Relations and Functions Test

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Relations and Functions Test - 8

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Weekly Quiz Competition

Question 1
1 / -0.25

Let R = {(P, Q) : OP = OQ , O being the origin} be an equivalence relation on A. The equivalence class [(1, 2)] is

A
{(x, y) : x² + y² = 5}

B
{(x, y) : x² = y² }

C
{(x, y) : x² + y² = 1}

D
{(x, y) : x² + y² = 4}

Solution

Correct Answer : a
Explanation : A = {(x,y) : x² + y² = 5}
=>(1,2) {(1)² + (4)² = 5}
=>{1 + 4 = 5}
=>{5 = 5}
Question 2
1 / -0.25

Let a relation T on the set R of real numbers be T = {(a, b) : 1 + ab <0, a, ∈R}. Then from among the ordered pairs (1, 1), (1, 2), (1, -2), (2, 2), the only pair that belongs to T is________.

A
(2, 2)

B
(1, 1)

C
(1, -2)

D
(1, 2)

Solution

Since T is a set of real number for it 's given ordered pairs we have a condition provided that is 1 + ab is less than zero. so in the given option c if we put order of a and b then it satisfy our given condition
Question 3
1 / -0.25

For real number x and y, we write _x R_y ⇔x-y + √2 is an irrational number. Then the relation R is:

A
Reflexive

B
Symmetric

C
Transitive

D
Equivalence

Solution

_x R_y =>x - y + √2 is an irrational number.
Let R is a binary relation on real numbers x and y.
Clearly, R is reflexive relation
As _x R_x iff x –x +√2 = √2 ,which is an irrational number.
Here R is not symmteric if we take x =√2 and y =1 then x –y + √2 is an irrational
number but y –x + √2 = 1, which is not irrational number
Now, R is transitive iff for all (x, y) ∈R and (y, z) ∈R implies (x, z) ∈R
But here R is not transitive as we take x = 1, y = 2 √2, z=√2
Given, _x R_y =>x - y + √2 is irrational ............1
and _y R_z =>y - z + √2 is irrational ............2
Add equation 1 and 2, we get
(x - y + √2) + (y - z + √2)
= x - z + √2 = 1, which is not an irrational
Question 4
1 / -0.25

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

A
Not symmetric

B
Transitive

C
Reflexive

D
A function

Solution

Let R={(1,3),(4,2),(2,4),(2,3),(3,1)} be a relation on the set A={1,2,3,4}, then
(a) Since (2,3) ∈R but (3,2) ∈/′R, so R is not symmetric.
(b) Since (1,3) ∈ R and (3,1) ∈/ R but (1,1) ∈/ R, so R is not transitive:
(c) Since (1,1) ∈/ R, so R is not reflexive.
(d) Since (2,4) ∈R and (2,3) ∈R, so R is not a function.
Question 5
1 / -0.25

If A = {1, 3, 5, 7} and we define a relation R = {(a, b), a, b ∈A: |a - b| = 8}. Then the number of elements in the relation R is

A
2

B
1

C
3

D
0

Solution

Clearly there is no pair in set A whose difference is 8 or -8.
so D is the correct option.
Question 6
1 / -0.25

If A = {1, 3, 5, 7} and define a relation, such that R = {(a, b) a, b ∈A : |a + b| = 8}. Then how many elements are there in the relation R

A
8

B
16

C
1

D
4

Solution

Number of relations would be 4 as.. 1 + 7,7 + 1, 3 + 5, 5 + 3 all are equal to 8
Question 7
1 / -0.25

In the set N x N, the relation R is defined by (a, b) R (c, d) ⇔ a + d = b + c. Then R is

