Unit 01 Sets, relations and functions Eng.doc

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SETS AND RELATIONS
SET
A set is a well-defined collection of distinct objects. Well-defined collection means that there
exists a rule with the help of which it is possible to tell whether a given object belongs or
does not belong to given collection. Generally sets are denoted by capital letters A, B, C, X,
Y, Z etc.
REPRESENTATION OF A SET
Usually, sets are represented in the following ways
ROASTER FORM OR TABULAR FORM
In this form, we list all the member of the set within braces (curly brackets) and separate
these by commas. For example, the set of all even numbers less than 10 and greater than
0 in the roster form is written as: A = {2,4, 6,8}
SET BUILDER FORM OR RULE FORM
In this form, we write a variable (say x) representing any member of the set followed by a
property satisfied by each member of the set. A = {x| x  5, x  N} the symbol ‘|’ stands for
the words” such that”.
1. TYPES OF SETS
NULL/ VOID/ EMPTY SET
A set which has no element is called the null set or empty set and is denoted by  (phi). The
number of elements of a set A is denoted as n (A) and n () = 0 as it contains no element.
For example the set of all real numbers whose square is –1.
SINGLETON SET
A set containing only one element is called Singleton Set.
FINITE AND INFINITE SET
A set, which has finite numbers of elements, is called a finite set. Otherwise it is called an in
finite set. For example, the set of all days in a week is a finite set whereas; the set of all
integers is an infinite set.
UNION OF SETS
Union of two or more sets is the set of all elements that belong to any of these sets. The symbol
used for union of sets is ‘’ i.e. AB = Union of set A and set B = {x: x  A or xB (or both)}
Example: A = {1, 2, 3, 4} and B = {2, 4, 5, 6} and C = {1, 2, 6, 8}, then ABC = {1, 2, 3, 4, 5, 6, 8}

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INTERSECTION OF SETS
It is the set of all the elements, which are common to all the sets. The symbol used for
intersection of sets is ‘’ i.e. A  B = {x: x  A and x B}
Example: If A = {1, 2, 3, 4} and B = {2, 4, 5, 6} and C = {1, 2, 6, 8}, then A  B  C = {2}
DIFFERENCE OF SETS
The difference of set A to B denoted as A  B is the set of those elements that are in the set
A but not in the set B i.e. A  B = {x: x A and x  B}
Similarly B  A = {x: xB and x A}
In general AB  BA
Example: If A = {a, b, c, d} and B = {b, c, e, f} then AB = {a, d} and BA = {e, f}.
Symmetric Difference of Two Sets:
For two sets A and B, symmetric difference of A and B is given by (A – B)  (B – A) and is
denoted by A  B.
SUBSET OF A SET
A set A is said to be a subset of the set B if each element of the set A is also the element of
the set B. The symbol used is ‘’ i.e. A  B  (x A  x  B).
Each set is a subset of its own set. Also a void set is a subset of any set. If there is at least
one element in B which does not belong to the set A, then A is a proper subset of set B and is
denoted as A  B. e.g If A = {a, b, c, d} and B = {b, c, d}. Then BA or equivalently AB (i.e
A is a super set of B). Total number of subsets of a finite set containing n elements is 2n.
Equality of Two Sets:
Sets A and B are said to be equal if AB and BA; we write A = B.
DISJOINT SETS
If two sets A and B have no common elements i.e. if no element of A is in B and no element
of B is in A, then A and B are said to be Disjoint Sets. Hence for Disjoint Sets A and B n
(A  B) = 0.
Some More Results Regarding the Order of Finite Sets:
Let A, B and C be finite sets and U be the finite universal set, then
(i). n (A  B) = n (A) + n (B) – n (A  B)
(ii). If A and B are disjoint, then n (A  B) = n (A) + n (B)
(iii). n (A –B) = n (A) – n (A  B) i.e. n (A) = n (A – B) + n (A  B)
(iv). n (A  B  C) = n (A) + n (B) + n (C) – n (A  B) – n (B  C) – n (A  C) + n (A  B
 C)
(v). n (set of elements which are in exactly two of the sets A, B, C)
= n (A B)+n (B  C) + n (C  A) –3n(A  B  C)
(vi). n(set of elements which are in atleast two of the sets A, B, C)
= n (A  B) + n (A  C) + n (B  C) –2n(A  B  C)

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(vii). n (set of elements which are in exactly one of the sets A, B, C)
= n (A) + n (B) + n (C) – 2n (A  B) – 2n (B  C) – 2n (A  C) + 3n (A  B  C)
Illustration -1: If A and B be two sets containing 3 and 6 element respectively,
what can be the minimum number of elements in A  B? Find also, the
maximum number of elements in A  B.
Solution: We have, n (A  B) = n(A) + n(B) – n(A  B)
This shows that n (A  B) is minimum or maximum according as
n (A  B) is maximum or minimum respectively.
Case 1: When n (A  B) is minimum, ie. n (A  B) = 0. This is possible only
when A  B = . In this case,
n(A  B) = n (A) + n (B) – 0 = n(A) + n (B) = 3 +6 = 9
n (A  B)max = 9
Case 2: When n (A  B) is maximum
This is possible only when A  B.
In this case n (A  B) = 3
 n (AB) = n(A) + n(B) – n (A B) = (3+6-3)=6
n (A  B)min = 6.
Illustration 2: In a group of 1000 people, there are 750 who can speak Hindi and 400
who can speak Bengali. How many can speak Hindi only? How many
can speak Bengali? How many can speak both Hindi and Bengali?
Solution: Total number of people = 1000
n (H) = 750
n (B) = 400
n (H  B) = n (H) + n (B) – n (H  B)
n (H  B) = 750 + 400 – 1000
= 150 speaking Hindi and Bengali both.
People speaking only Hindi = n (H) – n (H  B) = 750 – 150 = 600
People speaking only Bengali = n (B) – n (H  B) = 400 – 150 = 250.
Illustration 3: A survey shows that 63% of the Americans like cheese whereas
76% like apples. If x% of the Americans like both cheese and apples,
find the value of x.
Solution: Let A denote the set of Americans who like cheese and let B denote those
who like apples. Let the population of America be 100. Then,
n(A) = 63, n(B) = 76
Now, n(A  B) = n(A) + n(B) – n(A  B)
 n(AB) = 63+76-n(A  B)
 n (A  B) = 139 – n(A  B)
But n(AB)  100  n (A  B )  39 …(i)
Now, A  B  A and A  B  B
n(A  B )  n (A) and n (A  B)  n (B)
n (A  B)  63 …(ii)
From (i) and (ii), we have 39  n (A B )  63  39  x  63.
UNIVERSAL SET

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A non-empty set of which all the sets under consideration are subsets is called the
universal set. In any application of set theory, all the sets under consideration will likely to
be subsets of a fixed set called Universal Set. As name implies it is the set with collection of
all the elements and usually denoted by ‘U’.
e.g. (1) set of real numbers R is a universal set for the operations related to real numbers.
COMPLEMENTARY SET
The complement of a set A with respect to the Universal Set U is difference of U and A.
Complement of set A is denoted by A (or AC) (or A). Thus A is the set of all the elements
of the Universal Set which do not belong to the set A.
A = U – A = {x: x  U and x  A}
we can say that A  A = U (Universal Set) and A  A =  (Void Set)
Some of the useful properties/operations on sets are as follows:
 A  U = U
 A   = 
 C = U
 UC = 
Algebra of Sets:
Idempotent Law: For any set A,
 A  A = A
 A  A = A
Identity Law: For any set A,
 A   = A
 A  U = A
Commutative Law: For any two sets A and B
 A  B = B  A
 A  B = B  A
Associative Law: For any three sets A, B and C
 (A  B)  C = A  (B  C)
 A  (B  C) = (A  B)  C
Distributive Law: For any three sets A, B and C
 A  (B  C) = (A  B)  (A  C)
 A  (B  C) = (A  B)  (A  C)
De Morgan’s Law: For any two sets A and B
 (A  B) = A  B
 (A  B) = A  B
POWER SET
The set of all subsets of a given set A is called the power set A and is denoted by P (A). P
(A) = {S: S  A}
For example, if A = {1, 2, 3}, then
P(A) = { ,{1},{2},{3},{1},{1,2},{1,3},{2.3},{1,2,3}}

