unit 01 sets, relations and functions eng

SETS AND RELATIONS

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 SETA 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}

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)

(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

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 (or AC) (or A(). Thus is the set of all the elements of the Universal Set which do not belong to the set A.

= U A = {x: x ( U and x ( A}

we can say that A ( = U (Universal Set) and 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 Morgans 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}}

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

SubjectsNo. of students

Mathematics100

Physics70

Chemistry46

Mathematics and Physics30

Mathematics and Chemistry28

Physics and Chemistry23

Mathematics, Physics and Chemistry18

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

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

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.

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 R1 ( B ( A is defined by

R1 = {(b, a): (a, b) ( R, a ( A, b ( B}. It is clear that

a R b ( b R1 a

dom R1 = range R and range R1 = dom R

(R1)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) R1 = {(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 = R1 o S1, 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:

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 R1 = 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) ( R4Note:

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 ~.

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 {} 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 ( is real, show that R is an equivalence relation.

Solution:(i) z1 R z2 ( is real

( R is reflexive

( ( z1 ( C ( 0 is real.

(ii) z1 R z2 ( is real ( is real

( 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

now z1 R z2 ( is real

( is real ( is real

( is real

( is real

( (a1 + a2) (b1 b2) (a1 a2) (b1 + b2) = 0 (for purely real, imaginary part = 0)

( 2a2b1 2b2a1 = 0

( (

similarly, z2 R z3 (

z1 R z2 and z2 R z3 (

and ( ( 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)

( ( = where ( ( I+

( 8x = 14( + 6 ( x =

( x = =

x = ( + (( + 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

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.}

INTEGERSThe 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 NUMBERSAll numbers of the form p/q where p and q are integer and q ( 0, are called rational. Thus 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 NUMBERSReal 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 and

REAL NUMBERSThe complete set of rational and irrational numbers is the set of real numbers and is denoted by R. Thus. 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

COMPLEX NUMBERS

A number of the form x + iy, where x and y are real numbers and , 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. . Illustration 9:If a and b are two rational numbers such that

, find a and b.

Solution:

{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 is also rational

is also rational which is not true

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 INTERVALSOpen Interval:(a, b) = {x|a < x < b}

Close Interval:

Open-close Interval:

Close-open Interval:

INFINITE INTERVALS

4. inequalities

If p, q and r are real numbers, then

and

and equality holds for a = 1

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, ().

Illustration 12:Find the maximum value of where x.

Solution:Given that y = = .

This will be maximum when denominator will be minimum. We know

. Minimum value of is 2.

Hence the maximum value of y is.

5. wavy curve method

In order to solve the inequalities of the form

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

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 step3 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 doesnt 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

Solution:

Step - 1 ( make on real line all x at which numerator becomes zero with dark circles.

Step - 2 ( mark point of discontinuity (value of x at which denominator becomes zero) with empty circles

Step - 3 ( Check ((x) for x > 7, ((8) > 0

Exponents of factors of 1, 3, 4 is even, hence wave will not change the

direction at these points.

Hence

EMBED Equation.DSMT4 Illustration 14:Let f(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 .

Solution:

( 0

( ( 0

( ( 0

( ( 0 ( ( 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 > 1.

III.Solve the inequation .

Solution

(I) Given (x2 + x + 1)x 1

or

Taking Union of (1) and (2)

(II) Case I: When x 3 > 1.(1)

So, in this case solution set

Case II:

So, in this case solution of

So, complete solution set

Case III:

6. absolute value

Let x ( R.

Then the magnitude of x is called its absolute value and in general, denoted by (x( and is defined as .

Since the symbol always denotes the nonnegative square root of a, an alternate definition of (x( is .

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)

(v)

(vi)

. Here the equality sign holds if x and y either both are non-negative or non-positive in other words x.y0.

(vii)

Here the equality sign holds if x and y either both are non-negative or non-positive in other words x.y0.

The last two properties can be put in one compact form namely,

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:

(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

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 all the points whose distance from 3 on real line is greater than 5

Hence

(II)Given

(III)Given

(IV)Given

We Know

Hence

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, a1 and b>0. From the definition of the logarithm of the number b to the base a, we have an identity

This is known as Fundamental Logarithmic Identity.

