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QUESTION: 1

R = {(1, 1), (2, 2), (1, 2), (2, 1), (2, 3)} be a relation on A = {1, 2, 3}, then R is

Solution:

A relation R on a non empty set A is said to be reflexive if fx Rx for all x ∈ R, Therefore , R is not reflexive.

A relation R on a non empty set A is said to be symmetric if fx Ry ⇔ yRx, for all x, y ∈ R Therefore, R is not symmetric.

A relation R on a non empty set A is said to be antisymmetric if fx Ry and y Rx ⇒ x = y, for all x, y ∈ R. Therefore, R is not antisymmetric.

QUESTION: 2

If R is a relation from a set A to a set B and S is a relation from B to C, then the relation S ^{∘} R.

Solution:

If R is a relation from A→B and S is a relation fromB→C, then S ^{∘ }R is a composite function from A to C.

QUESTION: 3

Let A be the set of all students of a boys school. The relation R in A given by R = {(a, b) : a is sister of b} is

Solution:

Because, the relation is defined over the set A which is the set of all students of a boys school.

QUESTION: 4

The domain of definition of the function

Solution:

y is defined If -x ≥ 0, i.e.

QUESTION: 5

Solution:

By definition, The Signum function =

QUESTION: 6

Let A = {1, 2, 3}. Which of the following is not an equivalence relation on A ?

Solution:

A relation R on a non empty set A is said to be reflexive iff xRx for all x ∈ R . A relation R on a non empty set A is said to be symmetric iff xRy ⇔ yRx, for all x , y ∈ R .

A relation R on a non empty set A is said to be transitive iff xRy and yRz ⇒ xRz, for all x ∈ R. An equivalence relation satisfies all these three properties.

None of the given relations satisfies all three properties of equivalence relation.

QUESTION: 7

The binary operation * defined on the set of integers as a∗b = |a−b|-1 is:

Solution:

Here * is commutative as b*a = |b−a|−1 = |a−b|−1 = a∗b.

Because ,|−x| = |x| for all x ∈ R.

QUESTION: 8

Let T be the set of all triangles in a plane with R a relation in T given by R = {(T_{1}, T_{2}) : T_{1} is congruent to T_{2}}. Then R is

Solution:

Let T be the set of all triangles in a plane with R a relation in T given by R = {(T_{1}, T_{2}): T_{1} is congruent to T_{2}}. (T_{1}T_{2}) ∈ R iff T_{1 }is congruent to T_{2}.

Reflexivity :T_{1}≅T_{1} ⇒ (T_{1}T_{1}) ∈ R . Symmetry :(T_{1}, T_{2})∈ R ⇒T_{1}≅T_{2} ⇒ T_{2}≅T_{1} ⇒ (T_{2}, T_{1}) ∈ R

Transitivity :(T_{1},T_{1}) ∈ R and (T_{2}, T_{3}) ∈ R ⇒ T_{1}≅T_{2} and T_{2}≅T_{3} ⇒ T_{1}≅T_{3} ⇒ (T_{1}, T_{3}) ∈ R .

Therefore, R is an equivalence relation on T.

QUESTION: 9

The range of the function

Solution:

The given function is defined as Signum function. i.e.

domain is R and Range is {-1 , 0 , 1}.

QUESTION: 10

The function f (x) = x^{2}+ sin x is

Solution:

For even function : f(-x) = f(x) , therefore, f(- x) = (− x)^{2}+ sin (− x)

= x^{2}−sin x ≠ f(x).

For an odd function : f(-x) = - f(x) , therefore, f(-x) = (− x)^{2}+ sin (− x)

= x^{2}− sin x

≠ − f(x).

Therefore f(x) is neither even nor odd.

QUESTION: 11

If A is a finite set containing n distinct elements, then the number of relations on A is equal to

Solution:

QUESTION: 12

R is a relation from { 11, 12, 13} to {8, 10, 12} defined by y = x – 3. The relation R^{−1}

Solution:

R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x – 3. The relation is given by x = y + 3,from{8, 10, 12} to {11, 12, 13} ⇒ relation = {(8,11),(10,13)}.

