Table Of ContentRussian Ministry of Education and Science
GUBKIN RUSSIAN STATE UNIVERSITY OF OIL AND GAS
Department of High Mathematics
Yu.O. KOROLEVA
A.V. KOROLEV
BRIDGING COURSE
IN MATH
Educational Handbook
Moscow 2015
ББК 51(075)
Review by:
Prof. V.V. Kalinin
Prof. S.V. Shaposhnikov
Koroleva Yu. O., Korolev A. V.
Bridging course in Math: Educational Handbook. – М.: Publishing
Center Gubkin Russian State University of Oil and Gas, 2015. – 65 p.
The present book deals with some background knowledge in
mathematics that covers school level. Many exercises are solved
with detailed explanations. The book is oriented for students of any
specialization and those who listens the preparatory course.
This publication is the owned by Gubkin Russian State University of
Oil and Gas and reproduction without the agreement of the university is
prohibited.
© Yu.O. Koroleva, A.V. Korolev, 2015
© Gubkin Russian State University
of Oil and Gas, 2015
INTRODUCTION
Independently what profession you chose for your future education, any
university program always includes a math course. In order to have a good start
for studying Math at a High school it is very useful to refresh the school math
program to “build the bridge” from school to University. The present educa-
tional manuscript contains all necessary background which must know every
student to successfully study and pass the university math course. The book col-
lects theoretical material that generalizes the school math topics as well a great
number of practical exercises with detailed solutions. There are lists of exercises
at the end of each section to practice more yourself.
We hope the presented educational material will be useful to a certain au-
dience. In particular, the book could be interesting for those who are preparing
for graduated school exams. We wish good luck to everyone in the studying of
mathematics!
Yu. O. Koroleva, A.V. Korolev
The work was partially supported by the grant of President of Russian Fed-
eration supporting young Russian scientists (project MK-4615.2015.1).
3
Content
I. NUMERICAL EXPRESSIONS……………………………………………………....5
How to compare numbers ............................................................................................ 8
Properties of integer powers ........................................................................................ 9
Short multiplication formulas ..................................................................................... 10
Exercises ........................................................................................................................ 11
II. FUNCTIONS ................................................................................................................ 12
Domain and codomain ................................................................................................. 12
Some important functions ............................................................................................ 13
1. Linear function .................................................................................................. 14
2. Quadratic function ............................................................................................ 15
3. Inversely proportional function ....................................................................... 18
4. Exponential function ......................................................................................... 19
5. Logarithmic function......................................................................................... 20
6. The root function ......................................................................................... 21
7. The absolute value function ........................................................................ 21
8. The piece-vice fun√ct𝒙𝒙ion...................................................................................... 22
Shifts and stretches of functions .....|.х..|.......................................................................... 23
Drawing of graph for fractional- linear function ...................................................... 25
Composition of functions ............................................................................................. 27
Exercises ........................................................................................................................ 29
III. POLYNOMIALS ........................................................................................................ 30
How to divide polynomials ........................................................................................... 31
Factorization of polynomials ....................................................................................... 33
Equations ....................................................................................................................... 36
n = 2 or about the quadratic polynomial .................................................................... 37
Method of intervals ....................................................................................................... 38
Exercises ........................................................................................................................ 41
IV. EXPONENTIAL AND LOGARITHMIC EXPRESSIONS ................................... 42
Exponential equations and inequalities…………………………………………….. 43
Logarithms .................................................................................................................... 44
The simplest logarithmic equations and inequalities ................................................ 45
Exercises ........................................................................................................................ 48
V. TRIGONOMETRY ...................................................................................................... 49
The main formulas of trigonometry............................................................................ 49
Trigonometric equations .............................................................................................. 54
Inverse trigonometric functions .................................................................................. 54
The simplest trigonometric equations ........................................................................ 55
Homogeneous trigonometric equations………………………………………….......57
Graphs for trigonometric functions ............................................................................ 58
................................................................................... 59
................................................................................... 60
Exer1c.i yse(s𝒙𝒙 .)..=..s..i.n.. .𝒙𝒙..., . . .𝒚𝒚...(.𝒙𝒙..)..=...c..o..s.. .𝒙𝒙..................................................................................... 61
2. y(𝒙𝒙)=𝒕𝒕𝒈𝒈 𝒙𝒙, 𝒚𝒚(𝒙𝒙)=𝒄𝒄𝒕𝒕𝒈𝒈 𝒙𝒙
Bibliography ................................................................................................................ 63
4
I. NUMERICAL EXPRESSIONS
Numbers 1,2,3,..., are called natural. The set of all natural numbers is de-
noted by . The set {0, is the set of all integer numbers. Ob-
viously, natural numbers are subset of integer, i.e. any natural number is integer
ℕ ±1,±2,±3,….}
one as well.
