney
73
ii) a)
sinhx =
ex − e−x
2
cosh x =
ex + e−x
2
b)
RHS = 2 sinhx coshx
= 2
(
ex − e−x
2
)(
ex + e−x
2
)
=
1
2
(
e2x − e−2x)
= sinh(2x)
= RHS
i.e. sinh(2x) = 2 sinhx cosh x.
iii)
∫ pi
3
0
x sin(2x)dx =
[
x
(
−cos(2x)
2
)]pi
3
0
−
∫ pi
3
0
1
(
−cos(2x)
2
)
dx
=
π
3
(
−cos
(
2π
3
)
2
)
− 0 + 1
2
[
sin(2x)
2
]pi
3
0
=
π
12
+
1
4
√
3
2
=
π
12
+
√
3
8
.
iv)
(
ATA
)−1 (
ATA
)T
=
(
ATA
)−1 (
ATA
)
= I
v) a)
Π =

 01
3

+ λ1



 31
4

−

 01
3



+ λ2



 32
4

−

 01
3



 , λ1, λ2 ∈ R
=

 01
3

+ λ1

 30
1

+ λ2

 31
1

 , λ1, λ2 ∈ R.
c©2020 School of Mathematics and Statistics, UNSW Sydney
74
b)
n = ~AB × ~AC
=



 31
4

−

 01
3



×



 32
4

−

 01
3




=

 30
1

×

 31
1


=
∣∣∣∣∣∣
e1 e2 e3
3 0 1
3 1 1
∣∣∣∣∣∣
= e1(0× 1− 1× 1)− e2(3× 1− 1× 3) + e3(3× 1− 0× 3)
=

 −10
3

 .
c) Π in cartesian coordinates is
−1(x− 0) + 0(y − 1) + 3(z − 3) = 0
−x+ 3(z − 3) = 0.
d)
Area ABC =
1
2
∣∣∣ ~AB × ~AC∣∣∣
=
1
2
|n|
=
1
2
√
(−1)2 + (0)2 + (3)2
=
√
10
2
.
e)
~PA =

 01
3

−

 53
0

 =

 −5−2
3

 .
c©2020 School of Mathematics and Statistics, UNSW Sydney
75
min distance =
∣∣∣ ~PA n∣∣∣
|n|
=
∣∣∣∣∣∣

 −5−2
3



 −10
3


∣∣∣∣∣∣∣∣∣∣∣∣

 −10
3


∣∣∣∣∣∣
=
5 + 0 + 9√
(−1)2 + (0)2 + (3)2
=
14√
10
.
vi)
F = 2A+B + C
I = 2A+A2 +A3
A−1 = 2I +A+A2
= 2I +A+B
= 2

 1 0 00 1 0
0 0 1

+

 0 1 −21 −1 1
1 −1 0

+

 −1 1 10 1 −3
−1 2 −3


=

 1 2 −11 2 −2
0 1 −1

 .
3. i) a) DIAGRAM
b) |F| =
√
(
√
2− 1)2 + (√2− 1)2 =
√
6− 4√2;
Equivalently, |F| = 2−√2 (these two expressions are equal)
and F has direction North-East.
ii) a) ∣∣∣∣∣∣
1 −1 1
2 1 3
−1 2 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣
1 −1 1
0 3 1
0 1 2
∣∣∣∣∣∣ = −
∣∣∣∣∣∣
1 −1 1
0 1 2
0 3 1
∣∣∣∣∣∣ = −
∣∣∣∣∣∣
1 −1 1
0 1 2
0 0 −5
∣∣∣∣∣∣ = 5
b) N−1 =
1
3× 2− 1× 4
(
2 −1
−4 3
)
=
1
2
(
2 −1
−4 3
)
iii) a) p(1) = 14 + 2 · 12 − 3 = 0 and p(−1) = (−1)4 + 2(−1)2 − 3 = 0
b) p(z) = (z2 − 1)(z2 + 3).
c) ±1,±√3 i.
d) p(z) = (z − 1)(z + 1)(z −√3 i)(z +√3 i)
iv) Substitute x = 3 + λ, y = 2 + 2λ, and z = 1 + 3λ into the Cartesian equation:
0 = 6(3 + λ) + 8(2 + 2λ)− 9(1 + 3λ) = 25− 5λ .
c©2020 School of Mathematics and Statistics, UNSW Sydney
76
We see that λ = 5,
so the point of intersection is represented by ~x =

32
1

+ 5

12
3

 =

 812
16

.
v) a) Proof. By De Moivre’s Theorem,
cos 3θ + i sin 3θ = ei3θ = (eiθ)3 = (cos θ + i sin θ)3
= cos3 θ + i3 cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ .
By comparing the real parts of the equation, we see that
cos 3θ = cos3 θ − 3 cos θ sin2 θ = cos3 θ − 3 cos θ(1− cos2 θ) = 4 cos3 θ − 3 cos θ .
Hence, 4 cos3 θ = 3cos θ + cos 3θ.
b) By part a) with θ =
π
9
,
q
(
2 cos
π
9
)
= 8cos3
π
9
−6 cos π
9
−1 = 2
(
3 cos
π
9
+cos
3π
9
)
−6 cos π
9
−1 = 21
2
−1 = 0 .
Hence, 2 cos
π
9
is a root of the polynomial q(z) = z3 − 3z − 1.
vi) a) ~u · ~v = 1− β, so the vectors are orthogonal when β = 1.
b)
projv u =
~u · ~v
|~v|2 ~v =

