MATH1013 University Mathematics II Note
Billy Leung
June 2024
Introduction
This is a fundamental course in Mathematics. Topics covered in this course
include trigonometry, complex numbers, limits, continuities, derivatives, integration, ordinary differential equations and matrices. If you have studied Mathematics Extended Part M1 or M2 in the HKDSE curriculum, you might have
learned most of the content in this course.
This note will only list some important examples that help you better remember the concepts. We suggest readers to read this note for exam revision
only. If you have insufficient high school calculus knowledge and still hope to
score high in this course, there are plenty of online resources for you (e.g. just
search "basic calculus course", or read this note).
Recommended reading: George B. Thomas, Maurice D. Weir and Joel Hass:
Thomas’ Calculus (12th edition, Addison Wesley)
Contents
1 Trigonometry
1
2 Continuity
2
3 Differentiation
2
4 Integration
3
5 Ordinary Differential Equations
7
6 Matrices
8
1
Trigonometry
In high school, we learned the sine, cosine and tangent functions. Now, we
have three more trigonometric functions, the secant, cosecant and cotangent
1
functions:
1
cos(x)
1
csc(x) =
sin(x)
1
cot(x) =
tan(x)
sec(x) =
2
Continuity
A function f defined on a region E in R mapping each element from E to a
number in R is said to be continuous if for every x ∈ E, f (y) approaches to f (x)
as y is approaches x. We write this mathematically as
lim f (y) = f (x)
y→x
It is a must for you to memorize what the "lim" symbol means if you want to
pass this course.
Example 2.1. Let f : (0, +∞), f (x) = x1 . Is f continuous?
( It is tempting to think that f is not continuous because we know that g(x) =
1
for x ∈ R and x ̸= 0
x
is discontinuous at 0 for any c ∈ R. However, note
c
for x = 0
that the domain of f is (0, +∞). Looking back to the definition of continuity,
what we need to check in this scenario is whether the values of f (y) approach
the value of f (x) for x ∈ (0, +∞) as the input values y ∈ (0, +∞) approach x.
This can be verified as follows:
3
1
1
x
x−y
= lim
y→x xy
−h
= lim
(take y = x + h, as y → x, h → 0)
h→0 x(x + h)
0
=
x(x + 0)
=0
lim f (y) − f (x) = lim
y→x
y→x y
−
Differentiation
(x)
A function f : (a, b) → R is differentiable at x ∈ (a, b) if the limit limy→x f (y)−f
y−x
is finite (i.e. the right- and left- limits exist and are equal).
2
A differentiable function defined on X ⊂ R is automatically a continuous
function on X. To perform well in the exam, it is necessary to remember the
following in your calculation of derivatives:
f ′ (x) (f ′ : the derivative of f )
f (x)
c (constant)
0
xn (n ≥ 1 or n ≤ −1)
nxn−1
1
ln x
x
x
e
ex
sin x
cos x
cos x
− sin x
tan x
sec2 x
csc x
− csc x cot x
sec x
sec x tan x
cot x
− csc2 x
In addition, you ought to remember
d
(f (x) + g(x)) = f ′ (x) + g ′ (x)
dx
d
(f (x)g(x)) = f ′ (x)g(x) + g ′ (x)f (x)