Rational numbers are numbers that can be expressed as irreducible frac-
tion, here m is an integer number, and n is natural. For example,
ℚ
𝑚𝑚 2 3 1
By the way, integer numbers are rational as well, because any integer can be
𝑛𝑛, 5 ,11,−2.
represented as The set of rational numbers is bigger than : there
𝑚𝑚
exist many fractions which are not integers, for example: .
1 , 𝑚𝑚 ∈ ℤ. ℚ ℤ
21 3
The fraction is just a symbolic notation of a number. The same number
5 ,8
can be represented by different ways: by common fractions as well as decim-
als. The decimal is a positional form of the number which looks as follows:
, for example, 1,263. A part that comes before the comma
is the integer part of the number, the part coming after comma is the fractional
1 2 𝑛𝑛 1 2
±𝑎𝑎 𝑎𝑎 …𝑎𝑎 ,𝑏𝑏 𝑏𝑏 …
one.
In order to convert the rational fraction to the decimal one needs to divide
the nominator by the denominator, for example,
1 5 1
In particular, the obtained decimal can have infinite number of decimals af-
2 = 10 = 0,5 𝑜𝑜𝑜𝑜 3 = 0,3333...
ter the comma, then it is called the infinite decimal.
To convert the decimal into the rational fraction, one needs to rewrite its
fractional part as natural number divided by the corresponding power of 10.
Then one shall write the integer part in front of the result. For example,
.
145 145
One shall to learn how to convert the decimal numbers to the rational frac-
23,145 = 23 + 1000 = 231000
tions since it is more convenient to make operations with the last ones. Indeed, it
is not clear at all how to add two decimals with infinite number of digits after
the comma. With such approach one could to add the averages of the original
decimals, however, this leads to not precise answer.
The main property of the fraction is as follows. If one multiply the numera-
tor and denominator of the fraction by the same number, then the fraction keeps
the same value, while the fractions can be different, for example:
5
.
3 9 12
And vice verse, if both the nominator and denominator of the fraction have
4 = 12 = 16
the common factor, then both parts can be divided by that. This operation is
called reducing by a factor.
For example: – here both nominator and denominator is re-
12 12:4 3
duced by common factor 4.
16 = 16:4 = 4
The sum, multiplication and division of fractions are hold by the rules:
,
𝑚𝑚 𝑝𝑝 𝑚𝑚𝑞𝑞±𝑛𝑛𝑝𝑝 𝑚𝑚 𝑝𝑝 𝑚𝑚𝑝𝑝 𝑚𝑚 𝑝𝑝 𝑚𝑚𝑞𝑞
Fractions with 𝑛𝑛id±en𝑞𝑞tic=al d𝑛𝑛e𝑞𝑞nomina𝑛𝑛to∙ rs𝑞𝑞 a=re𝑛𝑛 v𝑞𝑞e,r y s𝑛𝑛im: 𝑞𝑞pl=e t𝑛𝑛o𝑝𝑝 a.dd, it is enough to
add nominators, while denominator keeps the same. For example, it is evident
that
1 5 6 3
It is much more difficult to add, for example, since denominators are dif-
8 + 8 = 8 = [𝑜𝑜𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑏𝑏𝑏𝑏 𝑡𝑡ℎ𝑟𝑟 𝑟𝑟𝑜𝑜𝑚𝑚𝑚𝑚𝑜𝑜𝑛𝑛 𝑓𝑓𝑎𝑎𝑟𝑟𝑡𝑡𝑜𝑜𝑜𝑜 𝑓𝑓𝑜𝑜𝑜𝑜 𝑛𝑛𝑟𝑟𝑚𝑚𝑏𝑏𝑟𝑟𝑜𝑜𝑛𝑛 6 𝑎𝑎𝑛𝑛𝑟𝑟 8] = 4.