QUESTION: 13

If A = {(1, 2, 3}, then the relation R = {(2, 3)} in A is

Solution:

A = {1,2,3}

B = {(2,3)} is not reflexive or symmetric on A but it is transitive

∵ if (a,b) exists but (b,c) does not exist then (a,c) does not need to exist and the relation is still transitive.

QUESTION: 14

The range of the function f(x) = [sin x] is

Solution:

The only possible integral values of sin x are {-1 ,0, 1}.

QUESTION: 15

The period of the function f(x) = sin^{2}x + tan x is

Solution:

We have the function f(x) = sin^{2}x + tan x, therefore, f(π+x)

= {sin^{2}(π+x)+tan(π+x)}

= {sin^{2}x+tanx} = f(x).

This implies that period of the function f(x) is π.

QUESTION: 16

Let A = {1, 2, 3}, then the domain of the relation R = {(1, 1), (2, 3), (2, 1)} defined on A is

Solution:

Since the domain is represented by the x- co ordinate of the ordered pair (x , y).Therefore, domain of the given relation is {1, 2}.

QUESTION: 17

Given the relation R = {(1, 2), (2, 3)} on eht set {1, 2, 3}, the minimum number of ordered pairs which when added to R make it an equivalence

Solution:

To make the relation an equivalence relation , the following ordered pairs are required (1,1),(2,2),(3,3)(2,1)(3,2)(1,3),(3,1)

QUESTION: 18

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relations on the set A = {1, 2, 3, 4}. the relation R is

Solution:

Given, R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}.

(a) Since, (1, 1), (2, 2), (3, 3), (4, 4) ∉ R. So, R is not reflexive.

(b) Since, (1, 3) ∈ R and (3, 1) ∈ R but (1, 1) ∉ R. So, R is not transitive.

(c) Since, (2, 3) ∈ R but (3,2) ∉ R. So, B is not symmetric.

(d) Since, (2, 4) ∈ R and (2, 3) ∈ R. So, R is not a function.

QUESTION: 19

The relation R in the set Z of integers given by R = {(a, b): 2 divides a – b}(a, b) where a, b ∈ Z is

Solution:

QUESTION: 20

The range of the function f(x) = x – [x] is

Solution:

Since [x] ≤ x, therefore , x – [x] ≥ 0. Also, x – [x] <1,∴0⩽x−[x]<1. Therefore ,Rf = [0,1).

QUESTION: 21

Let A contain n distinct numbers. How many bijections from A to A can be defined?

Solution:

A bijection from A to A is infact an arrangement of its n elements, taken all at a time, which can be done in ways. Hence, the number of bijections from A to A is

QUESTION: 22

Number of relations that can be defined on the set A = {a, b, c, d} is

Solution:

No. of elements in the set A = 4. Therefore , the no. of elements in A × A = 4 × 4 = 16. As, the no. of relations in A × A = no. of subsets of A × A = 2^{16}

QUESTION: 23

A relation R in a set A is called universal relation, if

Solution:

The relation R = A x A is called Universal relation.

QUESTION: 24

Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Then R is

Solution:

Consider any a ,b , c ∈A .

1) Since both a and a must be either even or odd, so (a , a) ∈R ⇒ R is reflexive.

2) Let (a ,b) ∈R ⇒ both a and b must be either even or odd, ⇒ both b and a must be either even or odd, ⇒ (b ,a) ∈R. Thus , (a ,b) ∈R ⇒ (b ,a) ∈R ⇒ R is symmetric.

3) Let (a ,b) ∈R and (b ,c) ∈R ⇒ both a and b must be either even or odd, also ,both b and c must be either even or odd, ⇒ all elements a, b and c must be either even or odd, ⇒ (a ,c) ∈R . Thus, (a ,b) ∈R ⇒ (b,c) ∈R ⇒ (a ,c) ∈R ⇒ R is transitive.

QUESTION: 25

If f(x) = x – x^{2}, then f(a + 1) – f(a – 1) , a ∈ R is :

Solution:

f(a+1)−f(a−1)

=(a+1)−(a+1)2−{(a−1)−(a−1)2} = 2−4a(a−1)2} = 2−4a

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