dx
d
(f (g(x)) = f ′ (g(x)) + g ′ (x)
dx
g(x)f ′ (x) − f (x)g ′ (x)
d f (x)
(
)=
dx g(x)
g(x)2
whenever the derivatives are defined (and for the fourth rule involving division,
we of course need g(x) ̸= 0).
There might also be some "analysis" stuff in the exams that will be demonstrated later in the exam questions section.
Here is a rule that you might not have heard of in high school but is examined
in this course:
Theorem 3.1 (L’Hôpital’s Rule, simple version). Assume f, g : I → R are
differentiable functions defined on an open interval I ⊂ R, and c ∈ I with
(limx→c f (x), limx→c g(x)) = (0, 0) or (∞, ∞) (∞ can be +∞ or −∞). If
′
(x)
limx→c fg′ (x)
exists and is finite, then
f (x)
f ′ (x)
= lim ′
x→c g(x)
x→c g (x)
lim
This rule tells us, for instance, limx→0 sinx x = 1
4
Integration
If you know that integration means integrating small pieces of things into a
large piece of object, you should have no problems in understanding this topic.
3
To do well in tests and exams, it just suffices to calculate integrals effectively.
Here are some tricks in calculating integrals:
Integration by Part
Example
4.1. 1. Evaluate the following integrals using integration by parts.
R
(a) R x sec2 xdx
(b) sin(ln x)dx
Solution. (a) Note that (tan x)′ = sec2 x. Integration-by-parts to get
Z
Z
2
x sec xdx = xd tan x
Z
Z
sin x
=x tan x − tan xdx = x tan x −
dx.
cos x
To evaluate the last integral, let u = cos x. Then du = sin xdx and
Z
Z
du
sin x
dx =
= ln | cos x| + C.
cos x
u
To round up,
Z
(b) Let I =
x sec2 xdx = x tan x − ln | cos x| + C.
R
sin(ln x)dx. Integration by parts twice to get
Z
Z
I = x sin(ln x) − xd(sin(ln x)) = x sin(ln x) − cos(ln x)dx
Z
= x sin(ln x) − x cos(ln x) − xd(cos(ln x))
Z
= x sin(ln x) − x cos(ln x) + sin(ln x)dx
I=
1
x(sin(ln x) − cos(ln x)) + C.
2
Using partial fraction
Example 4.2. Evaluate the following integrals. (a)
R
2
(b) x3x+x+x−4
2 +x+1
R 2x2 −19x+54
(c) x3 −7x2 +8x+16
Solution. (a) Factorize the denominator to get
R
4x−1
x2 +x−2
q(x) := x2 + x − 2 = (x + 2)(x − 1).
Suppose the partial fraction decomposition of the integrand is
f (x) :=
4x − 1
x2 + x − 2
4
=
A
B
+
.
x−1 x+2
Multiply through by q(x) and group the like terms to get
4x − 1 = A(x + 2) + B(x − 1) = (A + B)x + (2A − B).
Comparing the coefficients to arrive at a system of linear equations
(
A + B =4
2A − B = − 1.
Solve the system to get A = 1 and B = 3. Thus,
Z
Z
Z
1
3
f (x)dx =
dx +
dx
x−1
x+2
= ln |x − 1| + 3 ln |x + 2| + C = ln (x − 1)(x + 2)3 + C.
(b) Factorize the denominator to get
q(x) := x3 + x2 + x + 1 = (x + 1) x2 + 1 .
Suppose the partial fraction decomposition of the integrand is
f (x) :=
x2 + x − 4
A
Bx + C
=
+ 2
.
3
2
x +x +x+1
x+1
x +1
Multiply through by q(x) and group the like terms to get
x2 + x − 4 = A x2 + 1 + (Bx + C)(x + 1) = (A + B)x2 + (B + C)x + (A + C).
Comparing the coefficients to arrive at a system of linear equations