 01
−1

 ·

21
0


∣∣∣∣∣∣

21
0


∣∣∣∣∣∣
2

21
0

 = 1
5

21
0


c)
1√
2
= cos
π
4
=

 01
−1

 ·

21
β


∣∣∣∣∣∣

 01
−1


∣∣∣∣∣∣
∣∣∣∣∣∣

21
β


∣∣∣∣∣∣
=
1− β√
2
√
5 + β2
,
so 5 + β2 = (1− β)2 = β2 − 2β + 1. Hence, β = −2.
4. i) Let F (t) =
∫
cos
(
1
t
)
dt so F ′(t) = cos
(
1
t
)
. By the fundamental theorem of calculus∫ x3
x2
cos
(
1
t
)
dt = F (x3)− F (x2).
Then
d
dx
∫ x3
x2
cos
(
1
t
)
dt =
d
dx
(
F (x3)− F (x2))
= F ′(x3)
dx3
dx
− F ′(x2) dx
2
dx
= cos
(
1
x3
)
3x2 − cos
(
1
x2
)
2x.
c©2020 School of Mathematics and Statistics, UNSW Sydney
77
ii) a) The MAPLE session defines the function
f(x) =
175x2 − 350x + 10
x2 − 2x+ 2 .
and you are given that f is differentiable for all real x.
The MAPLE session gives the following information
• f(0) = 5 and f(4) = 141.
• f has two zeros where f(x) = 0, close to x = 0.029 and x = 1.971.
• f has one stationary point f ′(1) = 0 where f(1) = −165.
Thus a sketch of the function f over the interval [0, 4] is
0 0.5 1 1.5 2 2.5 3 3.5 4
−200
−150
−100
−50
0
50
100
150
b) ∣∣∣∣175x2 − 350x+ 10x2 − 2x+ 2
∣∣∣∣ = |f(x)|
is the absolute value of the function f(x) from part a). As |f(x)| is continuous on
the closed interval [0, 4] the (global) minimum and maximum both exist, and |f(x)|
has
• a minimum value of 0 which occurs at the two points x = 1±√1155/35
• a maximum value of 165 at x = 1.
iii) As cos2 x ≥ 0 for all x,
0 ≤ 1
e2x + cos2 x
≤ 1
e2x
.
Hence
0 ≤ K =
∫ ∞
0
dx
e2x + cos2 x
≤
∫ ∞
0
dx
e2x
.
Now ∫ ∞
0
dx
e2x
= lim
b→∞
∫ b
0
e−2x dx = lim
b→∞
[
−1
2
e−2x
]b
0
= lim
b→∞
−1
2
e−2b +
1
2
=
1
2
.
As K ≤ ∫∞0 dxe2x which converges, the comparison test implies that
K =
∫ ∞
0
dx
e2x + cos2 x
also converges.
c©2020 School of Mathematics and Statistics, UNSW Sydney
78
iv) The polar curve r = 1− cos θ and has the following values
θ 0 π2 π
3π
2
r = 1− cos θ 0 1 2 1
x = r cos θ 0 0 −2 0
y = r sin θ 0 1 0 −1
Noting that the curve is symmetric about θ = 0 (y = 0), a plot in the xy-plane is
−2.5 −2 −1.5 −1 −0.5 0 0.5 1
−1.5
−1
−0.5
0
0.5
1
1.5
x
y
r = cos(θ)
v) a) Mean Value Theorem: If f is continuous on the closed interval [a, b] and differentiable
on the open interval (a, b), there there exists a c ∈ (a, b), such that
f(b)− f(a)
b− a = f
′(c).
b) Let f(t) = sin−1 t, so
f ′(t) =
1√
1− t2 .
Moreover for any t with −1 < t < 1, we have 0 < 1− t2 ≤ 1, so f ′(t) ≥ 1.
Using the mean value theorem on [x, y] where −1 < x < y < 1, there exists a c with
−1 < x < c < y < 1 such that
sin−1 y − sin−1 x
y − x = f
′(c) =
1√
1− c2 ≥ 1.
As multiplying both sides by y − x > 0 preserves the inequality, this gives
sin−1 y − sin−1 x ≥ y − x.
vi) a) Let the length of EP = L1 and the length of FP = L2. As the triangles AEP and
BFP are right-angled, by Pythagoras L21 = a
2 + x2 and L22 = b
2 + (c− x)2, so
L = L1 + L2 =
√
a2 + x2 +
√
b2 + (c− x)2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
79
b) Differentiating L =
√
a2 + x2 +
√
b2 + (c− x)2 gives
dL
dx
=
1
2
2x√
a2 + x2
+
1
2
−2(c− x)√
b2 + (c− x)2
=
x
L1
− (c− x)
L2
.
Hence
dL
dx
= 0 =⇒ cos θ = x
L1
=
(c− x)
L2
= cosφ.
c) As the triangles AEP and BFP are similar when dLdx = 0 (right angles, θ = φ)
x
a
=
c− x
b
=⇒ x = ac
a+ b
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
80
MATH1131 June 2015 Solutions
1. i) lim
x→∞
x+ sinx
2x
= lim
x→∞
1 + sinxx
2
=
1
2
.
ii) a) f(32) =
5
4
b)
y = (x− 2)2 + 1
y = 3− x
f(x)
1
2
3
1 2 3-1-2-3 x
c)
lim
x→1
f(x) = f(1) = 2.
f ′(x) = −1 for 0 < x < 1, and f ′(1) = 1 for −1 < x < 0, hence f ′(0) does not exist.
iii) Let x = t2, t > 0, then dxdt = 2t and
I1 =
∫
2t
1 + t
dt =
∫
2− 2
1 + t
dt = 2t− 2 log(1 + t) + C
= 2
√
x− 2 log(1 +√x) + C.
iv) Since the limit is of indeterminate form, we apply L’Hopital’s rule.
lim
x→0
1− cos (x2)
x2
= lim
x→0
1
2 sin
(
x
2
)
2x
= lim
x→0
1
4 cos
(
x
2
)
2
=
1
8
.
v) a) z − w = 2 + 3i.
b) 10w/(z − 2) = 11 + 3i.
c) Since |z| = √26 and |w| = √13. Hence |(z/w)8| = 24 = 16.
d) zw = 13(1 + i) and so Arg(zw) = π4 .
e) |zw|8 = 138 × 16 and Arg((zw)8) = 8× π4 = 2π. Hence (zw)8 = 138 × 16.
vi)
x =

 11
4

+ λ

 10
−3

+ µ

 −23
−1

 , λ, µ ∈ R.
vii) a) From the MAPLE, A3 − 6A2 − 9A = I. Hence A−1 exists and A−1 = A2 − 6A− 9I
=


−1 −2 1
−1 −7 2
1 3 −1


c©2020 School of Mathematics and Statistics, UNSW Sydney
81
b) (AT )2 = (A2)T = BT
=


14 5 25
4 2 9
19 8 38


2. i)
40 + 42i = z2
= (a+ ib)2
= a2 − b2 + i2ab
i.e. a2 − b2 = 40
2ab = 42
b =
21
a
by inspection, or
a2 − 21
2
a2
= 40
a4 − 441 = 40a2
a2 = 20± 1
2
√
3364
= −9, 49
a = ±3i,±7
but a, b ∈ R so a = ±7, and
b =
21
±7 = ±3.
So z = 7 + 3i,−7− 3i.
ii) Converting to augmented form,

 1 2 0 b11 1 −1 b2
2 1 −3 b3


Row reducing, 
 1 2 0 b10 −1 −1 b2 − b1
0 −3 −3 b3 − 2b1



 1 2 0 b10 1 1 b1 − b2
0 0 0 b3 − 2b1 − 3(b2 − b1)


In order for a solution to exist, b3 − 2b1 − 3(b2 − b1) = 0, i.e. b3 + b1 − 3b2 = 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
82
iii)
Re(z)
Im(z)
−1 1 2 3
−3
−2
−1
1
2
3
iv) a)
u.v = |u||v| cos θ
2− 1 + 2 = √4 + 1 + 1√1 + 1 + 4 cos θ
3 = 6 cos θ
cos θ =
1
2
θ =
π
3
.
b)
u× v =
∣∣∣∣∣∣
i j k
2 1 −1
1 −1 −2
∣∣∣∣∣∣
= i(−2− 1)− j(−4 + 1) + k(−2− 1)
=

 −33
−3

 = 3

 −11
−1

 .
v) a)
f(x) =
{
1 + ax2, x ≤ 1,
bx+ 2x3, x > 1,
a)
limx→1− f(x) = limx→1− bx+ 2x3 = b+ 2
limx→1+ f(x) = limx→1+ 1 + ax2 = 1 + a
]
i.e. b+ 2 = 1 + a for lim
x→1
f(x) to exist.
a = b+ 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
83
b)
f ′(x) =
{
2ax, x < 1,
b+ 6x2, x > 1,
limx→1− f ′(x) = limx→1− b+ 6x2 = b+ 6
limx→1+ f ′(x) = limx→1+ 2ax = 2a
]
i.e. b+ 6 = 2a for lim
x→1
f ′(x) to exist.
Also require f to be continuous at x = 1, so additionally a = b+ 1 from part (a).
b+ 6 = 2a
b+ 6 = 2(b+ 1)
b = 4
a = b+ 1 = 5.
So a = 5 and b = 4 for f to be differentiable at x = 1.
vi)
I2 =
∫
x3 lnxdx
=
x4
4
lnx−
∫
x4
4
1
x
dx
=
x4
4
lnx− 1
4
∫
x3dx
=
x4
4
lnx− 1
4
x4
4
+ C
=
x4
4
(
lnx− 1
4
)
+ C.
vii) a) (x, y) = (0, 1) =⇒ (r, θ) = (1, π2 ).
x = r cos θ = (1 + cos θ) cos θ and
y = r sin θ = (1 + cos θ) sin θ.
dy
dx
=
dy
dθ
dx
dθ
=
d
dθ ((1 + cos θ) sin θ)
d
dθ ((1 + cos θ) cos θ)
=
cos θ + cos2 θ − sin2 θ
− sin θ − 2 cos θ sin θ
=
2cos2 θ + cos θ − 1
− sin θ − 2 cos θ sin θ
When θ = π2 ,
dy
dx
=
−1
−1 = 1.
i.e. the slope of the tangent at (x, y) = (0, 1) is 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
84
b)
dy
dx
= 0 =⇒ dy
dθ
= 2cos2 θ + cos θ − 1 = 0
(2 cos θ − 1)(cos θ + 1) = 0
cos θ =
1
2
,−1
θ =
π
3
,
5π
3
, π.
However, at θ = π, both dydθ = 0, and
dx
dθ = 0.
If we calculate the limiting value of dydx at θ = π we get 0, indicating a cusp point.
So the points where the tangent is horizontal are (r, θ) =
(
3
2 ,
π
3
)
and
(
3
2 ,
5π
3
)
.
c)
c©2020 School of Mathematics and Statistics, UNSW Sydney
85
3. i) Consider the equation 2|z − 3i| = |z + 3i|. Squaring both sides and using the identity
|z|2 = zz¯ yields
4 (zz¯ + 3i(z − z¯) + 9) = zz¯ − 3i(z − z¯) + 9
⇒ 3zz¯ + 15i(z − z¯) + 18 = 0
⇒ zz¯ + 5i(z − z¯) + 9 = 0
⇒ zz¯ + 5i(z − z¯) + 25 = 16
⇒ |z − 5i|2 = 16
⇒ |z − 5i| = 4 .
This is the equation of a circle, centre (0, 5) (on an Argand diagram) with radius 4.
ii) a) Since z = 2i is a root of a polynomial with real coefficients then another root of the
polynomial is the complex conjugate, i.e., z = −2i. Hence a quadratic factor of the
polynomial is z2 + 4. Using long division yields
z4 − 3z3 + 6z2 − 12z + 8 = (z2 + 4)(z2 − 3z + 2)
= (z2 + 4)(z − 1)(z − 2) .
b) Thus the four roots of p(z) are ±2i, 1, 2.
iii) a) To find a point-normal form for the equation of the plane Π we need a point Q
on the plane (to then create the position vector
−−→
OQ where O is the origin) and a
normal vector n to the plane. A point on the plane Π is