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Clearly, if A has n elements, then its power set P(A) contains exactly 2n elements.
Some More Results:
 n (set of elements neither in A nor in B) = n (A  B) = n(A  B) = n (U) – n (A  B)
 n (A  B) = n(A  B) = n (U) – n (A  B)
 n (A  B) = n [(A – B)  (B – A)] = n [(A  B)  (A  B)] = n (A) + n (B) –
2n(A  B).
VENN DIAGRAM
The diagrams drawn to represent sets are called Venn diagrams
or Eule -Venn diagrams. Here we represent the universal set U by
points within rectangle and the subset A of the set U represented
by the interior of a circle. If a set A is a subset of a set B then the
circle representing A is drawn inside the circle representing B. If A and B are no equal but
they have some common elements, then to represent A and B by two intersecting circles.
Illustration 4: A class has 175 students. The following
table shows the number of students studying one or more of the
following subjects in this case
Subjects No. of students
Mathematics 100
Physics 70
Chemistry 46
Mathematics and Physics 30
Mathematics and Chemistry 28
Physics and Chemistry 23
Mathematics, Physics and Chemistry 18
How many students are enrolled in Mathematics alone, Physics alone
and Chemistry alone? Are there students who have not offered any one
of these subjects?
Solution: Let P, C, M denote the sets of students studying Physics, Chemistry and
Mathematics respectively.
Let a, b, c, d, e, f, g denote the number of elements (students) contained in
the bounded region as shown in the diagram then
a + d + e + g = 70
c + d + f + g = 100
b + e + f + g = 46
d + g = 30
e + g = 23
f + g = 28
g = 18
M
C
P
a
e
g d
f
b c
U
after solving we get g = 18, f = 10, e = 5, d = 12, a = 35, b = 13 and c = 60
 a + b + c + d + e + f + g = 153
A
A B
B

U

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So, the number of students who have not offered any of these three subjects
= 175 –153 = 22
Number of students studying Mathematics only, c = 60
Number of students studying Physics only, a = 35
Number of students studying Chemistry only, b = 13.
CARTESIAN PRODUCT OF SETS
The Cartesian product (also known as the cross product) of two sets A and B, denoted by
AB (in the same order) is the set of all ordered pairs (x, y) such that xA and yB. What
we mean by ordered pair is that the pair(a, b) is not the same the pair as (b, a) unless a = b.
It implies that AB  BA in general. Also if A contains m elements and B contains n
elements then AB contains mn elements.
Similarly we can define AA = {(x, y); xA and yA}. We can also define cartesian product
of more than two sets.
e.g. A1 A2A3  . . . . An = {(a1, a2, . . . , an): a1 A1, a2  A2, . . . , an  An}
Illustration 5: If A ={a, b, c} and B = {b, c, d} then evaluate
(i). AB , AB , AB and BA
(ii). AB and BA
Solution: (i) AB = {x: xA or xB}= {a, b, c, d}
AB = {x: xA and xB}= {b, c}
AB = {x: xA and x  B}= {a}
BA = {x: xB and xB}= {d}
(ii) AB = {(x, y): xA and yB}
= {(a, b), (a, c), (a, d), (b, b), (b, c), (b, d), (c, b), (c, c), (c, d)}
BA = {(x, y): xB and yA}
= {(b, a), (b, b), (b, c),(c, a), (c, b), (c, c),(d, a), (d, b), (d, c)}
Note that AB  BA.
RELATIONS
Let A and B be two non-empty sets then every subset of A  B defines a relation from A to
B and every relation from A to B is subset of A  B.
Let R  A  B and (a, b)  R. then we say that a is related to b by the relation R and write it
as
a R b. If (a, b)  R, we write it as a b.
Example Let A {1, 2, 3, 4, 5}, B = {1, 3}
We set a relation from A to B as: a R b iff a  b; a  A, b  B. Then
R ={(1, 1), (1, 3), (2, 3), (3, 3)}AB
Domain and Range of a Relation:
Let R be a relation from A to B, that is, let R  A  B. Then
Domain R = {a: a  A, (a, b)  R for some b  B}
i.e. domain of R is the set of all the first elements of the ordered pairs which belong to R.
Also Range R = {b: b  B, (a, b)  R for some a  A},
i.e. range R is the set of all second elements of the ordered pairs which belong to R.
Thus Dom. R  A, Range R  B.
R

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Total Number of Distinct Relations from A to B:
Suppose the set A has m elements and the set B has n elements. Then the product set A 
B i.e. P (A  B) will have 2mn elements. A  B has 2mn different subsets which are different
relations from A to B.
Inverse Relation:
Let R  A  B be a relation from A to B. Then inverse relation R–1  B  A is defined by
R–1 = {(b, a): (a, b)  R, a  A, b  B}. It is clear that
 a R b  b R–1 a
 dom R–1 = range R and range R–1 = dom R
 (R–1)–1 = R
Example: Let A = {1, 2, 3, 4}, B = {a, b, c} and R = {(1, a), (1, c), (2, a)}. Then
(i) dom R = {1, 2}, range R = {a, c}
(ii) R–1 = {(a, 1), (c, 1), (a, 2)}
Compositions of Relations:
Let R  A  B, S  B  C be two relations. Then compositions of the relations R and S
denoted by SoR  A  C and is defined by (a, c)  (S o R) iff  b  B such that (a, b)  R,
(b, c)  S.
Example: Let A = {1, 2, 3}, B = {a, b, c, d}, C = {, , }
R( A  B) = {(1, a), (1, c), (2, d)}
S ( B  C) = {(a, ), (a, ), (c, )}
Then S o R( A  C) = {(1, ), (1, ), (1, )}
One should be careful in computing the relation R o S. Actually S o R starts
with R and R o S starts with S. In general S o R  R o S
Also (S o R)–1 = R–1 o S–1, know as reversal rule
Relations in a Set:
Let R be a relation from A to B. If B = A, then R is said to be a relation in A. Thus relation in
a set A is a subset of A  A.
Identity Relation:
R is an identity relation if (a, b)  R iff a = b, a  A, b  A. In other words, every element of
A is related to only itself.
Universal Relation in a Set:
Let A be any set and R be the set A  A, then R is called the Universal Relation in A.
Void Relation in a Set:
 is called Void Relation in a set.
Properties of Relations in a Set:
Reflexive Relations:

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R is a reflexive relation if (a, a)  R,  a  A. It should be noted if there is at least one
element
a  A such that (a, a)  R, then R is not reflexive.
Example: Let A = {1, 2, 3, 4, 5}
R = {(1, 1), (3, 2), (4, 2), (4, 4), (5, 2), (5, 5)} is not reflexive because 3  A
and (3, 3)  R.
R = {(1, 1), (3, 2), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5)} is reflexive since
(a, a)  R,  a  A.
Symmetric Relations:
R is called a symmetric relation on A if (x, y)  R  (y, x)  R
That is, y R x whenever x R y.
It should be noted that R is symmetric iff R–1 = R
Let A = {1, 2, 3}, then R = {(1, 1), (1, 3), (3, 1)} is symmetric.
Anti-symmetric Relations:
R is called a anti-symmetric relation if (a, b)  R and (b, a)  R  a = b
Thus, if a  b then a may be related to b or b may be related to a, but never both.
Or, we have never both a R b and b R a except when a = b.
Example: Let N be the set of natural numbers. A relation R  N  N is defined by
x R y iff x divides y (i.e. x/y)
Then x R y, y R x  x divides y, y divides x  x = y
Transitive Relations:
R is called a transitive relation if (a, b)  R, (b, c)  R  (a, c)  R
In other words if a is related to b, b is related to c, then a is related to c.
Transitivity fails only when there exists a, b, c such that a R b, b R c but a c.
Example: Consider the set A = {1, 2, 3} and the relation
R1 = {(1, 2), (1, 3)}
R2 = {(1, 2)}
R3 = {(1, 1)}
R4 = {(1, 2), (2, 1), (1, 1)}
Then R1, R2 and R3 transitive while R4 is not transitive since in R4, (2, 1)  R4,
(1, 2)  R4 but (2, 2)  R4
Note:
 It is interesting to note that every identity relation is reflexive but every reflexive
relation need not be an identity relation. Also identity relation is reflexive, symmetric
and transitive.
Equivalence Relation:
A relation R in a set A is called an equivalence relation if
(i) R is reflexive i.e., (a, a)  R,  a  A
(ii) R is symmetric i.e., (a, b)  R  (b, a)  R
(iii) R is transitive i.e., (a, b), (b, c)  R  (a, c) R
The equivalence relation is usually denoted by the symbol ~.
R