A + B = 1
B+C =1


A + C = −4.
Solve the system to get A = C = −2 and B = 3. Thus,
Z
Z
Z
Z
−2
3x
2
f (x)dx =
dx +
dx −
dx
x−1
x2 + 1
x2 + 1
1
3
= ln
+ ln x2 + 1 − 2 arctan x + C.
2
(x − 1)
2
(c) Factorize the denominator to get
q(x) := x3 − 7x2 + 8x + 16 = (x + 1)(x − 4)2 .
Suppose the partial fraction decomposition of the integrand is
f (x) :=
2x2 − 19x + 54
A
B
C
=
+
+
.
x3 − 7x2 + 8x + 16
x + 1 x − 4 (x − 4)2
5
Multiply through by q(x) and group the like terms to get
2x2 − 19x + 54 = A(x − 4)2 + B(x + 1)(x − 4) + C(x + 1)
= (A + B)x2 + (−8A − 3B + C)x + (16A − 4B + C).
Comparing the coefficients to arrive at a system of linear equations
Subtract (2) from (3) to get
24A − B = 73
From (1) and (4), we get A = 3 and B = −1; and then it follows from (3)
that C = 2. Thus,
Z
Z
Z
Z
3
1
2
dx
f (x)dx =
dx −
dx +
x+1
x−4
(x − 4)2
2
= ln |x + 1|3 − ln |x − 4| −
+C
x−4
(x + 1)3
2
−
+ C.
= ln
x−4
x−4
Finding integrals recursively
R
Example 4.3. Let In = (ln t)n dt for any integer n ≥ 1. (a) Use integration
by parts to obtain a formula relating In and In−1 for any n ≥ 2.
(b) Find I1 . Use your
R formula in (a) to compute I2 and I3 .
Solution. Let In = (ln t)n dt for any integer n ≥ 1.
(a) Integrate by parts to get
Z
Z
n
n
n
In = t(ln t) − td ((ln t) ) = t(ln t) − n (ln t)n−1 dt = t(ln t)n − nIn−1
(b) Try to deduce I1 = t ln t − t + C. Using the formula in (a),
I2 = t(ln t)2 − 2I1 = t(ln t)2 − 2t ln t + 2t + C
I3 = t(ln t)3 − 3I2 = t(ln t)3 − 3 t(ln t)2 − 2t(ln t − 1) + C
= t(ln t)3 − 3t(ln t)2 + 6t ln t − 6t + C
We close this section byRthe following table:
f (x)
f (x)dx
0
C
xn+1
n
x (n ̸= −1)
n+1 + C
1
ln
x+C
x
ex
ex + C
ln x
x ln x − x + C
sin x
− cos x
cos x
sin x
tan x
ln | sec x| + C
csc x
− ln | csc x + cot x| + C
sec x
ln | sec x + tan x| + C
cot x
ln | sin x| + C
6
5
Ordinary Differential Equations
Ordinary differential equations can be a broad topic. But for elementary math
course, it is enough to memorize the methods to solve certain types of ordinary
differential equations. We now give some examples.
Example 5.1. Solve the IVP: y ′′ − 4y ′ + 4y = 0, y(0) = 4, y ′ (0) = 5.
Solution. Solving the characteristic equation
z 2 − 4z + 4 = (z − 2)2 = 0,
a general solution of the ODE is
y(x) = Ae2x + Bxe2x .
Impose the initial condition to get
y(0) = A = 4 and y ′ (0) = e2x (2(A + Bx) + B) x=0 = 2A + B = 5.
Solve this system of equations to get A = 4 and B = −3, i.e., the solution
of the IVP is y(x) = e2x (4 − 3x).
sin x
, y(0) = 1.
Example 5.2. Solve the IVP: y ′ = xyy+1
Solution. This ode is separable:
Z
Z
y+1
dy = x sin xdx
y
Z Z
1
1+
dy = xd(− cos x)
y
Z
y + ln |y| = −x cos x + cos xdx = −x cos x + sin x + C.
Impose the initial condition to get
1 + ln 1 = −0 cos 0 + sin 0 + C.
So the solution is y + ln |y| = −x cos x + sin x + 1.
Example 5.3. Solve the IVP: (x2 + 1)y ′ + 3x(y − 1) = 0, y(0) = 2.
Solution. This ode is linear.
x2 + 1 y ′ + 3xy = 3x
3x
3x
y= 2
y′ + 2
x +1
x +1
An integrating factor is
e
R
3x
dx
x2 +1
2
= eln(x +1)
7
3/2
= x2 + 1
3/2
.
Multiply the ode by this integrating factor to get
1/2
1/2
y ′ + 3x x2 + 1
y = 3x x2 + 1
Z
Z
1/2
3/2
3/2
3 1/2
u=x2 +1
2
2
u du = u3/2 + C = x2 + 1
+ C.
dx =
y = 3x x + 1
x +1
2
x2 + 1
3/2
Impose the initial condition to get
2 = 1 + C or C = 1.
So the solution is y = 1 + x2 + 1
6
−3/2
.
Matrices
Usually, the exam questions will be about calculating determinants or inverses.
We give an example.