1 1
ferent. Adding fractions, it is extremely important to be able to find the com-
8 + 4,
mon denominator that can be evaluated as LCM (the least common multiple) of
fractions. LCM of numbers a and b (denoted in the sequel by LCM(a ,b)) is the
lowest number that can be divided by both a and b. In our example
НОК(8 ,4)=8. Thus, one should represent the factor as to have the
1 1 2
common denominator. After that it remains only to add the nominators:
4 4 = 8,
1 1 1 2 3
Example. Evaluate:
8 + 4 = 8 + 8 = 8.
7 1 7 1 7 1
Solution.
+ , ∙ , : .
8 2 8 2 8 2
1. The ra-
7 1 7+1∙4 11
tio is the sum of fractions. Besides this, the answer can be written in the form
8 + 2 = ( 𝑡𝑡ℎ𝑟𝑟 𝑟𝑟𝑜𝑜𝑚𝑚𝑚𝑚𝑜𝑜𝑛𝑛 𝑟𝑟𝑟𝑟𝑛𝑛𝑜𝑜𝑚𝑚𝑑𝑑𝑛𝑛𝑎𝑎𝑡𝑡𝑜𝑜𝑜𝑜 𝑟𝑟𝑞𝑞𝑟𝑟𝑟𝑟𝑒𝑒𝑛𝑛 8 ) = 8 = 8 .
11
“integer part plus the remain”:
8
7 1 11 3
2.
8 + 2 = 8 = 18.
7 1 7
3.
8 ∙ 2 = 16.
7 1 7∙2 7
Example. To evaluate
8:2 = 8∙1 = [𝑟𝑟𝑟𝑟𝑡𝑡 𝑛𝑛𝑟𝑟𝑚𝑚𝑏𝑏𝑟𝑟𝑜𝑜𝑛𝑛 2𝑎𝑎𝑛𝑛𝑟𝑟 8 𝑏𝑏𝑏𝑏 𝑡𝑡ℎ𝑟𝑟𝑑𝑑𝑜𝑜 𝑟𝑟𝑜𝑜𝑚𝑚𝑚𝑚𝑜𝑜𝑛𝑛 𝑓𝑓𝑎𝑎𝑟𝑟𝑡𝑡𝑜𝑜𝑜𝑜 2] = 4.
2
Solution. One can see that the task is to add fractions of different types:
3 + 0,01.
the rational fraction and decimal. The first step one needs to do is convert the
terms to the unique type. It is always more convenient to work with rational
6
fractions, because this way allows to find the common denominators more easy.
Thus, represent the number Hence,
1 2 2 1
0,01 𝑎𝑎𝑛𝑛 100. 3 + 0,01 = 3 + 100 =
2∙100+3 203
|𝑜𝑜𝑟𝑟𝑟𝑟E𝑟𝑟x𝑟𝑟a𝑟𝑟m 𝑡𝑡𝑜𝑜p l𝑡𝑡eℎ. 𝑟𝑟E 𝑟𝑟v𝑜𝑜al𝑚𝑚ua𝑚𝑚te𝑜𝑜 t𝑛𝑛h e𝑟𝑟 𝑟𝑟v𝑛𝑛al𝑜𝑜u𝑚𝑚e 𝑑𝑑 𝑛𝑛𝑎𝑎𝑡𝑡𝑜𝑜𝑜𝑜| = 3∙100 = 300.
1
1 1 2
Solution. Let us calculate it by steps. Fir.stly, one needs to compute the value
−� − �
2 4 3
in parenthesis:
1 2 1∙3−2∙4 −5
Then
− = = .
4 3 12 12
1 1 1 12 1
= = [𝐿𝐿𝐿𝐿𝐿𝐿 (12 ,2 ) = 12] = = = 1 .