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Equivalence Classes of an Equivalence Relation:
Let R be equivalence relation in A ( ). Let a  A.
Then the equivalence class of a denoted by [a] or { a } is defined as the set of all those
points of A which are related to a under the relation R. Thus [a] = {x : x  A, x R a}
It is easy to see that
(i) b  [a]  a  [b]
(ii) b  [a]  [a] = [b]
(iii) Two equivalence classes are either disjoint of identical.
as an example we consider a very important relation
x  y (mod n) iff n divides (x –y), is fixed positive integer. Consider n = 5 then
[0] = {x : x  0(mod 5)} = {5p : p  z} = {0, 5, 10, 15,....}
[1] = {x : x  1(mod 5)} = {x : x –1 = 5k, k  z} = {5k + 1: k  z} = {1, 6, 11, ...., –4, –9,....}
one can easily see that there are only 5 distinct equivalence classes viz. [0], [1], [2], [3] and
[4] when n = 5.
Illustration 6: N is the set of natural numbers. The
relation R is defined on N  N as follows:
(a, b) R (c, d)  a + d = b + c
Prove that R is equivalence relation.
Solution: (i) (a, b) R (a, b)  a + b = b + a
 R is reflexive.
(ii) (a, b) R (c, d)  a + d = b + c
 c + b = d + a
 (c, d) R (a, b)
 R is symmetric.
Now (iii) (a, b) R (c, d) and (c, d) R (e, f)  a + d = b + c & c + f = d + e
 a + d + c + f = b + c + d + e
 a + f = b + e  (a, b) R (e, f)
 R is transitive. Thus R is an equivalence relation on N  N.
Illustration 7: A relation R on the set of complex
numbers is defined by z1 R z2 
2
1
2
1
z
z
z
z


is real, show that R is an
equivalence relation.
Solution: (i) z1 R z2 
2
1
2
1
z
z
z
z


is real
 R is reflexive
  z1  C  0 is real.
(ii) z1 R z2 
2
1
2
1
z
z
z
z


is real  – 









2
1
1
2
z
z
z
z
is real
 









2
1
1
2
z
z
z
z
is real  z2 R z1  z1, z2  C
 R is symmetric.
(iii) Let z1 = a1 + ib1, z2 = a2 + ib2 and z3 = a3 + ib3
where a1, b1, a2, b2, a3, b3  R

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now z1 R z2 
2
1
2
1
z
z
z
z


is real

   
   
2
2
1
1
2
2
1
1
ib
a
ib
a
ib
a
ib
a






is real 
   
   
2
1
2
1
2
1
2
1
b
b
i
a
a
b
b
i
a
a






is real

   
   
   
   
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
b
b
i
a
a
b
b
i
a
a
b
b
i
a
a
b
b
i
a
a













is real

           
 
   2
2
1
2
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
b
b
a
a
a
a
b
b
a
a
b
b
i
b
b
b
b
a
a
a
a














is real
 (a1 + a2) (b1 –b2) –(a1 – a2) (b1 + b2) = 0 (for purely real, imaginary part = 0)
 2a2b1 –2b2a1 = 0

2
1
2
1
b
b
a
a
 
2
2
1
1
b
a
b
a

similarly, z2 R z3 
3
3
2
2
b
a
b
a

z1 R z2 and z2 R z3 
2
2
1
1
b
a
b
a

and
3
3
2
2
b
a
b
a
 
3
3
1
1
b
a
b
a
  z1 R z3
 R is transitive
Hence R is equivalence relation.
CONGRUENCES
Let m be a positive integer, then the two integer a and b said to be congurent modulo m' if
a –b is divisible by m i.e. a –b = m where  is an positive integer.
The congruent modulo m' is defined on all a, b  I by a  b (mod m) iff a –b = ,   I+
Illustration 8:Find all congruent solutions of 8x  6 (mod 14)
Solution: given 8x  6 (mod 14)
  =
14
6
x
8 
where   I+
 8x = 14 + 6  x =
8
6
14 

 x =
4
3
7 

=
 
4
1
3
4 



x =  +
4
3
( + 1) where   I+
and here greatest common divisor of 8 and 14 is 2 so, there are two required
solutions
for  = 3 and  = 7, x = 6, 13
2. NUMBER SYSTEM
NATURAL NUMBERS

11
The numbers 1, 2, 3, 4…. Are called natural numbers, their set is denoted by N. Thus N =
{1, 2, 3, 4, 5….}
WHOLE NUMBERS
The numbers 0, 1, 2, 3, 4…… Are called whole numbers, their set is denoted by W. Thus
W = {0, 1, 2, 3, 4….}
INTEGERS
The numbers. …– 3, –2, –1, 0, 1, 2, 3….are called integers and their set is denoted by I & Z
 Set of positive integers denoted by I+ and consists of {1, 2, 3…}, also known as set of
natural numbers.
 Set of negative integers denoted by I– and consists of {…– 3, – 2, – 1}
 Set of non-negative integers is {0,1,2, 3…} also known as set of whole numbers
 Set of non-positive integers is {…– 3, – 2, – 1, 0}
RATIONAL NUMBERS
All numbers of the form p/q where p and q are integer and q  0, are called rational. Thus
:
p
Q = p,q I and q 0 andHFC of p,qis1
q


 
 
 
it may be noted that every integer is a
rational number it can be written as p/1. Examples are 1/3, – 4/9 and 57
The rational numbers are precisely the real numbers with decimal expansions that are either
 Terminating (ending in an infinite string of zeros), for example 3/4 = .75000… = .75
or
 Non-ermating Repeating (ending with a block of digits that repeats over and over).
For example 23/11 = 2.090909… = 2.09. The bar indicates the block of repeating digits.
IRRATIONAL NUMBERS
Real numbers that are not rational are called irrational numbers. They are precisely the real
numbers with decimal expansions that are non-terminating non-repeating. Their set is
denoted by Qc (i.e. complementary set of Q) Examples are 3
, 2, 5
 and 10
log 3
REAL NUMBERS
The complete set of rational and
irrational numbers is the set of real
numbers and is denoted by R.
Thus c
R = Q Q
 . The real numbers can
also be expressed in terms of position of
a point on the real line (is the number line
wherein the position of a point relative to
the origin (i.e. 0) represents a unique real
number and vice versa). All the numbers
defined so far follow the order property
i.e. if there are two numbers a & b then
either a < b or a = b or a > b
-3 -2 -1 0 1 2 3
2 
Real line

12
COMPLEX NUMBERS
A number of the form x + iy, where x and y are real numbers and i = -1, is called a
complex number. It is usually denoted by z. i.e. z = x + iy, x is called the real part and y the
imaginary part of z. C denotes the set of complex number.
Note: The system of complex numbers includes the system of real numbers,
i.e. R C
 .
Illustration 9: If a and b are two rational numbers such that
2 2
(a - b + 1) + 3 (a + b - 3) = 0 , find a and b.
Solution: 2 2
(a - b +1) + 3 (a + b -3) = 0
{Rational} {Irrational}
 a2 – b + 1 = 0 and a2 + b – 3 = 0
 a = 1 and b = 2
Illustration 10:Prove irrationality of the number 50.
Solution: Let tan50 be a rational number then
0
0
2 0
2tan5
tan10 =
1- tan 5
is also rational
0 3 0
0
2 0
3tan10 - tan 10
tan30 =
1-3tan 10
 is also rational which is not true
0
tan5
 is an irrational number.
3. INTERVALS
A subset of the real line is called an interval. Intervals are important in solving inequalities
or in finding domains etc. If there are two numbers a, b R such that a < b, following type of
intervals can be defined
FINITE INTERVALS
Open Interval: (a, b) = {x|a < x < b}
Close Interval:    
a,b = x a x b
 
Open-close Interval:    
a,b x | a x b
  
a b
a b
a b

13
Close-open Interval:   
a, b x a x < b
  

INFINITE INTERVALS
   
,
a x x a
  
   
,
a x x a
  
a
   
,b x x b
  

  
,b x x b
  
   
, R set of allrealnumber
  
4. INEQUALITIES
If p, q and r are real numbers, then
 > or =
p q eitherp q p q
 
     
p q p r q r
 p < q and q<r p <r


> > 0
< < 0
kp kq if k
kp kq if k
p q

 