1 1 2
Example 6.1. Find the determinant of A =  2 3 4 . Hence find its
1 0 1
inverse.
Solution. The determinant of A is 1 × 3 × 1 + 1 × 4 × 1 + 2 × 2 × 0 − 2 × 3 ×
1 1 2
1 − 1 × 2 × 1 − 1 × 4 × 0=−1. Since 2 3 4 = −1 ̸= 0, A is invertible and
1 0 1
its inverse is

T
2 3
2 4
3 4
−


T 

1 0
1 1
0 1


−3
3
2 −3

1 
1
1
1
2
1
2
 = −  −1 −1 1  =  −2
 −
−
0 1
1 1
1 0 
−1 


3
−2 0
1


1 1
1 2
1 2
−
2 3
2 4
3 4
Exam Questions
We use the 2018 Dec Exam to illustrate how to apply what you have learned to
answer exam questions.
Part I. Multiple Choice Questions
(42 marks) In each of the questions below, there is ONLY ONE correct answer.
Write your answer on the multiple-choice answer sheet. Each question carries
equal marks and there is no penalty for incorrect answers. No steps are required.
8
1
1
−1

2
0 
−1
√
1. Let z = a + −1. If −1/2 < a < 0, then z 3 is in the
(a) Quadrant I
(b) Quadrant II
(c) Quadrant III
(d) Quadrant IV
2. Let C be the graph of the complex equation arg 2 − z3 = 5π
6 . Which of
the following MUST
be
TRUE?
√
(I) C ⊂ {x + y −1 : x < 0, y > 0}, the second quadrant of the Argand
plane.
(II) C is contained in a circle whose radius is greater than one.
(a) (I) only
(b) (II) only
(c) All of them
(d) None of them
3. Refer to the figure below (may not be drawn in scale), sin(3θ) =
1
(a) − √1+b
2
2
2−b
(b) 2(1+b
2 )3/2
2
b −3
(c) 3(1+b
2 )3/2
2
3b −1
(d) (1+b
2 )3/2
4. Suppose that sin α and sin β are zeros of the polynomial x2 + ax − b where
α, β ∈ (−π/2, π/2) and a, b > 0. Which of the following MUST be TRUE?
(I) cos(α + β) > 0
(II) α + β < 0.
(a) (I) only
(b) (II) only
(c) All of them
(d) None of them
In questions 5 and 6,
9
(
f (x) =
sin2
2
x +x−12
x2 −4x+3 π
2
if x ∈ R\{1, 3}
if x = 1
5. limx→3 f (x)
(a) is 1
(b) is 2
(c) is +∞
(d) None of above
6. limx→1+ f (x)
(a) is 1
(b) is 2
(c) is +∞
(d) None of above
2
x ln |x| if x ̸= 0
7. Let f (x) =
. Then f ′ (0)
0
if x = 0
(a) is 0
(b) is −∞
(c) is a nonzero number
(d) None of above
3
2
−19x−6
.
In questions 8 and 9, f (x) = 3x +5x
x2 +x−6
8. Which of the following is a vertical asymptote of the graph of f ?
(I) x = −3
(II) x = 2.
(a) (I) only
(b) (II) only
(c) All of them
(d) None of them
9. Which of the following is an oblique asymptote of the graph of f ?
(a) y = 3x
(b) y = 3x − 1
(c) y = 3x + 2
(d) None of them
10. limx→+∞ ln(x) π2 − arctan x
(a) = 0
(b) > 0
(c) = +∞
(d) None of them
10
In questions 11 and 12 , consider the graph of the equation
2
2
ex −y + y + 2xy 2 = 2
Given that it passes through the point P (1, −1) and y = f (x) is the local
solution of this equation at P .
11. The equation of the tangent line at P is
(a) 3y = −4x + 1
(b) 3y = 4x − 7
(c) y = 2x − 3
(d) y = 4x − 5
12. Let Q be a point of the graph near P with coordinates (0.99, β). If the
second Taylor polynomial of f at the reference point 1 is used to approximate β, then β ≈
(a) -1.035
(b) -0.954
(c) -0.825
(d) -0.753
13. Let g : R → R be differentiable and g ′ be continuous. If 0 ≤ g(x) ≤ 1 + x2
R +∞
for all x, then 0 e−x (g(x) − g ′ (x)) dx
(a) diverges to +∞ no matter what g is
(b) depends on g(0) only
(c) depends on both g(0) and limx→+∞ g(x)
(d) none of above
In questions 14 − 16,
p
f (x) = ln x + 1 + x2
14. What are the domain and the range of f ?
(a)
(b)
(c)
(d)
domain
(0, +∞)
(0, +∞)
(−∞, +∞)
(−∞, +∞)
range
(0, +∞)
(−∞, +∞)
(−∞, +∞)
(0, +∞)
15. Which of the following MUST be TRUE?
(a) f has no inverse function
(b) f has inverse function g but g is not differentiable
(c) f has inverse function g and g ′ (ln 3) = 5/3
(d) f has inverse function g and g ′ (ln 3) = 3/5
16. Let a > 0. Then
Ra
a+f (x)
dx
−a a2 +x2
11
(a) is 0
(b) is 1.5708 (4 decimal places)
(c) exists but its value depends on a
(d) is Undefined
17. Given that the slope of the graph of a function y = f (x) at (x, y) is
2x2 +4xy+y 2
. If P (−1, 3) and Q(−3/4, k) are points on the graph of f ,
x2
then the slope of the graph at Q is
(a) -1
(b) 2
(c) 3
(d) 5
18. If y = ϕ(x) is the solution of the initial value problem
y ′′ − 2y ′ + 5y = 0, y(0) = 1, y ′ (0) = −3
then ϕ(π) =
(a) −eπ
(b) eπ
(c) 2eπ
(d) None of them
T
2
0 −2
β 6 −7
19. Let A =
and B =
. If AB is invert−4 −2 1
5 −5 2
ible, then β
(a) ̸= −1/4
(b) ̸= −1/2
(c) ̸= 3
(d) None of them