10
0

 and hence a position
vector to Q on the plane Π is
−−→
OQ =

10
0

. The coefficients of the cartesian form
of the plane Π yield the normal vector n =

 1−3
2

. Hence a point-normal form for
the plane Π is given by
(
x−−−→OQ
)
· n = 0 ⇒

x− 1y
z

 ·

 1−3
2

 = 0 .
b) To show that the line ℓ is parallel to the plane Π we can show that the normal n to
the plane Π and the direction vector v for the line ℓ are perpendicular, i.e., n ·v = 0.
The direction vector v for the line ℓ is v =

 20
−1

. Hence

 1−3
2

 ·

 20
−1

 = 2 + 0− 2 = 0 .
Thus the plane Π and line ℓ are parallel.
c©2020 School of Mathematics and Statistics, UNSW Sydney
86
c) The shortest distance between a point P and a plane is given by the length of the
projection of a vector from a point on the plane, say Q, and P , i.e.,
−−→
QP and a
normal vector n to the plane. From part a) we have n =

 1−3
2

 and −−→QP =

32
2

.
Thus
|projn
−−→
QP | =
∣∣∣∣∣∣

32
2

 ·

 1−3
2


∣∣∣∣∣∣√√√√√

 1−3
2

 ·

 1−3
2


=
1√
14
.
Hence the shortest distance between the point P (4, 2, 2) and the plane Π is
1√
14
.
iv) a) Using elementary row operations to calculate the determinant yields∣∣∣∣∣∣
1 −1 3
3 a 2
2 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣
1 −1 3
0 a+ 3 −7
0 3 −5
∣∣∣∣∣∣ R2 → R2 − 3R1, R3 → R3 − 2R1
=
∣∣∣∣a+ 3 −73 −5
∣∣∣∣
= 6− 5a .
b) The matrix A will not have an inverse if det(A) = 0, i.e., a =
6
5
.
c) If matrix B is an inverse for matrix A then AB = I where I is the 3 × 3 identity
matrix, i.e., 
1 −1 33 a 2
2 1 1



−1 4 −51 b 7
1 −3 4

 =

1 0 00 1 0
0 0 1

 .
If we multiply row 2 of the first matrix by the 1st column of the second matrix yields
−3 + a + 2 = 0 with a = 1. Also if we multiply the 1st row of the first matrix by
the 2nd column of the second matrix yields 4 − b − 9 = 0 with b = −5. (A quick
check is to multiply the 2nd row of the first matrix by the 2nd column of the second
matrix to verify it is equal to 1). Thus
A =

1 −1 33 1 2
2 1 1

 and A−1 = B =

−1 4 −51 −5 7
1 −3 4

 .
v) By definition, XX−1 = I, where I is the n× n identity matrix. Hence
det(XX−1) = det(I) = 1
⇒ det(X) det(X−1) = 1 since det(AB) = det(A) det(B)
⇒ 1× det(X−1) = 1 since det(X) = 1
⇒ det(X−1) = 1 as required.
c©2020 School of Mathematics and Statistics, UNSW Sydney
87
4. i) a)
coshx =
ex + e−x
2
, sinhx =
ex − e−x
2
.
b)
d(cosh(6x))
dx
=
d
dx
(
e6x + e−6x
2
) = 6(
e6x − e−6x
2
) = 6 sinh(6x).
c) Consider the definition in a) and note that
sinhx =
ex − e−x
2
< ex/2.
Taking logs of both sides yields
ln(sinh(x)) < ln
(
ex
2
)
= x− ln 2.
ii) Using the fact that for x > 0,
lnx < x
it implies that
lnx
x3
<
1
x2
.
Integrating both sides of the inequality from 1 ti ∞ yields∫ ∞
1
lnx
x3
dx <
∫ ∞
1
dx
x2
.
To determine the integral, we write it terms of a proper integral, that is
lim
R→∞
∫ R
1
dx
x2
= lim
R→∞
[
1− 1
R
]
= 1.
Thus, the improper integral is bounded by 1,∫ ∞
1
lnx
x3
dx < 1.
Alternatively,
we can tackle this question directly by parts:
∫ ∞
1
lnx
x3
dx = lim
R→∞
([
− lnx
2x2
]R
1
+
∫ R
1
−1
2x3
dx
)
= 1/4.
iii) (a) Using the Fundamental Theorem of Calculus,
d
dx
(∫ x3
0
cos(t2) dt
)
= 3x2 cos(x6).
iv) a) Let p(x) := x3 + 3x+ 1.
Note that on the interval [−1, 0] that p(−1) = −3 and p(0) = 1. The function p is
a continuous function and thus by the Intermediate Value Theorem there is a point
c ∈ (−1, 0) such that p(c) = 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
88
b) Notice that the function is always increasing since
p′(x) = 3x2 + 3 > 0.
This means that the function is 1− 1 and has an inverse function g(x).
c) Using the chain rule,
p′(x)g′(p(x)) = 1.
That is,
g′(p(x)) =
1
p′(x)
.
Now, if we solve the equation
p(x) = 1
then x = 0. Thus,
g′(1) = g′(p(0)) =
1
3
.
v) a) If f is differentiable on the open interval (a, b) and continuous on the [a, b], then
there exists a point c ∈ (a, b) such that
f(b)− f(a)
b− a = f
′(c).
b) Let f(t) = cos t and consider the interval [x, y] then by the MVT, we have
cos(y)− cos(x)
y − x = sin(c).
Now, the | sin c| ≤ 1 for any c ∈ (x, y). If we take the absolute value of both sides,
we obtain ∣∣∣∣cos(y)− cos(x)y − x
∣∣∣∣ = | sin(c)| ≤ 1.
Rearranging, proves the inequality
| cos(y)− cos(x)| ≤ |y − x|.
c©2020 School of Mathematics and Statistics, UNSW Sydney
89
PAST HIGHER EXAM SOLUTIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
90
MATH1141 June 2011 Solutions
2. i) a) No solutions requires both 6− a2 − a = 0 and 2− a 6= 0. i.e.
−a2 − a+ 6 = 0
−(a− 2)(a + 3) = 0
a = 2,−3
AND
2− a 6= 0
a 6= 0.
Putting these together we have, for no solutions, a = −3.
b) For a unique solution, we require 6− a2 − a 6= 0, which (using part a) gives
a 6= 2 and a 6= −3.
c) For infinite solutions we require both 6− a2 − a = 0 and 2− a = 0.
i.e. to satisfy both a = 2.
ii) ∣∣∣∣∣∣
2 0 −1
−1 3 0
5 7 3
∣∣∣∣∣∣ = 2
∣∣∣∣ 3 07 3
∣∣∣∣− 0
∣∣∣∣ −1 05 3
∣∣∣∣− 1
∣∣∣∣ −1 35 7
∣∣∣∣
= 2(9− 0)− 0(3 − 0) + (−1)(7 − 15)
= 18 + 8
= 26.
Alternatively an upper triangular form could be used, multiplying the pivots and ad-
justing for the number of swaps of rows performed to get to the reduced form.
iii)
λ