< >
> <
r r
r r
p q if r 0
0
p q if r 0


  


p q
 +
1
+ 2 R
a a
a
   and equality holds for a = 1
 R
1
a + 2 a
a

    and equality holds for a = –1
Inequation Involving Exponential Expression:
 If k > 0, then kx > 0 for all real x.
 If k > 1, then kx > 1, when x > 0
 If 0 < k < 1, then kx < 1, when x > 0 and kx > 1, when x < 0.
Illustration 11: Solve the inequality (x)x – 2 > 1.
Solution: Case I: When x > 1, xx – 2 > 1
 x – 2 > 0  x > 2
So, solution set is x  (2, ).
Case II: When 0 < x < 1, xx – 2 > 1
 x – 2 < 0  x < 2.
So, solution in this case is x  (0, 1)
So, solution set is x  (0, 1)  (2, ).
a
b
b
a b

14
Illustration 12: Find the maximum value of 2
2x
x + 5x +1
where x +
∈
R .
Solution: Given that y = 2
2x
x +5x +1
=
2
1
x + +5
x
.
This will be maximum when denominator will be minimum. We know
1
x + 2 for x R
x
   . Minimum value of
1
x +
x
is 2.
Hence the maximum value of y is
2
7
.
5. WAVY CURVE METHOD
In order to solve the inequalities of the form
 
1 2 k
p
1 2
n n n
1 2 k
n
n n
1 2 p
(x a ) (x a ) ......(x a )
(x) 0 0, 0, 0
(x b ) (x b ) ......(x b )
  
    
  
f
where n1, n2, ……. , n k , m1, m2, ……. , mp are real numbers and a1, a2, ……. , ak, b1, b2,
……., bp are any real number such that ai  bj where i = 1, 2, 3, ….k and j = 1, 2, 3, ….p.
Method:
Step - 1 First arrange all values of x at which either numerator or denominator is
becomes zero, that means a1, a2,….., ak, b1, b2, ….bp in increasing order say
c1, c2, c3,……. cp + k. Plot them on real line
c1
c2
c3
cp+k-2
cp+k-1 cp+k
Step -2  Value of x at which numerator becomes zero should be marked with dark
circles.
Step - 3  All pints of discontinuities (x at which denominator becomes zero) should be
marked on number line with empty circles. Check the value of (x) for any
real number greater than the right most marked number on the number line.
Step - 4  From right to left draw a wavy curve (beginnings above the number line in
case of value of (x) is positive in step–3 otherwise from below the number
line), passing thoroughly all the marked points. So that when passes through
a point (exponent whose corresponds factor is odd) intersects the number
line, and when passing thoroughly a point (exponent whose corresponds
factor is even) the curve doesn’t intersect the real line and remain on the
same side of real line.
Step - 5  The appropriate intervals are chosen in accordance with the sign of
inequality (the function (x) is positive wherever the curve is above the
number line, it is negative if the curve is found below the number line). Their
union represents the solution of inequality
Illustration 13: Let  0

2 3 4
2 5
(x +1) (x - 2) (x - 4)
(x - 3) (x -7)
Solution: Step - 1  make on real line all x at which numerator becomes zero
with dark circles.

15
-1 2 4
Step - 2  mark point of discontinuity (value of x at which denominator
becomes zero) with empty circles
-1 2 4
3 7
Step - 3  Check (x) for x > 7, (8) > 0
- 1 2 3 4 7
Exponents of factors of –1, 3, 4 is even, hence wave will not change the
direction at these points.
Hence 
 
  
x , 1 1
,2 7,
      
Illustration 14: Let f(x) =
   
  
7
x
1
x
5
x
2
x
3
x





. Find intervals where f(x) is positive or
negative.
Solution: Here f(x) will possibly change
sign at –5, – 2, – 1, 3 and 7
numbers. Also note that f(x)
is not defined at x = – 1 and
7. For all x > 7, all the factors
in f(x) are > 0and so f(x)> 0;
for 3 < x < 7 all factors except
(x – 7) are > 0 and hence f(x)
< 0. We can continue like this
and we will have alternate sign changes.
Thus we have the following wavy curve:
f(x) > 0  x  (– 5, – 2)  (– 1, 3) (7, )
and
f(x) < 0  x  (– , – 5)(– 2, – 1)(3, 7)
Illustration 15: Solve the inequality
2
1
12
x
x
3
x
2
2




.
Solution:
2
1
12
x
x
3
x
2
2




 0

)
12
x
x
(
2
)
12
x
x
(
)
3
x
2
(
2
2
2






 0

)
12
x
x
(
2
18
x
3
x
2
2





 0
-5 -2 -1 3 7

16

)
12
x
x
(
2
18
x
3
x
2
2




 0 
( 3)( 6)
2( 4)( 3)
x x
x x
 
 
 0
Using wavy curve method
x  (–, –4)  [–3, 3)  [6, ).
Illustration 16:
I. Solve the inequation (x2 + x + 1)x  1.
II. Solve the inequality
2
2x -7x
(x - 3) > 1.
III. Solve the inequation 0

3 2
2x -1
2x + 3x + x
.
Solution (I) Given (x2 + x + 1)x
 1
x 0 ..................(1)
 
or 2
x 1 1 and x 0
x
   
 
1
,0 and x 0
x   
 
1
,0 ...............(2)
x
  
Taking Union of (1) and (2)  
[ 1, 0) 0,
  
(II) Case I: When x – 3 > 1……….(1)
   
2
3 7 2 7
3 1 2 7 0 , 0 , .........(2)
2
x x
x x x x
  
         
 
 
So, in this case solution set  
x 4,
 
Case II:  
0 < x 3 < 1 3, 4
x
  
2 7
2 7x< 0 0, but x (3, 4)
2
x x
 
   
 
 
So, in this case solution of
7
x 3,
2
 
 
 
So, complete solution set  
7
x 3, 4,
2
 
  
 
 
Case III:
3 2
2 1 2 1
0 0
1
2 3
2 ( 1)
2
x x
x x x
x x x
 
  
   
 
 
 
 
1 1
, 1 , 0 ,
2 2
x
   
        
   
   
6. ABSOLUTE VALUE
Let x  R.
Then the magnitude of x is called it’s absolute value and in
general, denoted by x and is defined as
0
< 0
x, x
x
x, x


 


.
Since the symbol a always denotes the nonnegative
square root of a, an alternate definition of x is 2
x x
 . O
x
y

17
Geometrically, x represents the distance of number ‘x’ from the origin, measured along the
number line. Similarly x – a represents the distance between x and a. There is another way
to define |x| as |x| = max {x, – x}.
Basic Properties
(i) x = x
(ii) x > a  x > a or x < – a if a R + and x  R if a  R–
(iii) x < a  – a < x < a if a R + and no solution if a  R– U {0}
(iv) x y = x y
(v) , 0
x
x
=
y y

y
(vi) x + y x + y
 . Here the equality sign holds if x and y either both are non-negative or
non-positive in other words x.y0.
(vii) y
~
x
y
x 
 Here the equality sign holds if x and y either both are non-negative or
non-positive in other words x.y0.
The last two properties can be put in one compact form namely,
y
x
y
x
y
~
x 



Illustration 17: Show that  x – 2  1
Solution:
Method 1:
Case I: If x  2…….(i)
x – 2  1  x  3…..(ii)
Taking intersection of (i) and (ii)
 x  [2, 3]………..(v)
Case II: If x < 2….(iii)
2 – x  1 x  1……...(iv)
Taking intersection of (iii) and (iv)
[1, 2)……...(vi)
Taking union of (v) and (vi)
x  [1, 3]
Method 2:
 x – 2  1, geometrically this represent
all the points on real line whose distance
from 2 is less than equal to 1
 x  [1, 3]
Method 3:
2 3
1 2
1 2 3

18
(Graphical method)
Firstly plot the graph of x –2 and y = 1
We have to find x for which x – 2 1
 Value of x for which graph of x – 2
lies below the graph of y = 1
 x  [1, 3]
Method 4: Since LHS and RHS in the given inequality are non-
negative, on squaring both the sides
 