4
5
3
20. Given that A =  0 −3 −5  is invertible. The (2, 3)-entry of
−1 3
5
−1
1
is
8A
(a) −17/2
(b) −1/8
(c) 5/32
(d) 10




1 3 2
1+a 3+b 2+c
b
c =
21. Given that det  −1 1 4  = 3. Then det  a
a b c
−2
2
8
(a) -6
(b) -3
(c) 2
(d) 6
12
Part II. Long Questions
(58 marks) Write all your solution to the questions of this part in the answer
book provided separately. For full credits, you should give concise and precise
reasoning to your solution; and show your steps clearly.
3/2
22. (24 marks) Let f (x) = |x+3|
|x−4| . (a) Write f as a case-defined function.
(b) Determine all vertical, horizontal/oblique asymptotes of the graph of
f , if any.
(c) Determine all the critical points of f .
(d) Determine all the intervals of increase and the intervals of decrease of
f . Hence, determine all the local and absolute extrema of f .
√
r
23. (16 marks) Let f (x) = (1 − sin x)1/ x , x > 0, and we adopt that
√ 0 =0
for any positive real number r. For instance, f (π/2) = (1 − 1)1/ π/2 = 0.
(a) Determine whether x = 0 is a vertical asymptote to the graph of f .
(b) Show that f is continuous at x = 2nπ + π/2 where n ∈ N ∪ {0}.
(c) Determine whether f has infinitely many critical points.
x
24. (18 marks) Let f (x) = x2 +2x+5
. Given that
′
f (x) =
5 − x2
(x2 + 2x + 5)
2,f
′′
(x) =
2 x3 − 15x − 10
(x2 + 2x + 5)
3
and f ′′′ (x) < 0 over [−1, 0].
(a) Show that
Z
1
x+1
2
f (x)dx =
ln (x + 1) + 4 − arctan
+C
2
2
by employing suitable technique(s) of integration. (WARNING: You get
zero point if you simply verify the equation by the definition of indefinite
integral/antiderivative.)
R +∞
(b) Determine whether −1 f (x)dx exists.
Rx
(c) Define g(x) = −1 f (t)dt for any x ≥ −1. Approximate g(0) by the
degree two Taylor polynomial of g with reference point a = −1. Also, find
an upper bound for the absolute error committed.
Solution. 1:d 2:b 3:d 4:c 5:a 6:d 7:a 8:b 9:c 10:a 11:d 12:a 13:d 14:c 15:c 16:b
17:b 18:b 19:a 20:d 21:a
22 (a)