 12
3

 =

 10
0

+ µ

 11
0

+ ν

 10
−1


i.e.
λ = 1 + µ+ ν
2λ = µ
3λ = −ν
Substituting the last 2 equations into the first we have
λ = 1 + 2λ− 3λ
2λ = 1
λ =
1
2
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
91
i.e.
x =
1
2

 12
3

 =

 121
3
2

 ,
Alternatively you could rearrange
λ

 12
3

− µ

 11
0

− ν

 10
−1

 =

 10
0



 1 −1 −12 −1 0
3 0 1



 λµ
ν

 =

 10
0


and put the problem in an augmented matrix:
 1 −1 −1 12 −1 0 0
3 0 1 0


This can be row-reduced and back-substituted to find (λ, µ, ν) which can be substituted
into the line (or plane) to find x.
Another alternative is to convert the plane to cartesian form and then substitute the
line and solve for λ.
iv) a)
lim
h→0
e−1/h2
h
= lim
h→0
e−1/h2
h2
h
= lim
h→0
e−1/h
2
h2
lim
h→0
h.
Letting x = 1
h2
, when h→ 0 then x→∞. i.e.
lim
h→0
e−1/h2
h
= lim
h→0
e−1/h2
h2
lim
h→0
h
= lim
x→∞xe
−x lim
h→0
h
= 0 · 0
= 0.
Alternatively, you could use
lim
h→0
e−1/h
2
h
= lim
h→0
h−1
e1/h
2
= ”
∞
∞” =⇒ use L’Hopital’s rule
= lim
h→0
−h−2
−2h−3e1/h2
= lim
h→0
h
2e1/h
2
=
0
∞
= 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
92
b) Firstly, you should check f(x) is continuous at x = 0.
lim
x→0+
f(x) = lim
x→0+
e−1/x
2
= 0
lim
x→0−
f(x) = lim
x→0+
e−1/x
2
= 0
f(0) = 0.
i.e. f(x) is continuous at x = 0.
The definition of the derivative at x = 0 is
f ′(0) = lim
h→0
f(0 + h)− f(0)
h
= lim
h→0
e−1/h2 − 0
h
= lim
h→0
e−1/h
2
h
which from part a)
= 0.
Thus f(x) is differentiable at x = 0 (with f ′(0) = 0).
v) a) The derivative at points other than the split points is
f(x) =


− 2
x2
for x < −1,
2x for − 1 < x < 6π ,
− 1
x2
(
72
π2
− 2) cos 1x for x > 6π .
This is a split funciton, so the positions of the splits are possible critical points.
x = −1:
lim
x→−1−
f(x) =
2
−1 = −2
lim
x→−1+
f(x) = (−1)2 − 1 = 0
i.e. limx→−1− f(x) 6= limx→−1+ f(x) so x = −1 is a critical point due to the
function being discontinuous (derivative undefined) at the is point.
x = 6π :
lim
x→ 6
pi
−
f(x) =
(
6
π
)2
− 1 = 36
π2
− 1
lim
x→ 6
pi
+
f(x) =
(
72
π2
− 2
)
sin
π
6
=
(
72
π2
− 2
)
1
2
=
36
π2
− 1
i.e. lim
x→ 6
pi
− f(x) = lim
x→ 6
pi
+ f(x) so the function is continuous at x = 6π . This
point is, however the global maximum of the function.
c©2020 School of Mathematics and Statistics, UNSW Sydney
93
Checking the derivative:
lim
x→ 6
pi
−
f ′(x) = 2
6
π
=
12
π
lim
x→ 6
pi
+
f ′(x) = −π
2
36
(
72
π2
− 2
)
cos
π
6
= −π
2
36
(
72
π2
− 2
) √
3
2
i.e. lim
x→ 6
pi
− f ′(x) 6= lim
x→ 6
pi
+ f ′(x), i.e. the derivative is not continuous at x = 6π ,
so x = 6π is critical point.
Now look for any stationary points in the intervals:
x < −1:
f ′(x) = − 2
x2
6= 0 for any x ∈ R. i.e. no stationary points.
−1 < x < 6π :
f ′(x) = 2x = 0 for x = 0, so there is a stationary point at x = 0. As f ′′(0) = 2 > 0,
this is a local minimum. Thus x = 0 is also a critical point.
x > 6π :
f ′(x) = − 1
x2
(
72
π2
− 2
)
cos
1
x
= 0 for x→ ±∞ or cos 1
x
= 0. For the latter case this
means x =
2
(2k + 1)π
, for k ∈ Z. However, we are restricted to x > 6π , i.e.
2
(2k + 1)π
>
6
π
1
2k + 1
> 3
2k + 1 > 3(2k + 1)2
0 > 12k2 + 10k + 2
0 > (2k + 1)(3k + 1)
i.e. k ∈ (−1/2,−1/3), but k ∈ Z, so there is no solution. i.e. there are no stationary
points for x > 6π .
Putting it altogether, we have 3 critical points:
x = −1, where the function is discontinuous, and the derivative is undefined,
x = 0, where we have a local minimum, and
x =
6
π
, where the derivative is undefined.
b)
lim
x→−∞ f(x) = limx→−∞
2
x
= 0
lim
x→∞ f(x) = limx→∞
(
72
π2
− 2
)
sin
1
x
= 0
Thus there is a horizontal asymptote at f(x) = 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
94
c)
f(x)
x
36
π2
− 1
|
8
|
6
|
4
|
6/π
|
1
|
−1
|
−2
|
−4
|
−6
|
−8
−2
−1
1
2
3. i) a) We recognise that S is the sum of a geometric progression with common ratio e2iθ/3.
Therefore
S = eiθ
∞∑
k=0
(
e2iθ
3
)k
=
eiθ
1− e2iθ/3
=
3eiθ
3− e2iθ ·
3− e−2iθ
3− e−2iθ
=
3
(
3eiθ − e−iθ)
(3− cos(2θ))2 + sin2(2θ)
=
3
(
3eiθ − e−iθ)
10− 6 cos(2θ)
as required.
b) We observe that T = Im(S), and hence using part (a) we calculate
T =
3(3 sin(θ)− sin(−θ))
10− 6 cos(2θ)
=
3× 4 sin(θ)
10− 6 cos(2θ)
=
6 sin(θ)
5− 3 cos(2θ) .
ii) To show that u− v is perpendicular to u+ v, we show that the dot product of the two
vectors is zero. Now, using properties of the dot product and the fact that u and v have
c©2020 School of Mathematics and Statistics, UNSW Sydney
95
the same length, we have
(u− v) · (u+ v) = u · u+ u · v − v · u− v · v
= u · u− v · v
= ‖u‖2 − ‖v‖2
= 0,
as required.
Alternatively, note that u− v and u+ v are the diagonals of the rhombus spanned by
the vectors u and v, as shown in the figure.
u
v
u+ v
u− v
We know that the diagonals of a rhombus are perpendicular, completing the proof.
iii) a) Recall that ej is the n× 1 vector with a 1 in the jth position and zeros everywhere
else. Therefore
Aej =


a1j
a2j
...
anj

 ,
which is the jth column of A. Similarly, eTi is a 1 × n matrix with a 1 in the ith
column and zeros everywhere else. Hence
eTi Aej = e
T
i


a1j
a2j
...
anj

 = aij ,
since aij is the ith entry of the jth column of A. This completes the proof.
b) If A is symmetric then AT = A, and hence
(Ax)Ty = xTATy = xTAy,
as required.
c) Suppose that xTAy = (Ax)Ty for all vectors x,y ∈ Rn. Rewriting this gives
xTAy = xTATy (∗)
for all x,y ∈ Rn. Now let x = ei and y = ej, where i, j are arbitrary integers in
{1, . . . , n}. Applying part (a) to condition (∗) gives
aij = (A
T )ij = aji.
This proves that A is symmetric, since i and j were arbitrary.
c©2020 School of Mathematics and Statistics, UNSW Sydney
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iv) a) To show thatM reflects the vector a to the vector −a, we must show thatMa = −a.
We calculate
Ma =
1
9