2
x 2 1
       
x 1 x 3 0 x 1
,3
    
Illustration 18: Solve the inequality |x – 1| + |x – 2|  3.
Working Rule: First of all equate the expression to zero whose modulus occur in the
given inequation and from this find the values of x. These values of x will
divide the interval (–, ) into several parts. Then solve the inequation in
all these parts separately.
Solution: Given |x – 1| + |x – 2|  3,
Case I: x  1
Thus (–x + 1) – (x – 2)  3
–x + 1 – x + 2  3
–2x  0  x  0
So, solution is x  [0, 1] ……(1)
Case II: 1 < x  2
x – 1 – x + 2  3
So, solution set is x  (1, 2] ……(2)
Case III: x > 2
x – 1 + x – 2  3
2x – 3  3  2x  6  x  3
So, solution is x  (2, 3]
Combining all the three solution x  [0, 3].
Illustration 19:
I. Solve the following inequalities for real values of x:
(a) |x –3| > 5 (b) |2x –3| < 1 (c) 0 < |x –1|  3
II. Solve for x if |x2+x+1| = x2+x+1
Solution: (I) Given 3 5
x   all the points whose distance from 3 on real line
is greater than 5
-2 0 3
Hence    
, 2 3,
   
(II) Given 2 3 1
x  
1 2 3
y = 2 - x
y = x - 2
y = 1

19
1 2 3 1 1 2
x x
       
(III) Given 0 1 3
x
  
1 0 and 1 3
x x
   
1and 3 1 3
2 4
x R x
x
      
  
 
2,4 {1}
x
   
(IV) Given 2 2
1 1
x x x x
    
We Know
2
2 1 3
1 0
2 4
x x x
 
     
 
 
Hence 2 2
1 ,
x x x x x R
     
7. LOGARITHMIC FUNCTION
The logarithm of a given number b to the base ‘a’ is the exponent indicating the power to which
the base ‘a’ must be raised to obtain the number b. This number is designated as log a b.
Hence log a b = x  ax = b, a > 0, a  1 and b>0. From the definition of the logarithm of the
number b to the base ‘a’, we have an identity

a
log b
a =b,a >0,a 1&b>0
This is known as Fundamental Logarithmic Identity.
GRAPH OF LOGRITHMIC FUNCTION
y
x
O
logax (a > 1)
logax (0<a < 1)
1
PROPERTIES OF LOGARITHMIC FUNCTION
 The expression a
log b is meaningful for b >0 and for either 0 < a < 1 or a > 1.
 Let a > 1, then





a
positive if b>1
log b= zeroif b=1
negative if 0<b<1
and
If 0 < a < 1, then





b
negative if b>1
log a = zeroif b=1
positiveif 0 <b<1
 log a(mn) = log a m + log a n

20
 c
a
c
log b
log b =
log a
, c > 0 and c  1.
 log a
m
n
 
 
 
= log a m – log a n
 log a mn = n log a m
 b
a
1
log a =
log b
provided both a and b are non-unity.
 loga1 = 0
 logaa = 1
 1 2
b 1 b 2
1 2
a a if b 1
log a log a
a a if 0 b 1
 

  
  

Illustration 20: Solve for x: 4 log x = log (15 x2 + 16)
Solution: x4 – 15 x2 – 16 = 0
 (x2 + 1)(x2 – 16) = 0
 x =  4
But log x is not defined when x = – 4, the x = 4 is the only answer.
Remark: The students often forget to test for positive values of argument for which only
log has defined.
Illustration 21: Solve for x: log ½ (x – 2) > 4
Solution: In such type of questions first we make the base same.
Given that log ½ (x – 2) > 4 log ½
½
log ½ (x – 2) > log ½
1/16
x – 2 < 1 / 16
x < 33/16
also x – 2 > 0  x > 2
hence x  (2, 33/16)
Illustration 22:
I. Solve the inequality loge (x2 – 16) < loge (4x – 11).
II. If log1/2 (x – 1) > 0, then find the interval in which x lies.
Solution: (I) Given log e (x2 –16) < log e (4x –11)
Since base is greater than one x2 – 4x – 15 < 0 and (x – 4) (x + 4)
> 0 and
11
4
x 
 
11
1
,5 and(x 4)( 4) 0and x >
4
x x
     
     
11
1
, 5 andx , 4 4, and x >
4
x
       
 
4, 5
x
 
(II) Given 
x-
1 x-
1
1/2 1/2 1/2
log >0 log >log
 
x -1<1 x < 2
Also x –1 > 0 Here x (1,3)


21
OBJECTIVE ASSIGNMENT
1. If A = { 1, 3, 5, 7, 9, 11, 13, 15, 17}, B = { 2, 4, …, 18} and N is the universal set,
then A ((A B )  B) is
(A) A (B) N
(C) B (D) None of these
Sol: We have, (A  B)  B = A
((A  B )  B)  A = A  A  = N.
Hence (B) is the correct answer.
2. If X = {8n – 7 n – 1 n  N} and Y = {49 (n-1) n  N}, then
(A) x  Y (B) Y  X
(C) X = Y (D) None of these
Sol: We have,
8n – 7 n-1 = (7+1)n – 7n-1= (nC272 + nC373+…+nCn7n)
= 49 (nC2+nC37+…+nCn7n-2) for n  2
For n = 1, 8n – 7n-1 = 0
Thus, 8n-7n-1 is a multiple of 49 for n  2 and 0 for n = 1. Hence X consists of all
positive integral multiple of 49 of the form 49 Kn, where Kn = nC2+nC3 7+…+nCn7n-2
together with zero. Also Y consists of all positive integral multiple of 49 including
zero. Therefore, X  Y.
Hence (A) is the correct answer.
3. If X and Y are two sets, then x  (Y X) equals
(A) X (B) Y
(C)  (D) None of these
Sol: We have,
X  (Y  X) = X  (Y  X)  Y = (X  X) Y =   Y  = 
Hence (C) is the correct answer.
4. Let A = {x : x is a multiple of 3} and B = { x : x is a multiple of 5}. Then A  B is
given by
(A) {3, 6, 9 …} (B) { 5, 10, 15, 20,..}
(C) {15, 30, 45, …} (D) None of these
Sol: Since x  A  B  x  A and x B  x is a multiple of 3 and x is a multiple of 5  x
is a multiple of 15.
Hence A  B = {x x is a multiple of 15} = {15, 30, 45, …}.
5. Which of the following is the empty set?
(A) {x  x is a real number and x2 – 1 = 0}
(B) {x  x is a real number and x2 + 1 = 0}
(C) {x  x is a real number and x2 – 9 = 0}
(D) {x  x is a real number and x2 = x+2 }
Sol: Since x2 + 1 = 0  x =  i.

22
6. Two finite sets have m and n elements. The total number of subsets of the first
set is 56 more than the total number of subsets of second set. The values of m
and n are
(A) 7, 6 (B) 6, 3
(C) 5, 1 (D) 8, 7
Sol: we are given 2m –2n = 56
By trial m = 6 and n = 3
7. The Sols of 8x = 6 (mod 14) are
(A) [8], [6] (B) [8], [4]
(C) [6], [13] (D) [8], [4], [16]
Sol: Note that x = [6] = {...., –6, –8, 6, 20, 34...} satisfies the given equation since for
x = 6, we have 8x –6 = 48 –6 = 42 and 14/42. Similarly for x = 20, 8x –6 = 154 and
14/154 etc.
Similarly x = [13] = {....., –15, –1, 13, 27, 41, .....} is a Sol of the given equation since
for x = 13
8x – 6 = 98 and 14/98.
Also it can be seen that [8], [14] and [16] are not the Sols of the given equation.
8. Assume R and S are (non empty) relations in a set A. which of the relations
given below is false
(A) If R and S are transitive, then R  S is transitive
(B) If R and S are transitive, then R  S is transitive
(C) If R and S are symmetric, then R  S is symmetric
(D) If R and S are reflexive, then R  S is reflexive
Sol: for example on the set A = {1, 2, 3}, the relations R = {(1, 1), (1, 2)} and
S = {(2, 2), (2, 3)} are transitive but relations R  S = {(1, 1), (2, 2), (1, 2), (2, 3)} is
not transitive, since (1, 2)  R  S and (2, 3)  R  S but (1, 3)  R  S.
9. If R be a relation < from A = {1, 2, 3, 4} to B = {1, 3, 5} i.e. (a, b) R iff a<b, then
RoR–1 is
(A) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
(B) {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
(C) {(3, 3), (3, 5), (5, 3), (5, 5)}
(D) {(3, 3), (3, 4), (4, 5)}
Sol: We have R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
 R–1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
hence R O R–1 = {(3, 3), (3, 5), (5, 3), (5, 5)}
10. If R and R are symmetric relations (not disjoint) on a set A, then the relation
R  R is