1 1 2 1 −5 6+5 11 11
−� − � −
2 4 3 2 12 12
Example. Simplify the expression
1 1
�𝑎𝑎+𝑏𝑏�(𝑎𝑎−𝑏𝑏)
1 1
Solution. Let us simplify the expres�s𝑎𝑎i−o𝑏𝑏n� (b𝑎𝑎y+ 𝑏𝑏u)s.ing the fraction properties.
1 1 𝑎𝑎+𝑏𝑏
� + �(𝑎𝑎−𝑏𝑏) (𝑎𝑎−𝑏𝑏) (𝑎𝑎+𝑏𝑏)(𝑎𝑎−𝑏𝑏)𝑎𝑎𝑏𝑏 𝑎𝑎−𝑏𝑏
𝑎𝑎 𝑏𝑏 𝑎𝑎𝑏𝑏
= = = = −1.
1 1 𝑏𝑏−𝑎𝑎 𝑎𝑎𝑏𝑏(𝑏𝑏−𝑎𝑎)(𝑎𝑎+𝑏𝑏) 𝑏𝑏−𝑎𝑎
Irratio�na−l nu�m(𝑎𝑎b+er𝑏𝑏s ) are real( 𝑎𝑎nu+m𝑏𝑏b)er that can not be represented as fraction
𝑎𝑎 𝑏𝑏 𝑎𝑎𝑏𝑏
, (where m ), i.e. any irrational number is not the rational ones. It
𝑚𝑚
can be written as infinity nonperiodic decimal fraction. For example, the num-
𝑛𝑛 ∈ ℤ, 𝑛𝑛 ∈ ℕ
bers , are irrational ones.
The union of all rational numbers together with irrational ones is the set of all
√2 𝑟𝑟 and 𝜋𝜋
real numbers which is denoted by . The real line is illustrated on the figure.
It includes both natural, integer, rational and irrational numbers.
ℝ ℝ
The distance between the number and origin is called the absolute value
of the number and is denoted by According to such definition,
𝑎𝑎
𝒂𝒂 |𝑎𝑎|.
−𝑎𝑎, 𝑑𝑑𝑓𝑓 𝑎𝑎 < 0
�
|𝑎𝑎| = � .
𝑎𝑎, 𝑑𝑑𝑓𝑓 𝑎𝑎 ≥ 0 7
For example, since
Example. Find a number such that
|𝜋𝜋 − 3| = 𝜋𝜋 − 3, 𝜋𝜋 − 3 > 0.
Solution. One needs to find such point on the real line that the distance
𝑎𝑎, |𝑎𝑎 − 1| = 2.
from it to 1 (the point a=1) equals 2. There are two points satisfying such prop-
𝑎𝑎
erties: or
𝑎𝑎 = −1 𝑎𝑎 = 3.
How to compare numbers
Sometimes it is quite important to point the numbers on the real line in the
correct order with respect to each other. This is a simple task if the numbers are
integer or rational. In the case the numbers are irrational, this question is not
trivial: one needs to know how to compare numbers with each other without a
calculator!
Example. To compare numbers and
Solution. It is clear that «approximately» both numbers are greater than 3,
1 + 2√2 √14.
but less than 4. We need to prove that one number is greater than the other one.
During the compare procedure we will write the symbol «v» between the num-
bers that substitute the unknown sign “>”, “<” or “=”. We use the equivalent
conversion:
v v v 14
2
v 14 v 5 2 v 32 v 25.
1+2√2 √14 ⇔ (1+2√2 ) √14 ⇔ 1+4√2 +8 ⇔
The symbol «v» is «>» since 32 > 25. Ther2efore, 2
9+4√2 ⇔ 4√2 ⇔ (4√2 ) 5 ⇔
Example. Compare numbers and 1+2√2 > √14.
−1−√5
Solution. Denote by «^» the symbol opposite to «v». Then
2 −√2.
v v ^
−1−√5
^ 8 ^ 2 20 ^ 4.
2 −√2 ⇔ −1 − √5 −2√2 ⇔ 1 + √5 2√2 ⇔
The symbol «^» is «>» since Hence, the opposite unknown symbol
1 + 2√5 + 5 ⇔ 2√5 ⇔
«v» is «<». Thus, the result is as follows: <
20 > 4.
−1−√5
2 −√2.