 1 4 −84 7 4
−8 4 1



 2−1
2

 = 1
9

 2− 4− 168− 7 + 8
−16− 4 + 2

 =

−21
2

 = −a
as required.
b) Every vector in the plane is perpendicular to the vector a. Therefore the Cartesian
equation of the plane Π is
2x− y + 2z = 0.
c) Any nonzero vector u which is the position vector of a point in the plane Π satisfies
Mu = u. One such vector is u =

12
0

.
d) The shortest distance from B to the plane is equal to the length of the projection
of b onto a. We calculate
|proja(b)| =
|a · b|
‖a‖ =
|12 + 6 + 0|√
4 + 1 + 4
=
18
3
= 6.
The shortest distance from B to the plane Π is 6 units.
4. i) a) We apply L’Hoˆpital’s rule repeatedly (checking each time that he required conditions
are satisfied)
lim
x→∞
xn
ex
= lim
x→∞
nxn−1
ex
= ... = lim
x→∞
n!
ex
= 0.
b) By (a), e−xxn+2 → 0 as n → ∞ and so e−xxn+2 < 1 for sufficiently large n,
say n > M . Also by continuity, e−xxn+2 < K on the interval [1,M ]. Thus if
C =max{K, 1}, e−xxn+2 < C for all x ≥ 1 and the result follows.
c) ∫ ∞
1
e−xxn dx <
∫ ∞
1
C
x2
dx.
The latter integral converges by the p-test and so the given integral converges for
any n ∈ N.
ii) a) Applying the Mean Value Theorem to f on [0, 2], we have, for some c ∈ (0, 2),
f ′(c) =
f(2)− f(0)
2− 0 = 6.
b) Applying the Mean Value Theorem to f ′ on [0, c], we have, for some d ∈ (0, c),
f ′′(d) =
f(c)− f(0)
c− 0 =
6
c
> 3.
iii) a) Since coshx > 1 for all x > 0,the result follows.
b)
f ′(x) = 1− a sech2x > 0
if a ≤ 1. Thus f is strictly increasing on (0,∞). Also f(0) = 0, hence f(x) > 0 for
all x > 0 when a ≤ 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
97
c) For f to have a positive zero, f ′(x) = 1 − a sech2x = 0 for some x > 0 so a =
1
sech2x
> 1.
iv) a)
A(t) =
√
(s− a)(s − b)(s− c)(s − d)− 1
2
abcd(1 + cos t)
has a maximum when 1 + cos t = 0, that is, when α+ β = π.
b) Now 1 + cos t is decreasing on [0, π] and so −(1 + cos t) is increasing. Thus A(t) is
increasing and continuous on [0, π] and so has an inverse.
Furthermore, A′(t) =
1
2
abcd sin t
2A(t) so that A is differentiable. By the inverse function
theorem, B = A−1 is differentiable on (A(0), A(π)).
c) If t = π2 , then A(
π
2 ) =
√
(s − a)(s− b)(s − c)(s − d)− 12abcd = A0. By the inverse
function theorem,
B′(A0) =
1
A′(π2 )
=
1
abcd
4A(π)
=
4A0
abcd
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
98
MATH1141 June 2012 Solutions
2. i) a)
f (x) = x2 + cos
(
x2
)
, x ∈ (0, 2√π)
f
′
(x) = 2x
(
1− sin (x2))
Note f
′
(x) = 0, x ∈ (0, 2√π]
f
′′
(x) = 2
(
1− sin (x2) )+ 4x2cos (x2)
f
′
(x) = 0 ⇒
x = 0, which is not in the domain, or
x2 =
π
2
,
5π
2
x =
√
π
2
,
√
5π
2
. Note 3
√
π
2
is not in the domain.
f
′′
(√
π
2
)
= 0, f
′′
(√
5π
2
)
= 0.
So the critical points are x =
√
π
2 ,
√
5π
2 , the stationary points (both points of
inflection) and x = 2
√
π, the endpoint, and global maximum of the function.
b)
f
(
2
√
π
)
= 4π + 1
f (0) = 1
f
′
(x) = 2x
(
1− sin (x2)) ≥ 0 for x ∈ (0, 2√π]
So f is an increasing function on its domain.
It is also continuous (combination of elementary functions) and thus is invertible.
Range (f) = (1, 4π + 1], so Dom
(
f−1
)
= (1, 4π + 1].
From above we can see f
(√
5π
2
)
= 5π2 . Thusf
−1 (5π
2
)
=
√
5π
2 .
c) From the inverse function theorem, ddxf
−1(x) = − 1
f ′ (f−1(x))
. So f−1(x) is not differ-
entiable where f
′
(
f−1 (x)
)
= 0. i.e. f−1 (x) =
√
π
2 ,
√
5π
2 , or x =
π
2 ,
5π
2 . i.e. f
−1 (x)
is differentiable in (1, 4π + 1) , x 6= π2 , 5π2 .
ii) a)
f (x) =
∫ x2−9x
0
e−t
2
dt
f
′
(x) = e−(x
2−9x)2(2x− 9)
Stationary point: f
′
(x) = 0.
Mean value theorem: f (x) is continuous on [0, 9] and differentiable on (0, 9). There-
c©2020 School of Mathematics and Statistics, UNSW Sydney
99
fore
f (9)− f (0)
9− 0 = f
′ (x0) , x0 ∈ (0, 9) .
f (9) =
∫ 0
0
e−t
2
dt = 0
f (0) =
∫ 0
0
e−t
2
dt = 0
Therefore,
f
′
(x0) = 0, x0 ∈ (0, 9) .
b)
f
′
(x) = e−(x
2−9x)2(2x− 9) = 0
i.e. the stationary point is x = 92 .
f
′′
(x) =
(
2− 2 (x2 − 9x) (2x− 9)) e−(x2−9x)2
f
′′
(
9
2
)
=
(
2− 2
((
9
2
)2
− 99
2
)(
2
9
2
− 9
))
e
−
(
( 92)
2−9 9
2
)2
= 2e−(
81
4 )
2
> 0
iii) a) Since z lies on the unit circle, z = eiθ for some θ ∈ [0, 2π]. Hence z+ 1
z
= eiθ+e−iθ =
2cos θ ∈ R.
b) By (a), z +
1
z
has a maximum value of 2 (when θ = 0).
iv)
cos 4θ = 8cos4θ − 8cos2θ + 1
ei4θ = (cos θ + isin θ )4 by de Moivre
′
s theorem
= cos4θ + 4icos3θ sin θ − 6cos2θ sin2θ − 4icos θ sin3θ + sin4θ (binomial expansion)
cos 4θ
= Re
(
ei4θ
)
= cos4θ − 6cos2θ sin2θ + sin4θ
= cos4θ − 6cos2θ (1− cos2θ )+ (1− cos2θ )2
= cos4θ − 6cos2θ + 6cos4θ + 1− 2cos2θ + cos4θ
= 8cos4θ − 8cos2θ + 1
v) a)
x =