23
(A) reflexive (B) symmetric
(C) transitive (D) none of these
Sol: Since R  R are not disjoint, there is at least one ordered pair, say, (a, b) in
R  R.
but (a, b)  R  R  (a, b)  R and (a, b)  R
since R and R are symmetric relations, we get
(b, a)  R and (b, a)  R
and consequently (b, a)  R  R
similarly if any other ordered pair (c, d)  R  R, then we must also have,
(d, c)  R  R
hence R  R is symmetric.
11. Let  be the relation on the set R of all real numbers defined by setting a  b iff
|a –b| 
2
1
. Then  is
(A) reflexive and symmetric but not transitive
(B) symmetric and transitive but not reflexive
(C) transitive but neither reflexive nor symmetric
(D) none of these
Sol:  is reflexive since |a –a| = 0 <
2
1
for all a  R
 is symmetric since |a –b| <
2
1
 |b –a| <
2
1
 is not transitive. For if we take three numbers
8
1
,
3
1
,
4
3
, then
2
1
12
5
3
1
4
3


 and
2
1
24
5
8
1
3
1



but
2
1
8
5
8
1
4
3



Thus
3
1
4
3
 and
8
1
3
1
 but
4
3
(~)
8
1
Directions for questions 12 to 15
In a group of children, 35 play football out of which 20 play footballs only, 22 play
hockey; 25 play cricket out of which 11 play cricket only. Out of these 7 play cricket
and football but bot hockey, 3 play football and hockey but not cricket and 12 play
football and cricket both.
12. How many play all the three games?
(A) 5 (B) 6
(C) 8 (D) 10
13. How many play cricket and hockey but not football?
(A) 1 (B) 2
(C) 3 (D) 4

24
14. How many plays hockey only ?
(A) 8 (B) 10
(C) 12 (D) None of these
15. What is the total number of children in the group?
(A) 40 (B) 50
(C) 56 (D) 60
Solutions for Q. No. 12 to 15
12. Number of players playing all the games = 5.
13. number of players playing cricket and hockey
but not football = 2.
14. number of players playing hockey only = 12.
15. total number of children = 60.
Hence (D) is the correct answer.
20
7
5
3 2
11
12
Hockey
Football
Cricket
16. India today conducted a survey if people taking tea/ coffee. Out of 9000 people,
5550 people take tea, 3600 people take coffee and 1500 people take both tea and
coffee. How many people take neither tea nor coffee?
(A) 350 (B) 700
(C) 1120 (D) 1350
Sol: People taking neither tea nor coffee =
1350.
1350
4050
2100
1500
Tea
Coffee
17. If A =










 x
0
and
2
1
x
cos
:
x , B =










 x
3
and
2
1
x
sin
:
x
Then A  B =
(A)
5
,
6


 

 

(B)
7
,
6


 

 

(C) 




 

3
2
,
3
(D) 




 
6
5
,
0
Sol: A = {x : cos x > –1/2 and 0  x  }
B = {x : sin x > 1/2 and /3  x  }
cos x > –
2
1
 x <
3
2

25
A = 




 
3
2
,
0
sin x >
2
1
 x >
6

, x <
6
5
B = 




 

3
5
,
6
A  B = 




 

3
2
,
3
18. The relation R defined in A = {1, 2, 3} by aRb if  
2 2
a b 5. Which of the
following is false?
(A) R = {(1, 1) (2, 2), (3, 3), (2, 1), (1, 2), (2, 3), (3, 2)}
(B) R–1 = R
(C) Domain of R = (1, 2, 3)
(D) Range of R = {5}.
Solution : Let a = 1      
2 2 2
a b 5 1 b 5
    
2
b 1 5 b 1
,2.
Let     
2 2
a 2 a b 5
    
2
b 4 5 b 1
,2,3.
Let        
2 2 2
a 3 a b 5 9 b 5
       
2 2 2
a 3 a b 5 9 b 5
R {(1
,1),(1
,2), (2,1)(2,2),(2,3),(3,3)}
=
= R.
Domain of R = {x : (x, y) Î R} = {1, 2, 3}.
Range of R = {y : (x, y) Î R} = {1, 2, 3}.
19. The relation R defined on the set A = {1, 2, 3, 4, 5} by R = {(x, y) :
2 2
x y 16
- < is given by
(A) R = {(1, 1) (2, 2), (3, 3), (2, 1), (1, 2), (2, 3), (3, 2)}
(B) {(2, 2), (3, 2), (4, 2), (2, 4)}
(C) {(3, 3), (4, 3), (5, 4), (3, 4)}
(D) None of these.
Solution : We have R = {(x, y) : 2 2
x y 16}.
- <
      
2 2 2
Let x 1 x y 16 1 y 16
    
2
y 1 16 y 1
,2,3,4
      
2 2 2
Let y 2 x y 16 4 y 16
    
2
y 4 16 y 1
,2,3,4
      
2 2 2
Let y 3 x y 16 9 y 16

26
    
2
y 9 16 y 1
,2,3,4.
      
2 2 2
Let y 4 x y 16 16 y 16
    
2
y 16 16 y 1
,2,3,4,5.
      
2 2 2
Let y 5 x y 16 25 y 16
    
2
y 25 16 y 4,5.
R
{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3),
(4, 4), (4, 5), (5, 4), (5, 5)}.
The correct answer is (D).
20. If R = {(x, y) : x, y Î Z, x2 + y2 4} is a relation in Z, then domain of R is
(A) {0, 1, 2} (B) {–2, –1, 0}
(C) {–2, –1, 0, 1, 2} (D) None of these.
Solution : We have R = {(x, y) : x, y Z, x2 + y2
4}.
Let x = 0 x2 + y2  4      
2
y 4 y 0, 1
, 2.
Let         
2 2 2
x 2 x y 4 y 0 y 0.
R {(0,0), (0, 1), (0, 1), (0, 2),(0,2),( 1
,0),
= - - -
(1, 0), (1, 1), (1, –1), (–1, 1), (–1, –1), (2, 0), (–2, 0}.
Domain of R {x : (x,y) R} {0, 1
, 1
, 2, 2}
= Î = - -
21. If A = {a, b, d, e)}, B = {c, d, f, m} and C = {a, l, m, o), then C  (A  B) will be given
by
(A) {a, d, l, m} (B) {b, c, l, o}
(C) {a, l, m} (D) {a, b, c, d, f, l, m, o}
Sol: A  B = {a, b, c, d, e, f, l, m} ( A = {a, b, d, e, l}, B = {c, d, f, m})
C  (A  B) = {a, l, m}.
22. A relation R from C to R is defined by xRy iff x y.
= Which of the following
is correct?
(A) (2 + 3i) R 13 (B) 3 R (–3)
(C) (1 + i) R 2 (D) iR1.
Solution : 2 2
2 3i (2) (3) 13 13.
+ = + = ¹
(2 + 3i) R 13 is wrong.
2 2
3 (3) 0 9 3 3
= + = = ¹ -
3R (–3) is wrong
2 2
1 i (1) (1) 2 2
+ = + = ¹
(1 + i) R2 is wrong
2 2
1 i (1) (1) 2 2
+ = + = ¹
iR1 is correct.