(x+3)3/2

for x ∈ (4, +∞)
 x−4


+∞ (or undefined) for x = 4
f (x) = (x+3)3/2

for x ∈ [−3, 4)

4−x

 (−3−x)
3/2

for x ∈ (−∞, −3)
4−x
13
(b) Note that limx→4+ f (x) = +∞ and it is not hard to show that x = 4 is the
only vertical asymptote.
Next, the graph of f has no horizontal asymptote because
limx→+∞ f (x) = +∞ and limx→−∞ f (x) = +∞. The graph of f has no
oblique asymptote as well because one can easily show for any m ̸= 0 and
f (x)
f (x)
c ∈ R, limx→+∞ mx+c
= 0 and limx→−∞ mx+c
= 0.
(c) The critical points are those x such that f ′ (x) = 0.
For x > 4, f ′ (x) = (x−18)(x+3)
2(x−4)2
1/2
; for x ∈ (−3, 4), f ′ (x) = (18−x)(x+3)
2(x−4)2
1/2
; and
1/2
for x < −3, f ′ (x) = (x−18)(−3−x)
2(x−4)2
′
(try to get these yourself). Finally, one
also has f (−3) = 0. It is now clear that the critical points of f are -3 and 18.
x
x < −3 x = −3 −3 < x < 4 4 < x < 18 x = 18 x > 18
(d)
f ′ (x)
0
+
0
+
Intervals of increase: (−3, 4) and (4, 18)
Intervals of decrease: (−∞, −3) and (18, +∞)
Local minimum point: x = −3, x = 18
Local maximum point: none
Absolute minimum: x = −3
Absolute maximum point: none
√
√
23 (a) As limx→0+ f (x) = limx→0+ (1 − sin x)1/ x ≤ limx→0+ 11/ x = 1 (if the
limit exists), x = 0 is not a vertical asymptote to the graph of f .
(b) The trick is to apply the sandwich theorem to
√
1 − sin x ≤ (1 − sin x)1/ x ≤ (1 − sin x)r
for some r ∈ R+ and r < √1x . The details are left to the readers.
√
−1
(c) As f ′ (x) = √
(1 − sin x)1/ x−1 · cos x for x > 0 whenever it is defined,
x
f ′ (2πn) = 0 for all n ∈ N, so f has infinitely many critical points.
π
24 (a) The trick is to let x = −1 + 2 tan θ with θ ranging from −π
2 to 2 . Then
2
you should get dx = 2 sec θdθ. The remainder of the solution is left to the
readers.
R +∞
(b) The integral 1 f (x)dx exists and equals +∞ because
Z +∞
1
x+1 a
[ln((x + 1)2 + 4) − arctan
]
a→∞ 2
2 −1
f (x)dx = lim
1
π
while limx→+∞ ln((x + 1)2 + 4) = +∞ and limx→+∞ arctan x+1
2 = 2
(c) Let P2 (x) be the degree two Taylor polynomial of g with reference point
a = −1. Then
g ′′ (−1)(x + 1)2
2
f ′ (−1)(x + 1)2
= 0 + f (−1)(x + 1) +
2
for x ∈ [−1, 0] by Fundamental Theorem of Calculus.
P2 (x) = g(−1) + g ′ (−1)(x + 1)2 +
14
f ′ (−1)(0 + 1)2
2
(−1)3 − 15(−1) − 10
5 − (−1)2
+
=
((−1)2 + 2(−1) + 5)2
((−1)2 + 2(−1) + 5)3
5
=
16
g(0) ≈ P2 (0) = 0 + f (−1)(0 + 1) +
g ′′′ (c)(1)3
| for some c ∈ (−1, 0)
6
f ′′ (c)
| (fundamental theorem of Calculus)
=|
6
c3 − 15c − 10
=| 2
|
3(c + 2c + 5)
13 − 15 · (−1) + 10
<|
|
3(02 + 2 · (−1) + 5)
26
=
81
Absolute error committed = |
15