 12
0

+ λ1

 11
−3

+ λ2

 3−1
−1

 , λ1, λ2 ∈ R

 44
−7

 =

 12
0

+ λ1

 11
−3

+ λ2

 3−1
−1


c©2020 School of Mathematics and Statistics, UNSW Sydney
100

 32
−7

 = λ1

 11
−3

+ λ2

 3−1
−1



 1 31 −1
−3 −1
∣∣∣∣∣∣
3
2
−7

→

 1 30 −4
0 8
∣∣∣∣∣∣
3
−1
2

→

 1 30 −4
0 0
∣∣∣∣∣∣
3
−1
0


This system has unique solution, so the point lies on the plane.
b) If b is parallel to P then b=

 13
2

 = λ1

 11
−3

+ λ2

 3−1
−1



 1 31 −1
−3 −1
∣∣∣∣∣∣
1
3
2

→

 1 30 −4
0 8
∣∣∣∣∣∣
3
2
5

→

 1 30 −4
0 0
∣∣∣∣∣∣
3
2
9


i.e. no solution, and b is not parallel to P .
c)
c ·

 11
−3

 = 0, andc ·

 3−1
−1

 = 0.
Therefore c is orthogonal to P .
3. i) det(B) = −35.
ii) a) x =

40
5

+ λ

 1−1
1

 where λ ∈ R.
b) The quadratic distance from x to

00
0

 is
(4 + λ)2 + (−λ)2 + (5 + λ)2 = 3λ2 + 18λ + 41
which is minimal when λ = − 182·3 = −3.
The closest point is thus given by x =

40
5

− 3

 1−1
1

 =

13
2

.
iii) a) All a 6= 1,−2.
b) a = 1.
c) a = −2.
iv) a) For example,
(
0 1
0 0
)
b) If Q is nilpotent (of degree 2), then |detQ| =
√
|detQ2| =√|0| = 0,
so Q is not invertible.
c) Proof. S−1Q = S−1QI = S−1QSS−1 = S−1SQS−1 = IQS−1 = QS−1.
d) By c),
(S+Q)(S−1−S−2Q) = I+QS−1−S−1Q−QS−2Q = I−S−1QQS−1 = I−S−10S−1 = I .
Hence, S +Q is invertible, and k = 2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
101
v) Proof. We see that
a = x · e1 = |x||e1| cosα = |x| cosα ;
b = x · e2 = |x||e2| cos β = |x| cos β ;
c = x · e3 = |x||e3| cos γ = |x| cos γ .
Hence,
cos2 α+ cos2 β + cos2 γ =
a2
|x|2 +
b2
|x|2 +
c2
|x|2 =
|x|2
|x|2 = 1 .
vi) Proof. Since det(AT ) = det(A) 6= 0, every vector b ∈ R can be written uniquely as a
linear combination of columns of AT , or in other words, rows of A.
4. i) a) Since cos(−2θ) = cos θ the curve is symmetric about the x axis and since cos 2(π −
θ) = cos 2θ the curve is symmetric about the y axis.
b) Diagram as below
x
y θ = π4
θ = 5π4
θ = 3π4
θ = 7π4
1−1
−2
b
b
b
b
b
b
b
b
ii) a) tanhx =
ex − e−x
ex + e−x
.
b) Diagram as below
x
tanhx
1
−1
c) Since tanhx =
ex − e−x
ex + e−x
= 1− 2e
−x
ex + e−x
, (1− tanhx)/e−2x = 2e
x
ex + e−x
which
tends to 2 as x→∞.
d) Since
∫ ∞
0
e−2x dx converges, it follows from (c) by the limit comparison test that∫ ∞
0
(1− tanhx) dx converges also.
e)∫ ∞
0
(1− tanhx) dx = lim
R→∞
∫ R
0
(1− tanhx) dx = lim
R→∞
[x− log(cosh x)]R0
c©2020 School of Mathematics and Statistics, UNSW Sydney
102
= lim
R→∞
R− log(coshR) = lim
R→∞
log
2eR
eR + e−R
= log 2
iii) a) Integrating by parts with u = f(x), we have
∫ b
a
f(x) sinnx dx =
[
−cos(nx)
n
]b
a
+
1
n
∫ b
a
f ′(x) cos(nx) dx
=
f(a) cos(nx)− f(b) cos(nb)
n
+
1
n
∫ b
a
f ′(x) cos(nx) dx.
b) ∣∣∣∣
∫ b
a
f ′(x) cos nx dx
∣∣∣∣ ≤
∫ b
a
∣∣f ′(x) cos nx∣∣ dx ≤ ∫ b
a
|f ′(x)| dx ≤
∫ b
a
Ldx = L(b− a).
c)
lim
n→∞
∫ b
a
f(x) sinnx dx = lim
n→∞
f(a) cos(nx)− f(b) cos(nb)
n
+
1
n
∫ b
a
f ′(x) cos(nx) dx
≤ lim
n→∞K(n)/n+ L(b− a)/n = 0
since K(n) is bounded.
c©2020 School of Mathematics and Statistics, UNSW Sydney
103
MATH1141 June 2013 Solutions
2. i) a) Away from 0 f is differentiable. Since lim
x→0+
f(x) = lim
x→0−
f(x) = 0, f is continuous
at x = 0 and since the derivatives of the two constituent functions have the same
derivative at x = 0, by the ‘split function theorem’, f is differentiable at x = 0 and
hence everywhere.
f ′(x) =
{
3x2 if x < 0
2x if x ≥ 0.
b) Since lim
x→0+
f ′(x) = lim
x→0−
f ′(x) = 0, so f ′ is continuous at x = 0
c) For x < 0, lim
h→0−
f ′(0 + h)− f ′(0)
h
= 0, while for x > 0, lim
h→0+
f ′(0 + h)− f ′(0)
h
= 2.
Hence f ′ is not differentiable at x = 0.
ii) a)
LP (f) =
1
n
n∑
k=1
f
(
k
n
)
=
1
n
n∑
k=1
1
1 + kn
=
n∑
k=1
1
n+ k
.
b) Under the given assumption,
lim
n→∞
n∑
k=1
1
n+ k
=
∫ 1
0
1
1 + x
= log 2.
iii) a) 12 (a+ b).
b) x = c+ λ(12 (a+ b)− c), λ ∈ R.
c) In triangle ABC, usin the dot product formula, cosA = (b−a)·(c−a)|b−a| |c−a| =
1
2 and so
A = π/3. Similarly cosB = 12 so B = π/3. Hence ABC is equilateral.
iv) a)
x =

 4−2
0

+ λ

 −12
1

 , λ ∈ R.
b) This is the line of intersection of two planes.
c) Substituting the line into the thrid equation (which is a sphere), gives 3λ2−8λ+5 = 0
and so λ = 1, 5/3. The value λ = 1 gives the point

 30
1

 and λ = 5/3 gives the
point 13

 74
5

 .
3. i) A vector normal to the plane
n =

 02
−1

×

−11
3

 =

71
2

 .
c©2020 School of Mathematics and Statistics, UNSW Sydney
104
A vector from the point

12
4

 on the plane to p:

12
4

−

32
3

 =

−20
1

 .
The length of the projection of this vector onto n is the shortest distance from p to the
plane ∣∣∣∣∣∣