A
Symmetric and transitive but not reflexive

B
Reflexive and transitive but not symmetric

C
Equivalence relation

D
Partial order relation

Solution

(a, b) R (c, d) <=>a + d = b + c
Reflexive:
(a, b) R (a, b) <=>a + b = b + a
This is true for all (a, b) €N x N
Hence, it is reflexive.
Symmetric:
Let (a, b) R (c, d) <=>a + d = b + c
a + d = b + c
c + b = d + a (same)
By the above equation,
(c, d) R (a, b)
Hence,
(c, d) R (a, b) <=>c + b = d + a
Hence, it is symmetric
Transitive:
Let (a, b) R (c, d) <=>a + d = b + c, --- eqn 1
(c, d) R (e, f) <=>c + f = d + e --- eqn 2
for all a, b, c, d, e, f €N
eqn 1 : a + d = b + c
⇒ a - b = c - d
eqn 2 : c + f = d + e
⇒c - d = e - f
So, a-b = e-f
⇒a + f = b + e
⇒(a, b) R (e, f)
Hence, it is transitive.
This is an equivalence relation
Question 8
1 / -0.25

If A = {1, 2, 3, 4} and B = {1, 3, 5} and R is a relation from A to B defined by(a, b) ∈ element of R ⇔ a < b. Then, R = ?

A
{(2, 3), (4, 5), (1, 3), (2, 5)}

B
{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}

C
{(2, 3), (4, 5), (1, 3), (2, 5), (5, 3)}

D
{(5, 3), (3, 5), (5, 4), (4, 5)}

Solution

A = {1, 2, 3, 4}
B = {1, 3, 5}
(a, b) ∈element of R ⇔a < b for all a ∈ A, b ∈ B
(a, b) pairs satisfying the condition of R are:
(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)
So,
R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
Question 9
1 / -0.25

Let R be the relation on the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3,3), (3,2)}. then R is

A
R is reflexive and symmetric but not transitive.

B
R is symmetric and transitive but not reflexive.

C
R is an equivalence relation.

D
R is reflexive and transitive but not symmetric.

Solution

R be the relation in the set {1, 2,3, 4] given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
it is seen that (a, a) ∈R for every a ∈{1, 2, 3, 4}
so,R is reflexive.
it is seen that (a, b) = (b, a) ∈R
because, (1, 2)∈R but (2, 1) ∉R
so, R is not symmetric.
it is seen that (a, b), (b, c) ∈R ⇒(a, c) ∈R for all a, b, c ∈{1, 2, 3, 4}.
so, R is transitive.
Hence, R is reflexive and transitive but not symmetric.
Question 10
1 / -0.25

Let A = {1, 2, 3, 4, 5, 6, 7}. P = {1, 2}, Q = {3, 7}. Write the elements of the set R so that P, Q and R form a partition that results in equivalence relation.

A
{4, 5, 6}

B
{0}

C
{1, 2, 3, 4, 5, 6, 7}

D
{ }

Solution

P{1, 2} union Q {3, 7} union R{} = A {1, 2, 3, 4, 5, 6, 7}
therefore R = {4, 5, 6}

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Relations and Functions Test - 8

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Question 1

{(x, y) : x2 + y2 = 5}

{(x, y) : x2 = y2 }

{(x, y) : x2 + y2 = 1}

{(x, y) : x2 + y2 = 4}

Solution

Question 2

(2, 2)

(1, 1)

(1, -2)

(1, 2)

Solution

Question 3

Reflexive

Symmetric

Transitive

Equivalence

Solution

Question 4

Not symmetric

Transitive

Reflexive

A function

Solution

Question 5

2

1

3

0

Solution

Question 6

8

16

1

4

Solution

Question 7

Symmetric and transitive but not reflexive

Reflexive and transitive but not symmetric

Equivalence relation

Partial order relation

Solution

Question 8

{(2, 3), (4, 5), (1, 3), (2, 5)}

{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}

{(2, 3), (4, 5), (1, 3), (2, 5), (5, 3)}

{(5, 3), (3, 5), (5, 4), (4, 5)}

Solution

Question 9

R is reflexive and symmetric but not transitive.

R is symmetric and transitive but not reflexive.

R is an equivalence relation.

R is reflexive and transitive but not symmetric.

Solution

Question 10

{4, 5, 6}

{0}

{1, 2, 3, 4, 5, 6, 7}

{ }

Solution

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{(x, y) : x² + y² = 5}

{(x, y) : x² = y² }

{(x, y) : x² + y² = 1}

{(x, y) : x² + y² = 4}