8
Properties of integer powers
Let be some number. Then the number is called the
𝑛𝑛
power o𝑎𝑎f a number Number is called base𝑎𝑎 an≡d �𝑎𝑎 �∙i�s𝑎𝑎 �e∙x⋯�p�o∙�n𝑎𝑎ent. The main
𝑛𝑛 𝑡𝑡𝑑𝑑𝑚𝑚𝑟𝑟𝑛𝑛
properties are as follows:
𝑛𝑛 𝑎𝑎. 𝑎𝑎 𝑛𝑛
1.
0 1
2.
𝑎𝑎𝑚𝑚 = 1, 𝑎𝑎 = 𝑎𝑎.
𝑛𝑛
3. 𝑛𝑛 𝑚𝑚
𝑎𝑎 = √𝑎𝑎 , 𝑤𝑤ℎ𝑟𝑟𝑜𝑜𝑟𝑟 𝑚𝑚 𝑎𝑎𝑛𝑛𝑟𝑟 𝑛𝑛 𝑛𝑛𝑎𝑎𝑡𝑡𝑟𝑟𝑜𝑜𝑎𝑎𝑒𝑒 𝑛𝑛𝑟𝑟𝑚𝑚𝑏𝑏𝑟𝑟𝑜𝑜𝑛𝑛.
1
−𝑛𝑛
4. 𝑛𝑛
𝑎𝑎 = 𝑎𝑎
𝑛𝑛 𝑚𝑚 𝑛𝑛+𝑚𝑚
5.
𝑎𝑎𝑛𝑛 ∙ 𝑎𝑎 = 𝑎𝑎
𝑎𝑎
𝑛𝑛−𝑚𝑚
6. 𝑚𝑚
𝑎𝑎 = 𝑎𝑎
7. 𝑛𝑛 𝑚𝑚 𝑚𝑚 𝑛𝑛 𝑛𝑛𝑚𝑚
(𝑎𝑎 ) = (𝑎𝑎 ) = 𝑎𝑎
𝑛𝑛 𝑛𝑛 𝑛𝑛
8. .
(𝑎𝑎𝑏𝑏) = 𝑎𝑎 𝑏𝑏
𝑛𝑛 𝑛𝑛
𝑎𝑎 𝑎𝑎
𝑛𝑛
Exa�m𝑏𝑏�ple=. S𝑏𝑏implify the expression
2 1
1 3 1 −2
2 1 2
�√𝑎𝑎� � � �𝑎𝑎 �
𝑎𝑎
1 2 .
4
Solution. We use the properties o�f𝑎𝑎 po�wers:
2 1 1 1
12 1 3 12 −2 12 2 −1 23 12 −2 1∙1+(−1)∙2+1∙�−1�
�√𝑎𝑎� � � �𝑎𝑎 � �𝑎𝑎 � (𝑎𝑎 ) �𝑎𝑎 � 22 3 2 2
𝑎𝑎 𝑎𝑎
1 2 = 1∙2 = 1
4 2
4
𝑎𝑎 𝑎𝑎
�𝑎𝑎 �
1 2 1 2
− − −
4 3 4 3 2 1 7
𝑎𝑎 𝑎𝑎 −3−2 −6 1
= 1 = 1 = 𝑎𝑎 = 𝑎𝑎 = 7.
Example. To simplify 2 2 . 6
3𝑎𝑎 𝑎𝑎 8 7 𝑎𝑎
Solution. By using the prop3√e2rty √th2at and some more properties
�𝑎𝑎 :𝑎𝑎 ∙ �𝑎𝑎𝑚𝑚 ∙ √𝑎𝑎
𝑛𝑛
of powers, one deduce: 𝑛𝑛 𝑚𝑚
𝑎𝑎 = √𝑎𝑎
1 8
3 8 7 3√2 1+7 7 1 1
�𝑎𝑎3√2:𝑎𝑎√2 ∙ �𝑎𝑎 ∙ √𝑎𝑎 = 𝑎𝑎 3 :𝑎𝑎√2 ∙ 𝑎𝑎 8 = 𝑎𝑎√2 ∶ 𝑎𝑎√2 ∙ 𝑎𝑎8 = 𝑎𝑎√2−√2+7 = 𝑎𝑎7
7
= √𝑎𝑎.