−20
1

 ·

71
2


∣∣∣∣∣∣∣∣∣∣∣∣

71
2


∣∣∣∣∣∣
=
4√
6
.
ii) a) α1 = 1.
b) Complex roots come in conjugate pair because p has real coefficients. Hence there
are two complex roots and two real roots.
c) Assume that α2 is the second real root, and that α3 and α4 are the conjugate
complex roots (α4 = α¯3). Since α1 + · · ·+ α4 = 1 and α1 = 1, there holds
α2 + α3 + α4 = α2 + 2a = 0, (2)
where we denote by a the real part of α3. Noting that 0 is not a root, we deduce
that either α2 < 0 or a < 0.
d) If α is a root satisfying |α| ≤ 1/2 then from p(α) = 0 we deduce α4−α3−α2−α = −2,
so that
2 = |α4 − α3 − α2 − α| ≤ |α|4 + |α|3 + |α|2 + |α| ≤ 1
24
+
1
23
+
1
22
+
1
2
=
15
16
,
contradiction! Hence |αj | > 1/2 for j = 1, 2, 3, 4.
iii) a) False. Take for example
A =
(
0 1
0 0
)
and B =
(
1 1
0 0
)
b) True. (Note: det(AB) = det(A) det(B).)
c) False. Since det(AB) = det(A) det(B) and det(AC) = det(A) det(C) we need to
choose an example such that det(A) = 0. For example
A =
(
0 0
0 0
)
, B =
(
1 0
0 1
)
and C =
(
0 1
1 0
)
d) False. Choose an example such that A is not invertible, i.e., det(A) = 0 as in c).
(The statement is true if A is invertible.)
iv) a) {u1, . . . ,uk} is an orthonormal set if
ui · uj =
{
1 if i = j,
0 if i 6= j.
c©2020 School of Mathematics and Statistics, UNSW Sydney
105
b) Since the jth row of MT is vj , we have
MTM =


v1
v2
...
vn


(
v1 v2 · · · vn
)
=


v1 · v1 v1 · v2 · · · v1 · vn
v2 · v1 v2 · v2 · · · v2 · vn
...
...
. . .
...
vn · v1 vn · v2 · · · vn · vn

 = I
i.e., M is an orthogonal matrix. Hence
det(M)2 = det(MT ) det(M) = det(MTM) = det(I) = 1,
implying det(M) = ±1.
4. i) a)
lim
t→∞ t
2e−t
2
= lim
t→∞
t2
et2
= lim
t→∞
2t
2tet2
= lim
t→∞
1
et2
= 0.
b) Since et
2 ≥ t2 for t ≥ 1, we have that
∫ ∞
1
e−t
2
dt ≤
∫ ∞
1
1
t2
dt, which converges by
the p-test. Also
∫ 1
0
e−t
2
dt <
∫ 1
0
1 dt = 1 and so the original integral converges.
c) f ′(x) = 3x2e−x6 and so the only critical point is (0, 0) which is an inflection point.
As x→∞ f(x)→ I.
d) DIAGRAM
ii)
lim
x→c
f(x)− f(c)− f ′(c)(x− c)
(x− c)2 = limx→c
f ′(x)− f ′(c)
2(x− c) = limx→c
f ′′(x)
2
=
f ′′(c)
2
.
iii) a) Differentiating implicitly, 2(x2+y2)(2x+2yy′)−2(2x−2yy′) = 0, so y′ = x(1− (x
2 + y2))
y(1 + x2 + y2)
.
The tangents correspond to when the numerator is 0, and this occurs when x = 0,
i.e. on the y axis or when x2 + y2 = 1.
b) The condition x = 0 yields the point, (x, y) = (0,±
√
−1 +√b), provided b ≥ 1.
The condition x2 + y2 = 1 yields 4y2 = b; 0 < b ≤ 4 and hence the points (x, y) =
(
√
4−b
2 ,±
√
b
2 ).
iv) a) g(0) = f(0) ≥ 0 and g(2) = f(2)− 8 ≤ 0 and so by the IVT, g has at least one root
ξ in the interval [0, 2]. Thus f(ξ) = ξ3.
b) Applying the MVT to f on [0, 2] we have f(2)−f(0)2−0 = f
′(η) for some η ∈ (0, 2) and
the result follows.
c©2020 School of Mathematics and Statistics, UNSW Sydney
106
MATH1141 June 2014 Solutions
3. i)
g′(x) = 3 + 2 sin 2x = 2(1 + sin 2x) + 1 > 0.
Therefore, g(x) is an increasing function. It is also continuous (because it is a compo-
sition of elementary functions) so it is invertible. By the inverse function theorem, the
inverse is also continuous and differentiable.
By inspection, g(0) = −2 so h(−2) = 0. Thus
h′(−2) = 1
g′(h(−2)) =
1
g′(0)
=
1
3
.
ii) a) Suppose f is continuous on [a, b] and differentiable on (a, b). Then there exists at
least one real number c in (a, b) such that
f(b)− f(a)
b− a = f
′(c).
b) Let f(x) = tan−1 x (1 mark). Then f ′(x) = 1/(1+x2) and the Mean Value Theorem
states that there exists a c in (a, b) such that
f ′(c) =
tan−1 b− tan−1 a
b− a . (1 mark)
But 0 < 1/(1 + c2) ≤ 1 so
0 <
tan−1 b− tan−1 a
b− a ≤ 1
or
0 < tan−1 b− tan−1 a ≤ b− a.
c) Let I =
∫∞
1 g(t) dt where g(t) = tan−1(t+ t−2)−tan−1 t. From (b), 0 ≤ g(t) ≤ h(t)
where h(t) = t+ t−2 − t = t−2.
Since
∫∞
1 h(t)dt =
∫∞
1 t
−2dt converges (by the p-test) then so does
∫∞
1 g(t)dt (by
the comparison test).
iii) Let L = 2. Then
|f(x)− L| =
∣∣∣∣ excoshx − 2
∣∣∣∣
=
ex
cosh x
∣∣∣∣1− 2coshxex
∣∣∣∣
=
ex
cosh x
∣∣1− 1− e−2x∣∣
=
ex
cosh x
∣∣−e−2x∣∣
=
e−x
cosh x
≤ e−x,
since coshx ≥ 1.
Let ǫ > 0. Then e−x < ǫ if and only if −x < ln ǫ or x > − ln ǫ = ln ǫ−1.
Let M = ln ǫ−1. Then we have shown that if x > M then there exists an ǫ > 0 such
that |f(x)− L| < ǫ, as required.
c©2020 School of Mathematics and Statistics, UNSW Sydney
107
iv) a) r′ = −4 sin 4θ = 0 when θ = nπ/4 for n = 0, 1, 2, · · · . Thus:
r takes a maximum value of 2 at θ = 0, π/2, π, 3π/2, 2π.
r takes a minimum value of 0 at θ = π/4, 3π/4, 5π/4, 7π/4.
b) Diagram as below
v) a) f(x) = xx lnx = (elnx)x lnx = ex(lnx)
2
. So
f ′(x) =
d
dx
(
x(lnx)2
)
ex(lnx)
2
=
(
(lnx)2 + x
2 lnx
x
)
ex(lnx
2
=
(
(ln x)2 + 2 ln x
)
xx lnx.
b) Since xx lnx = ex(lnx)
2
> 0 for x > 0, the sign of f ′(x) will be determined by the
sign of g(x) = (lnx)2 + 2 ln x.
Let z = lnx. Then g = z2 + 2z = z(z + 2). This negative for −2 < z < 0. Thus,
f ′(x) is negative for e−2 < x < 1 and positive for x < e−2 or x > 1.
c) Diagram as below
c©2020 School of Mathematics and Statistics, UNSW Sydney
108
4. i) We first write the augmented matrix
 2 0 −4 b13 1 −2 b2
−2 −1 0 b3

 .
and row reduce to get: 
 2 0 −4 b10 2 8 2b2 − 3b1
0 0 0 2b3 + 2b2 − b1

 .
The system has a solution if the final column in a row reduced form of the augmented
matrix is not a leading column; this condition corresponds to the equation
−b1 + 2b2 + 2b3 = 0.
ii) The distance from the point represented by the vector

 xx
x

 to I (which is represented
by the vector

 10
0

) is given by the distance formula:
d =
√
(x− 1)2 + (x− 0)2 + (x− 0)2 =
√
3x2 − 2x+ 1.
The distance will be 1 if and only if the square of the distance is 1, so we get the equation
3x2 − 2x+ 1 = 1,
or
3x2 − 2x = 0,
c©2020 School of Mathematics and Statistics, UNSW Sydney
109
which has the solution
x = 0 or x =
2
3
.
To get a point distinct from the origin, we take x = 23 and the corresponding vector is
 232
3
2
3