GRAPH OF LOGRITHMIC FUNCTION

PROPERTIES OF LOGARITHMIC FUNCTION The expression is meaningful for b >0 and for either 0 < a < 1 or a > 1. Let a > 1, thenand

If 0 < a < 1, then

log a(mn) = log a m + log a n

, c > 0 and c ( 1.

log a = log a m log a n

log a mn = n log a m

provided both a and b are non-unity. loga1 = 0

logaa = 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 (x 2) > 4

Solution:In such type of questions first we make the base same.

Given that log (x 2) > 4 log

log (x 2) > log 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

(II)Given

Also x 1 > 0 Here

OBJECTIVE ASSIGNMENT1.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 theseSol: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.


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 1/2 and 0 ( x ( (}

B = {x : sin x > 1/2 and (/3 ( x ( (}

cos x >

( x <

A =

sin x > ( x > , x <

B =

A ( B =


18.The relation R defined in A = {1, 2, 3} by aRb if Which of the following is false?

(A) R = {(1, 1) (2, 2), (3, 3), (2, 1), (1, 2), (2, 3), (3, 2)}

(B) R1 = R

(C) Domain of R = (1, 2, 3)

(D) Range of R = {5}.Solution :

Let a = 1

Let

Let

= R.

Domain of R = {x : (x, y) R} = {1, 2, 3}.

Range of R = {y : (x, y) R} = {1, 2, 3}.

Hence (D) is the correct answer.

19.The relation R defined on the set A = {1, 2, 3, 4, 5} by R = {(x, y) : 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) :

{(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 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

Let

(1, 0), (1, 1), (1, 1), (1, 1), (1, 1), (2, 0), (2, 0}.


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 Which of the following is correct?

(A) (2 + 3i) R 13(B) 3 R (3)

(C) (1 + i) R 2(D) iR1.Solution :

(2 + 3i) R 13 is wrong.

3R (3) is wrong

(1 + i) R2 is wrong

iR1 is correct.


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) : 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) R} = {2, 3, 4, 5, 6}

Range of R = {y : (x, y) 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 = ex 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

(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.

Hence (A) is the correct answer.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 theseSol:(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

(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 N}

x + 2y = 8

...................................................................

...................................................................


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)S1oR1(B)R1oS1 (C)SoR(D)None of theseSol:We have (SoR)1 = R1oS1.Hence (B) is the correct answer.

37:

If an = {ax: x ( N}, then 3N 7N =

(A) 3N(B) 7N

(C) N(D) 21NSolution:

(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


38:Let A = {x: x ( R, (x( < 1}, B = {x: x ( R, (x1( ( 1} and AB = R D, then set D is:

(A)

(B)

(C)

(D) None of these

Solution:

We have A = {x: x ( R, 1 < x < 1}, B = {x: x ( R, x1 1 or x1 ( 1}

= {x: x ( R, x ( 0 or x ( 2}

( AB = R D

Where D = {x: x ( R, 1 ( X < 2}


39:

x may have values:

(A) 2, 3(B) 7

(C) 2, 3(D) 2, 3

Solution:

We have


40:

Number of positive integral values of x satisfies the equation

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

. 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.

.(1)

Also (2)

Taking intersection of (1) and (2)


42:

The set of real values of x for which is:

(A)

(B)

(C)

(D) none of these

Solution:

Given

( 5x (x + 2) > x2 ( 4x2 + 10x ( 0

( x ( 0 or x , 5/2

Also

( 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 is:

(A)( 2,0)(B) (0, ()

(C) ( 5,5)(D) ( 2,2)

Solution:

We have


45:If , then x belongs to:

(A) (2, 3)(B) (0,1)

(C) ( 1, 0)(D) (3,4)

Solution:

Cube both the sides, we get

. Cleary x lies between 0 and 1.


46:If (log5 x) 2 + log5 x < 2, then x belong to:

(A)

(B)

(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)

(B)

(C)

(D)

Solution :

Given set is singleton.


49.

The set 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

50.

If A = and B = , then which of the following is false ?