27
23. If universal set is the set of real numbers R and A = {x| –5 < x  2, x  Z}, B = {–3, 0,
1} and C = {a, b, c}, then (A  B)  C is given by
(A)  (B) {a, b, c, –3, 0, 1}
(C) {a, b, c} (D) none of these
Sol: (A  B) = R – {–3, 0, 1}
(A  B)  C = .
24. Let R be a relation in N defined by R = {(1 + x, 1 + x2) :  
x 5,x N}. Which
of the following is false ?
(A) R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 25)}
(B) Domain of R = {2, 3, 4, 5, 6}
(C) Range of R = {2, 5, 10, 17, 26}
(D) at least one is false.
Solution : R = {(1+1, 1 + 1), (1 + 2, 1 + 4), (1 + 3, 1 + 9),
(1 + 4, 1 + 16), (1 + 5, 1 + 25)
= {(2, 2), (3, 5), (4, 10), (5, 17), (6, 26)}
Domain of R = {x : (x, y) Î R} = {2, 3, 4, 5, 6}
Range of R = {y : (x, y) Î R} = {2, 5, 10, 17, 26}
25. Let A = {(x, y)  y = ex, x R} and B = {(x, y)  = e-x, x  R}. Then
(A) A  B =  (B) A  B 
(C) A  B = R2 (D) none of these
Sol: y = ex and y = e–x will intersect at one point
A  B  
26. If A and B are two sets, then A  (A  B) equals
(A) A (B) B
(C)  (D) none of these
Sol: A  (A  B) = A
27. If A = {, {}}, then the power set of A is
(A) A (B) {, {}, A}
(C) {, {}, {{}}, A} (D) none of these
Sol: Power set means set of all the subjects
 power set of A = {, {}, {{}}, A}
28. A set contains n elements. The power set contains
(A) n elements (B) 2n elements
(C) n2 elements (D) none of these
Sol: Number of subsets = 2n, where n is number of elements in the given set.
29. If A = {1, 2, 3} and B = {3, 8}, then (A  B)  (A  B) is

28
(A) {(3, 1), (3, 2), (3,3), (3,8)} (B) {(1, 3), (2,3), (3,3), (8,3)}
(C) {(1, 2), (2,2), (3,3), (8,8)} (D) {(8,3), (8,2), (8,1), (8,8)}
Sol: A = {1, 2, 3}, B = {3, 8}
A  B = {1, 2, 3, 8}
A  B = {3}
 (A  B)  (A  B) = {(1, 3), (2, 3), (3, 3), (8, 3)}
30. Let A = {p, q, r, s} and B ={1, 2, 3} . Which of the following relations from A to B is not
a function?
(A) R1 = {(p, 1), (q, 2), (r, 1), (s, 2) (B) R2 = {(p, 1), (q, 1), (r, 1), (s, 1)
(C) R3 = {(p, 1), (q, 2), (r, 2), (s, 2) (D) R1 = {(p, 2), (q, 3), (r, 2), (s, 2)
Sol: R3 = {(p, 1), (p, 2), (r, 2), (s, 2)}
is not a function.
31. The set of all integers x such that | x – 3 | < 2 is equal to
(A) {1, 2, 3, 4, 5} (B) {1, 2, 3, 4}
(C) {2, 3, 4} (D) {–4, –3, –2}
Sol: | x – 3 | < 2  3 – 2 < x < 3 + 2  1 < x < 5
 x = 2, 3, 4.
 Required set = {2, 3, 4}.
Alternative method
x = 1  | x – 3 | = | 1 – 3 | = | – 2 | = 2 <
/ 2
 (a) and (b) are not correct
x = – 4  | x – 3 | = | – 4 – 3 | = | – 2 | = 7 <
/ 2
 (D) is not correct.
32. Which of the following does not have a proper subset?
(A) {x : x Q} (B) {x : x N, 3 < x < 4}
(C) {x : x Q, 3 < x < 4} (D) None of these.
Sol: The set {x : x  N, 3 < x < 4} is empty because there is no natural number between 3
and 4.
 This set cannot have a proper subset.
33. The set (A  B  C)  (A  B'  C')'  C' is equal to
(A) B  C' (B) A  C
(C) B'  C' (D) None of these
Sol: (A  B  C)  (A  B'  C')'  C'
= (A  B  C)  (A'  B  C)  C'
= [(A  A')  (B  C)]  C'
= (  B  C)  C' = (B  C)  C'
= (B  C')  (C  C') = (B  C')   = B  C'.
34. Let R be a relation in N defined by R = {(x, y) : x + 2y = 8}. The range of R
is

29
(A) {2, 4, 6} (B) {1, 2, 3}
(C) {1, 2, 3, 4, 6} (D) None of these.
Solution : R = {(x, y) : x + 2y = 8, x, y Î N}
x + 2y = 8

 
8 x
y
2
       
7
x 1 y N; x 2 y 3 N
2
       
5
x 3 y N; x 4 y 2 N
2
       
3
x 5 y N; x 6 y 1 N
2
       
1
x 7 y N; x 8 y 0 N
2
...................................................................
...................................................................
R {(2,3),(4,2),(6,1)
=
Range of R {y : (x,y) R) {1
, 2, 3}.
= Î =
35. Let A = {1, 2, 3, ......., 45} and R be the relation 'is square of' in A. Which of the
following is false?
(A) R = {(1, 1), (4, 2), (9, 3), (16, 4), (25, 5), (36, 6)}
(B) Domain of R = {1, 4, 9, 16, 25, 36}
(C) Range of R = {1, 2, 3, 4, 5, 6}
(D) At least one is false.
Sol: We have (1)2 = 1, (2)2 = 4, (3)2 = 9, (4)2 = 16, (5)2 = 25, (6)2 = 36.
 R = {(1, 1), (4, 2), (9, 3), (16, 4), (25, 5), (36, 6)}
Domain of R = {x ; (x, y)  R} = {1, 4, 9, 16, 25, 36}
Range of R = {y ; (x, y)  R} = {1, 2, 3, 4, 5, 6}
36. If R  A  B and S  B  C be two relations, then (SoR)–1 is equal to
(A) S–1oR–1 (B) R–1oS–1
(C) SoR (D) None of these
Sol: We have (SoR)–1 = R–1oS–1.
37: If an = {ax: x  N}, then 3N  7N =
(A) 3N (B) 7N
(C) N (D) 21N
Solution: (D)
We have 3N = {3x: x  N} = {3, 6, 9, 12,…}
And 7N = {7x: x  N} = {7, 14, 21, 28, 35, 42…}
Hence 3N  7N = {21, 42, 63, ….} = {21x: xN} = 21N

30
38: Let A = {x: x  R, x < 1}, B = {x: x  R, x–1  1} and A B = R – D, then
set D is:
(A)  
x : 1 < x ≤2 (B)  
x : 1 ≤x < 2
(C)  
x :1 ≤x ≤2 (D) None of these
Solution: We have A = {x: x  R, –1 < x < 1}, B = {x: x  R, x–1  –1 or x–1  1}
= {x: x  R, x  0 or x  2}
 A B = R – D
Where D = {x: x  R, 1  X < 2}
39:
 
2
7
log x -4x+5
7 = x -1, x may have values:
(A) 2, 3 (B) 7
(C) –2, – 3 (D) 2, – 3
Solution: We have
 
2
7
log x -4x+5
7 = x -1
 
log
2
x 4x 5 x 1 x
a x
a
     
  
x 3 x 2 0 x 2,3
     
40: Number of positive integral values of x satisfies the equation
1 1
2 2
log | x +1 | ³log 4 is:
(A) 1 (B) 2
(C) 3 (D) 4
Solution: Since base of log both the sides is same and lying between 0 and 1 hence
inequality will change and |x + 1| < 4
4 x 1 4 5 x 3
         . Hence positive integral values of x are 1 and 2.
41: The solution set of the inequality log 10 (x2 – 16) log10 (4x – 11)
(A) (3, 5] (B) (4, 5]
(C) (6, 5] (D) none of these
Solution: Since base of log is same both the sides and greater than 1, hence inequality
will remain same.
2 2
16 4 11 4 5 0 1 5
x x x x x
            ……………….(1)
Also 2
16 0 and 4x-11 0 either x -4 or x 4
x       ……………(2)
Taking intersection of (1) and (2) …
 
4,5
x 
42: The set of real values of x for which 0.2
x + 2
log 1
x
 is:
(A)  
-5
0
2
 
 


-∞
, ∪ ,+∞ (B)
5
,+
2
 





(C)    
-∞
,-2 ∪0,+∞ (D) none of these
x < 4
x > 4
1x 5
- 4 -1 4 5

31
Solution: Given  
0.2
x +2 x +2 1
log 1 or 0.2 <1
x x 5
 
 5x (x + 2) > x2  4x2 + 10x  0
 x  0 or x , – 5/2
Also
x +2
> 0
x
 x (x + 2) > 0  x < –2 or x > 0
 the solution set is (–, –5/2] (0, + )
43: If |x – 1| + |x2 + x + 1| = |x2 + 2x|, then x belongs to:
(A) (2, ) (B) [1, )
(C) (– 1, ) (D) (– 2, )
Solution: Above is possible if
(x– 1)(x2 + x + 1)  0 ….(1)
Now for x2 + x + 1
Discriminant = 1 – 4 = – 3 < 0
 x2 + x + 1 > 0  x  R ….(2)
From (1) and (2), (x – 1)  0  x  [1, ).
44: Solution set of the inequality
1
x
x+2 1
5 >
25
 