.
iii) a) The inverse is given by
1
det(A)
(
α −i
−1− i 2
)
=
1
2α+ 1− i
(
α −i
−1− i 2
)
.
b) Suppose that det(A2) = −1. Then we have
−1 = det(A2) = (det(A))2,
so det(A) = ±i. On the other hand, we have
det(A) = 2α+ 1− i = ±i,
so we get the solutions
α = −1
2
or α =
2i− 1
2
.
iv) a) The roots of z9 − 1 are the ninth roots of unity: 1 and e± 2pik9 , k = 1, 2, 3, 4. From
the factorization
(z9 − 1) = (z3 − 1)(z6 + z3 + 1)
we see that the roots z6 + z3 + 1 are those ninth roots of unity which are not also
cube roots of unity, which leaves the six roots e±
2pik
9 , k = 1, 2, 4.
b) Dividing z6 + z3 + 1 = 0 by z3 gives the equation
z3 + 1 +
1
z3
= 0,
or
z3 +
1
z3
= −1.
We have
x3 = (z +
1
z
)3 = z3 +
1
z3
+ 3z +
3
z
= −1 + 3x,
so
x3 − 3x+ 1 = 0.
c) Let z1 = e
2pii
9 , z2 = e
4pii
9 , z3 = e
8pii
9 . Then by parts (a) and (b), the three numbers
z1+
1
z1
, z2 +
1
z2
, z3 +
1
z3
are the roots of the cubic polynomial x3− 3x+1. Since this
polynomial has no quadratic term, the sum of these roots must be 0, and we have
0 = z1 +
1
z1
+ z2 +
1
z2
+ z3 +
1
z3
= 2(cos
2π
9
+ cos
4π
9
+ cos
8π
9
).
c©2020 School of Mathematics and Statistics, UNSW Sydney
110
v) a) We can write any vector as x = |x|u, with u a unit vector. Then
|Mx| = |M(|x|u)| = |x| · |Mu| ≤ |x|||M ||
by the definition of ||M ||.
b) For any unit vector u, we have
|(MN)u| = |M(Nu)| ≤ ||M || · |Nu| ≤ ||M || · ||N ||,
where we have used part (a) for the first inequality and the definition of norm for
the second.
Therefore ||M || · ||N || is at least as big as the maximum of |(MN)u| for all unit
vectors u, which is the norm of MN .
c) The matrix given takes an arbitrary unit vector
(
u
v
)
to the vector
(
v
−2u
)
.
The second vector is clearly not more than twice as long as the first vector. On the
other hand, if v = 0 then is it exactly twice as long. Therefore the norm is 2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
111
MATH1141 June 2015 Solutions
2. i) a) If f is differentiable on the open interval (a, b) and continuous on the [a, b], then
there exists a point c ∈ (a, b) such that
f(b)− f(a)
b− a = f
′(c).
b) Let x be a real number, x ≤ 2. The function f satisfies the requirements of the
Mean Value Theorem on [x, 2] so
f(2)− f(x)
2− x = f
′(c)
for some c ∈ (2, x). Hence
f(2)− f(x)
2− x ≤ 1⇒ f(x) ≥ x.
ii) The limit satisfies the conditions for L’Hopital’s rule, so limx→0 g(x) = limx→0
f(x)−f(0)
2x =
limx→0
f ′(x)
2 =
f ′(0)
2 .
iii) a) Now
f(−x) =
∫ (−x)3
0
(t2 − 1)et2dt.
Replace −t with t in the integral, so replace dt with −dt, then
f(−x) = −
∫ x3
0
(t2 − 1)et2dt = −f(x).
Hence f is odd.
b) Set f ′(x) = 3x2[(x6 − 1)ex6 ] = 0 for a stationary point, giving x = 0, 1,−1.
c) Around x = 0, f ′(x) is positive so we have an inflection point at x = 0.
Around x = 1, f ′(x) moves from negative to positive so we have a minimum at
x = 1.
Around x = −1, f ′(x) moves from positive to negative so we have a maximum at
x = −1.
iv) a) Differentiating implicitly and solving for y′ we have
y′ = −y
x
3x2 + y2
x2 + 3y2
.
b)
dA
dx
= y + xy′ = 0
for a stationary point. Substituting and noting that x, y > 0 we have y = x.
Substituting this back into the equation fo curve we obtain x = y = α, where
α = 1√
42
.
c) Passing to polar coordinates, we have r4 sin θ cos θ = 1 or r4 sin(2θ) = 2.
Also A = xy = r2 sin θ cos θ = 1r2 .
c©2020 School of Mathematics and Statistics, UNSW Sydney
112
d) We can write A =
√
sin(2θ)√
2
which has a maximum when θ = π4 . Thus, y = x and
this is the line on which A has the maximum point x = y = α, where α = 1√
42
.
3. i) Let A be the point (1, 0, 1) and let v =

 01
1

; then −→AP =

 12
0

−

 10
1

 =

 02
−1

,
−→
AP · v = 1 and v · v = 2.
The distance is then
|−→AP − proj
v
−→
AP
| =
∣∣∣∣∣−→AP −
−→
AP · v
v · v v
∣∣∣∣∣ =
∣∣∣∣∣∣

 02
−1

− 1
2

 01
1


∣∣∣∣∣∣ =
3
2
∣∣∣∣∣∣

 01
−1


∣∣∣∣∣∣ =
3√
2
.
ii) a) |a+b|2+ |a−b|2 = (a+b) ·(a+b)+(a−b) ·(a−b) = 2a ·a+2a ·b = 2|a|2+2|b|2
b) Now, |a| and |b| are the side lengths of the parallellogram
whereas |a+ b| = |−−→AB +−−→AD| = |−−→AB +−−→BC| = |−→AC|
and |a− b| = |−−→AB −−−→AD| = |−−→AB +−−→DA| = |−−→DB| are the diagonal lengths.
Therefore, a) states that the sum of the squared diagonal lengths is twice the sum
of the squared side lengths.
b) Now, |a| and |b| are the side lengths of the parallellogram
whereas |a+ b| = |−−→AB +−−→AD| = |−−→AB +−−→BC| = |−→AC|
and |a− b| = |−−→AB −−−→AD| = |−−→AB +−−→DA| = |−−→DB| are the diagonal lengths.
Therefore, a) states that the sum of the squared diagonal lengths is twice the sum
of the squared side lengths.
iii) a)
cos 5θ + i sin 5θ = ei5θ = (eiθ)5 = (cos θ + i sin θ)5
= cos5 θ + 5i cos4 θ sin θ − 10 cos3 θ sin2 θ − 10i cos2 θ sin3 θ + 5cos θ sin4 θ + i sin5 θ
= cos5 θ − 10 cos3 θ sin2 θ + 5cos θ sin4 θ + i(5 cos4 θ sin θ − 10 cos2 θ sin3 θ + sin5 θ)
By comparing the imaginary parts, we see that
sin 5θ = 5cos4 θ sin θ − 10 cos2 θ sin3 θ + sin5 θ
= 5(1− sin2 θ)2 sin θ − 10(1 − sin2 θ) sin3 θ + sin5 θ
= 16 sin5 θ − 20 sin3 θ + 5 sin θ
= 16x5 − 20x3 + 5x .
b) Set θ = π10 . By a), p(x) = 16x
5−20x3+5x−1 = sin 5θ−1 = sin 5π10 −1 = 1−1 = 0.
c) 1, −1±
√
5
4 .
d) By parts b) and c), sin π10 must be one of the roots 1,
−1±√5
4 .
Since it is not 1 and is not negative, we see that
sin
π
10
=
−1 +√5
4
.
iv) det(B) = 0 and det(C) = −2(−1)2 det(A) = −14
since the third row of B is a linear combination of the first two rows
and since swapping rows twice on the transpose CT gives A with rows scaled by −1, 2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
113
v) a) Proof. |Qv|2 = (Qv) · (Qv) = vTQTQv = vT Iv = vTv = v · v = |v|2,
so |Qv| = |v|.
b) Proof. By a), |λ||v| = |λv| = |Qv| = |v|.
Since |v| 6= 0, we see that |λ| = 1, so λ = ±1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
114
THE UNIVERSITY OF NEW SOUTH WALES
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
=
1
a
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
c©2020 School of Mathematics and Statistics, UNSW Sydney