(A)

(B)

(C)

(D)

Solution :

A = {31, 32, 33, 34, 35, 36}

= {3, 9, 27, 81, 243, 729}

and B = {91, 92, 93, 94} = {9, 81, 729, 6561}

(a)

(b)

(c)

(d)

= {3, 27, 243}


51.If n(A) = 115, n(B) = 326, n(AB)=47, then n is equal to

(A) 373(B) 165

(C) 370(D) None of these.

Solution :

(a) n(A) = n(AB) + n implies

115 = 47 + n.

52.If A = is equal to

(A) {1. 1}(B) {1. 1. o. i)

(C) {i, i}(D) None of these.

Solution :

Hence (C) is the correct answer.53.If for , then the set is equal to

(A) 8 N(B) 48 N

(C) 12 N(D) 25 N.

Solution :

= {24, 48, ....} = 24N.

Alternatively, L.C.M. of 6 and 8 = 24.


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.


55.If A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, then (A B) (BC) 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)}

`

B C = {3, 4} {4, 5, 6} = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

So (A B) (BC)= {(3, 4)}


56.If A = and B = {(x, y) : y = x, x R}, then

(A)

(B)

(C)


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 .


57.If A = , then

(A)

(B)

(C)

(D)

.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 .


58.If and b, c are coprime, then

(A) b = cd(B) c = bd

(C) d = bc(D) None of these.

Solution :

We have , and

We have

and d = cn2 where n1, n2 N

b/d and c/d bc/d, because b and c are coprime.

Also bcbN and bc = cb cN


59.If A = { : 2 cos2 + sin 2} and B = , then is equal to

(A)

(B)

(C)


Let : 2 cos2 + sin 2 and .

Case I.

.

Case II.

.


60.Let be a relation in R. The relation R is :

(A) reflexive(B) symmetric

(C) transitive(D) anti-symmetricSolution :

We have R = {(x, y) : x2 + y2 = 1; x, y R}.

4 abd (4)2 + (4)2 = 32 1.

is not reflexive.

Let (x, y) R.

is symmetric.

(0, 1), (1, 0) R because

(0)2 + (1)2 = 1 and (1)2 + (0)2 = 1.

Also (0)2 + (0)2 = 0 1.

is not transitive.