 
 
is:
(A) (– 2,0) (B) (0, )
(C) (– 5,5) (D) (– 2,2)
Solution: We have
1
x+2
If 1, than
1
5
25 m n
a a m n
 
 
  
 
  
   
x a >
2
x 2 x
2
5 5
2
x 2
x
x 2x 2
0
x
1
0 x (0, )
x


 
   
 
 
 
 
 
    
45: If
 
1
3
7 +5 2
x =
2 + 2 2
, then x belongs to:
(A) (2, 3) (B) (0,1)
(C) (– 1, 0) (D) (3,4)
Solution: Cube both the sides, we get
3 7+5 2 1
x = =
8
8(7+5 2)

32
1
2
x
  . Cleary x lies between 0 and 1.
46: If (log5 x) 2 + log5 x < 2, then x belong to:
(A)
1
,5
25
 
 
 
(B)
1 1
,
5 5
 
 
 
(C) (1, ) (D) none of these
Solution: We have (log5 x)2 + log5 x < 2
Put log 5 x = a then a2 + a < 2
 a2 + a –2 < 0  (a + 2) (a – 1) < 0
 –2 < a < 1 or –2 < log 5 x < 1
 5 –2 < x < 5
i.e. 1/25 < x < 5
47: If a2 + b2 – ab – a – b + 1  0 , a, b  R, then a + b is equal to:
(A) 5 (B) 2
(C) 9 (D) 4
Solution: Given that a2 + b2 – ab – a – b + 1  0
 2a2 + 2b2 – 2ab – 2a – 2b + 2  0
(a – b)2 + (a – 1)2 + (b– 1)2  0
 a = 1 and b = 1
48. Which of the following is a singleton set ?
(A)  
(x : x 5,x N) (B)  
(x : x 6,x Z)
(C)    
2
(x : x 2x 1 0,x N) (D)  
2
(x : x 7,x N)
Solution :   
x 5 x 5 
( x N)
49. The set 
2
{x : x x} may be equal to
(A) {0} (B) {1}
(C) {3} (D) { }.
Solution : (D) Since there is no number which is not equal to itself, the set

 
{x : x x}
50. If A =  
n
{x : x 3 ,n N} and B =  
n
(x : x 9 ,n 4), 
n N} then which of the
following is false ?
(A)  
A B {3, 9, 27, 81
, 243, 729, 6561}
(B)  
A B {9, 81
, 729, 6561}
(C)  
A B {3, 27, 243}
(D)  
A B {3, 27, 343, 6561}
Solution : A = {31, 32, 33, 34, 35, 36}
= {3, 9, 27, 81, 243, 729}
and B = {91, 92, 93, 94} = {9, 81, 729, 6561}

33
(a)  
A B {3, 9, 27, 81
, 243, 729, 6561}
(b)  
A B {9, 81
, 729}
(c) A B {3, 27, 243}
- =
(d)     
A B (A B) (B A)
= {3, 27, 243} ∪
{ 6561} ={ 3,27,243,6561}
51. If n(A) = 115, n(B) = 326, n(A–B)=47, then n 
A B is equal to
(A) 373 (B) 165
(C) 370 (D) None of these.
Solution : (a) n(A) = n(A–B) + n  B)
( A implies
115 = 47 + n  B)
( A .
    
n B) 115 47 68.
( A
     
n B) n(A) n(B) n(A B)
( A
52. If A =      
2 4
{x C: x 1} and B {x C: x 1}, thenA B is equal to
(A) {–1. 1} (B) {–1. 1. o. –i)
(C) {–i, i} (D) None of these.
Solution :       
2
x 1 x 1
,1. A { 1
,1}
   
4 2
x 1 x 1
,1
      
x i, i, 1
,1. B { i, i, 1
,1}

          
A B (A B) (B A) { i,i} { }.
i,i
53. If for  
  
N, N {ax : x N} , then the set 6N 8N
Ç is equal to
(A) 8 N (B) 48 N
(C) 12 N (D) 25 N.
Solution :   
6N 8N {6,12,18,24,30,....) {8,16,24,32....}
= {24, 48, ....} = 24N.
Alternatively, L.C.M. of 6 and 8 = 24.
  
6N 8N 24N.
54. If (x + 3, 4 –y) = (1, 7), then (x – 3, 4 + y) is equal to
(A) (–2, –3) (B) (–5,1)
(C) (3, 4) (D) None of these.
Solution : (x + 3, 4 –y) = (1, 7)  x + 3 = 1, 4 – y = 7
 x = 1 – 3 = –2, y = 4 –7 = –3.
(x 3,4 y) ( 2 3,4 3) ( 5,1).
- + = - - - = -
55. If A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, then (A × B)  (B×C) is equal to
(A) {1, 4} (B) {3, 4}
(C) {(1, 4), (3, 4)} (D) None of these.
Solution : A × B = {1, 2, 3}× {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

34
` B × C = {3, 4}× {4, 5, 6} = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
So (A × B) Ç (B×C) = {(3, 4)}
56. If A =
 
  
 
 
1
(x,y): y ,0 x R
x
and B = {(x, y) : y = –x, x Î R}, then
(A)  
A B A (B)  
A B B
(C) 
 
A B (D) None of these.
Solution : Y = 1/X or XY = 1.
So A is the set of all points on the rectangular hyperbola xy = 1 with
branches in I and III quadrants. y = –x represents a line with slope –1 and c is
equal to 0.
Therefore B is the set of all points on this line. Since the graphs of xy = 1 and y
= –x are non intersecting, we have 
 
A B .
Y
xy=1
O
X
xY=1
o
135
y= - x
57. If A = x
{ ( x,y): y =e ,x ∈
R} and B ={ ( x, y): y = x, x ∈
R} , then
(A) 
A B (B) 
A B
(C) 
 
A B (D) 
 
A B .
Solution : A is the set of all points on the graph of y = ex. B is the set of all points on the
line y = x. Since the curves are non intersecting, we have A B
Ç = f .
Y
(0,1)
O X
Y=x
o
45
x
y e
58. If   
aN=
{ an: n N}andbN cN=dN,wherea,b,c N and b, c are coprime,
then
(A) b = cd (B) c = bd

35
(C) d = bc (D) None of these.
Solution : We have 
bN=
{ bn: n N}, 
cN=
{ cn: n N} and 
dN=
{ dn: n N}
We have 
bN cN=dN
  
d =d.1 bN cN
  1
d bn and d = cn2 where n1, n2 Î N
b/d and c/d bc/d, because b and c are coprime.
Also bcbN and bc = cb  cN
   
bc bN cN or bc dN
   
3 3
bc dn forn N d/bc
bc d
=
59. 2 2} and B =
 
 
 
π 3π
θ: ≤
θ
≤
2 2
, then A ∩
B is
equal to
(A)
 
 
 
 
5
θ: ≤θ
≤
2 6
(B)


 
 
 
3
θ: ≤θ
≤
2
(C)
  
 
 
 
5 3
θ: ≤
θ
≤ or≤
θ
≤
2 6 2
(D) None of these.
Solution : Let : 2 cos2
 2 and
p
3
2
.
 
   
2
2 2 sin sin 2
   
     
2
2 sin sin 0 sin (2sin 1) 0.
Case I.  
  
sin 0, 2sin 1 0
  
    
1 1
sin 0, sin sin
2 2
 

  
5
2 6
.
1
O 
sin 
-1
1
2
1
2
2
5
6
3
2
2
Case II.  
  
sin 0, 2sin 1 0
  
    
1
sin 0, sin sin 0
2

 
  
3
2
.
  
   
 
      
 
 
5 3
A B : or
2 6 2

36
60. Let  
 
  
2 2
R x,y : x y 1
,x,y R be a relation in R. The relation R is :
(A) reflexive (B) symmetric
(C) transitive (D) anti-symmetric
Solution : We have R = {(x, y) : x2 + y2 = 1; x, y  R}.
4Î abd (4)2 + (4)2 = 32  1.  
(4,4) R.
R
is not reflexive.
Let (x, y) Î R. 2 2
x y 1
+ =
    
2 2
y x 1 (y,x) R
R
is symmetric.
(0, 1), (1, 0) Î R because
(0)2 + (1)2 = 1 and (1)2 + (0)2 = 1.
Also (0)2 + (0)2 = 0  1.  
(0,0) R.
R
is not transitive.

Unit 01 Sets, relations and functions Eng.doc

Unit 01 Sets, relations and functions Eng.doc

Unit 01 Sets, relations and functions Eng.doc

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