9
Short multiplication formulas
Let us write down the main short multiplication formulas that could be use-
ful to simplify different algebraic expressions.
1.
2. 2 2 2
(𝑎𝑎 ± 𝑏𝑏) = 𝑎𝑎 ± 2𝑎𝑎𝑏𝑏 + 𝑏𝑏
Not2e! The2re is no analogous formula for sum !
𝑎𝑎 − 𝑏𝑏 = (𝑎𝑎 − 𝑏𝑏)(𝑎𝑎 + 𝑏𝑏)
3. 2 2
𝑎𝑎 + 𝑏𝑏
4. 3 3 2 2 3
(𝑎𝑎 ± 𝑏𝑏) = 𝑎𝑎 ± 3𝑎𝑎 𝑏𝑏 + 3𝑎𝑎𝑏𝑏 ± 𝑏𝑏
3 3 2 2
𝑎𝑎 ± 𝑏𝑏 = (𝑎𝑎 ± 𝑏𝑏)(𝑎𝑎 ∓ 𝑎𝑎𝑏𝑏 + 𝑏𝑏 )
Example. Evaluate
√5+√6
Solution. Multiply both nominator and denominator of the fraction by
√6−√5 − 2√30.
√6 + √5:
2
√5+√6 �√5+√6�(√6+√5) �√5+√6�
−2√30 = −2√30 = 2 2 −2√30
√6−√5 �√6−√5�(√6+√5) √6 −√5
2 2
√5 +2√5√6+√6
Answer: 11.
= −2√30 = 5+2√30+6−2√30 = 11.
6−5
Example. Simplify the expression
2 2 −1 2 2
𝑎𝑎 +𝑏𝑏 1 1 𝑎𝑎 𝑏𝑏
Solution. We proceed in a few steps�:� 𝑏𝑏 − 𝑎𝑎�:�𝑎𝑎 + 𝑏𝑏� � ∙ 𝑎𝑎3+𝑏𝑏3
1. Simplification in the first parenthesis:
.
2 2 2 2
𝑎𝑎 +𝑏𝑏 𝑎𝑎 +𝑏𝑏 −𝑎𝑎𝑏𝑏
𝑏𝑏 − 𝑎𝑎 = |𝑟𝑟𝑜𝑜𝑚𝑚𝑚𝑚𝑜𝑜𝑛𝑛 𝑟𝑟𝑟𝑟𝑛𝑛𝑜𝑜𝑚𝑚𝑑𝑑𝑛𝑛𝑎𝑎𝑡𝑡𝑜𝑜𝑜𝑜| = 𝑏𝑏
2. The second parenthesis:
|common denominator|
−1 −1
1 1 𝑎𝑎+𝑏𝑏 𝑎𝑎𝑏𝑏
�𝑎𝑎 + 𝑏𝑏� = = � 𝑎𝑎𝑏𝑏 � = 𝑎𝑎+𝑏𝑏.
3. Now we divide the result of the first step by the result of the second:
2 2 −1 2 2 2 2 3 3
𝑎𝑎 +𝑏𝑏 1 1 𝑎𝑎 +𝑏𝑏 −𝑎𝑎𝑏𝑏 𝑎𝑎𝑏𝑏 (𝑎𝑎 +𝑏𝑏 −𝑎𝑎𝑏𝑏)(𝑎𝑎+𝑏𝑏) 𝑎𝑎 +𝑏𝑏
2
� 𝑏𝑏 − 𝑎𝑎�:�𝑎𝑎 + 𝑏𝑏� = 𝑏𝑏 :𝑎𝑎+𝑏𝑏 = 𝑏𝑏∙𝑎𝑎𝑏𝑏 = 𝑎𝑎𝑏𝑏 .
4. It remains to multiply both fractions:
3 3 2 2
𝑎𝑎 +𝑏𝑏 𝑎𝑎 𝑏𝑏
Answer: a. 2 3 3
𝑎𝑎𝑏𝑏 ∙ 𝑎𝑎 +𝑏𝑏 = 𝑎𝑎.
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