Hence (B) is the correct answer. EMBED Word.Picture.8

EMBED Word.Picture.8


-3

-2

-1

0

1

2

3

2

Real line

EMBED Visio.Drawing.6












PAGE 1

_1202982510.unknown

_1306605429.unknown

_1306606374.unknown

_1306606638.unknown

_1306606929.unknown

_1306607271.unknown

_1306607333.unknown

_1306607818.unknown

_1306784870.unknown

_1306607346.unknown

_1306607305.unknown

_1306607320.unknown

_1306607292.unknown

_1306607128.unknown

_1306607199.unknown

_1306607232.unknown

_1306607251.unknown

_1306607214.unknown

_1306607155.unknown

_1306607173.unknown

_1306607143.unknown

_1306606991.unknown

_1306607087.unknown

_1306607113.unknown

_1306607023.unknown

_1306607009.unknown

_1306606964.unknown

_1306606976.unknown

_1306606944.unknown

_1306606756.unknown

_1306606831.unknown

_1306606875.unknown

_1306606886.unknown

_1306606843.unknown

_1306606781.unknown

_1306606795.unknown

_1306606768.unknown

_1306606698.unknown

_1306606728.unknown

_1306606743.unknown

_1306606709.unknown

_1306606673.unknown

_1306606685.unknown

_1306606656.unknown

_1306606500.unknown

_1306606579.unknown

_1306606610.unknown

_1306606622.unknown

_1306606593.unknown

_1306606538.unknown

_1306606564.unknown

_1306606519.unknown

_1306606443.unknown

_1306606472.unknown

_1306606485.unknown

_1306606459.unknown

_1306606403.unknown

_1306606416.unknown

_1306606390.unknown

_1306606097.unknown

_1306606222.unknown

_1306606304.unknown

_1306606330.unknown

_1306606353.unknown

_1306606318.unknown

_1306606271.unknown

_1306606289.unknown

_1306606240.unknown

_1306606160.unknown

_1306606187.unknown

_1306606207.unknown

_1306606174.unknown

_1306606125.unknown

_1306606146.unknown

_1306606110.unknown

_1306605758.unknown

_1306605949.unknown

_1306606037.unknown

_1306606063.unknown

_1306606082.unknown

_1306605965.unknown

_1306605998.unknown

_1306605812.unknown

_1306605847.unknown

_1306605776.unknown

_1306605554.unknown

_1306605590.unknown

_1306605738.unknown

_1306605572.unknown

_1306605515.unknown

_1306605534.unknown

_1306605497.unknown

_1306605076.unknown

_1306605280.unknown

_1306605362.unknown

_1306605394.unknown

_1306605409.unknown

_1306605378.unknown

_1306605317.unknown

_1306605335.unknown

_1306605296.unknown

_1306605184.unknown

_1306605223.unknown

_1306605263.unknown

_1306605203.unknown

_1306605137.unknown

_1306605162.unknown

_1306605119.unknown

_1204173591.unknown

_1204176478.unknown

_1204181230.unknown

_1204182023.unknown

_1204182522.unknown

_1306605031.unknown

_1204182936.unknown

_1204182106.unknown

_1204181317.unknown

_1204180449.unknown

_1204181099.unknown

_1204179348.vsd

_1204180182.unknown

_1204178885.vsd

_1204178241.vsd

_1204175678.unknown

_1204175737.unknown

_1204175975.unknown

_1204176322.unknown

_1204175860.unknown

_1204175554.unknown

_1204175604.unknown

_1204173791.unknown

_1204174096.unknown

_1204173668.unknown

_1202993602.vsd

_1204094638.unknown

_1204096596.unknown

_1204127355.unknown

_1204171982.unknown

_1204173489.unknown

_1204127691.unknown

_1204098638.unknown

_1204094758.unknown

_1204095191.unknown

_1204094020.unknown

_1204094218.unknown

_1204093676.unknown

_1202983518.unknown

_1202992522.unknown

_1202992532.unknown

_1202992538.unknown

_1202983534.unknown

_1202982551.unknown

_1202982571.unknown

_1202982550.unknown

_1177245854.unknown

_1177254713.unknown

_1177260528.unknown

_1202912827.unknown

_1202917053.unknown

_1202919409.unknown

_1202919454.unknown

_1202920483.unknown

_1202920566.unknown

_1202920575.unknown

_1202920531.unknown

_1202919503.unknown

_1202919441.unknown

_1202919345.unknown

_1202919386.unknown

_1202919200.unknown

_1202912854.unknown

_1202915618.unknown

_1202912843.unknown

_1202912026.unknown

_1202912058.unknown

_1202912442.unknown

_1202912036.unknown

_1190212617.unknown

_1202911643.unknown

_1202911710.unknown

_1190212639.unknown

_1202634357.unknown

_1190212402.vsd

_1190212430.vsd

_1190212561.vsd

_1190212582.vsd

_1190212437.vsd

_1190212421.vsd

_1190212355.unknown

_1177255285.unknown

_1177255891.unknown

_1177256293.unknown

_1177256363.unknown

_1177256636.unknown

_1177256319.unknown

_1177256130.unknown

_1177255398.unknown

_1177255551.unknown

_1177255336.unknown

_1177255041.unknown

_1177255151.unknown

_1177255198.unknown

_1177255139.unknown

_1177254916.unknown

_1177254994.unknown

_1177254796.unknown

_1177254040.unknown

_1177254414.unknown

_1177254698.unknown

_1177254699.unknown

_1177254463.unknown

_1177254202.unknown

_1177254345.unknown

_1177254121.unknown

_1177253793.unknown

_1177253913.unknown

_1177254033.unknown

_1177253807.unknown

_1177247766.unknown

_1177251246.unknown

_1177252275.unknown

_1177252318.unknown

_1177252158.unknown

_1177251513.doc

x < 4

x > 4

1x 5

- 4

-1

4

5

_1177251177.unknown

_1177251194.unknown

_1177248590.unknown

_1177246807.unknown

_1177247418.doc

1

2

3

y = 2 - x

y = x - 2

y = 1

_1177247731.doc

logax (a > 1)

y

x

O

logax (0

unit 01 sets, relations and functions eng

Documents

difference of set

null set

infinite set

void set

singleton set

super set of

proper subset of set

element of b