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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1 | Set 2

• Last Updated : 30 Apr, 2021

### (i) F = {(a, 3), (b, 2), (c, 1)}

Solution:

As, F = {(a, 3), (b, 2), (c, 1)} and S = {a,b,c} and T={1,2,3}

F: S→T is defined as

F(a) = 3, F(b) = 2 and F(c) = 1

F is one-one and onto.

Taking F-1, so F-1: T→S

a = F-1(3), b = F-1(2) and c = F-1(1)

F-1 = {(3,a),(2,b),(1,c)}

### (ii) F = {(a, 2), (b, 1), (c, 1)}

Solution:

As, F = {(a, 2), (b, 1), (c, 1)}

F: S→T is defined as

F(a) = 2, F(b) = 1 and F(c) = 1

Here, F(b) = F(c) but b ≠ c

Hence, F is not one-one.

So, F is not invertible and F-1 doesn’t exists.

### Question 12. Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗b = |a – b| and a o b = a, ∀ a, b ∈ R. Show that ∗ is commutative but not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a ∗ (b o c) = (a ∗ b) o (a ∗ c). [If it is so, we say that the operation ∗ distributes over the operation o]. Does o distribute over ∗? Justify your answer.

Solution:

Binary operations ∗ : R × R → R defined as a ∗b = |a – b|

a*b = |a-b|

b*a = |b-a| = |-(a-b)| = |a-b|

a*b = b*a

Hence, ∗ is commutative.

Now, let’s take a=1, b=2 and c=3 for better understanding

a*(b*c) = a*|b-c| = |a-|b-c|| = |1-|2-3|| = 0

(a*b)*c = |a-b|*c = ||a-b|-c| = ||1-2|-3| = 2

a*(b*c) ≠ (a*b)*c

Hence, ∗ is not associative.

Binary operations o : R × R → R defined as a o b = a, ∀ a, b ∈ R

a o b = a

b o a = b

a o b ≠ b o a

Hence, o is not commutative.

a o (b o c) = a o b = a

(a o b) o c = a o c = a

a o (b o c) ≠ (a o b) o c

Hence, o is associative.

Let’s check for a ∗ (b o c) = (a ∗ b) o (a ∗ c) a, b, c ∈ R

a ∗ (b o c) = a * b = |a-b|

(a ∗ b) o (a ∗ c) = |a-b| o |a-c| = |a-b|

Hence, a ∗ (b o c) = (a ∗ b) o (a ∗ c)

Now, let’s check for a o (b * c) = (a o b) * (a o c)

a o (b * c) = a

(a o b) * (a o c) = a * a = |a-a| = 0

Hence, a o (b * c) ≠ (a o b) * (a o c)

o does not distribute over ∗

### (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

Solution:

Set X, such that P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X)

φ*A = (φ-A) U (A-φ) = φ U A = A

A*φ = (A-φ) U (φ-A) = A U φ = A

Hence, φ is the identity element for the operation * on P(X)

A*A = (A-A) U (A-A) = φ U φ = φ

A = A-1

Hence, all the elements A of P(X) are invertible with A–1 = A.

### Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.

Solution:

Let the set x = {0, 1, 2, 3, 4, 5}

Let’s take i as identity element, where a*i = a = i*a ∀ a ∈ x

a*0 = a

0*a = a, when (a+0<6)

Hence, zero is the identity for this operation

An element a ∈ x is invertible if there exists b ∈ x such that a*b = b*a = 0 From above equations, we have

a = -b or b = 6-a

But, as x = {0, 1, 2, 3, 4, 5} and a,b∈ x. Then a≠-b

Hence, b = 6-a is the inverse of an element a∈ x

a≠0

a-1 = 6-a

### (Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g (a) ∀ a ∈ A, are called equal functions).

Solution:

Given, f, g : A → B be functions defined by f(x) = x2 – x, x ∈ A and g(x) = x ∈ A

At x = -1

f(0) = (-1)2 – (-1) = 2

g(0) = = 2

Here, f(-1) = g(-1) and 2=2

At x = 0

f(0) = 02 – 0 = 0

g(0) = = 0

Here, f(0) = g(0) and 0=0

At x = 1

f(1) = 12 – 1 = 0

g(1) = = 0

Here, f(1) = g(1) and 1=1

At x = 2

f(1) = 22 – 2 = 2

g(1) = = 2

Here, f(2) = g(2) and 2=2

For, every c∈ A, f(c) = g(c)

Hence, f and g are equal functions.

### (D) 4

Solution:

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

Reflexive : (1,1), (2,2), (3,3) ∈ R

Symmetric: (1,2), (2,1)∈ R and (1,3), (3,1) ∈ R

R is not Transitive because, (1,2), (1,3) ∈ R but (3,2) ∉R

So, if we will add (3,2) and (2,3) or both, then R will become transitive.

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

Hence, A is the correct option.

### Question 17. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1

(B) 2

(C) 3

(D) 4

Solution:

Smallest equivalence relations containing (1, 2):

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

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

Hence, B is the correct option.

### and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?

Solution:

Given, f : R → R and g : R → R

when x ∈ (0,1]

[x] = 1, when x=1

[x] = 0, when 0<x<1 Now, fog(x)=f(g(x)) = f([x])  And, Now gof(x) = g(f(x))

g(1) =  = 1

g(0) =  = 0

g(-1) = [-1] = -1

When x ∈ (0,1), fog = 0 and gof = 1. fog(1) ≠ gof(1)

Hence, fog and gof do not coincide in (0, 1].

### (D ) 8

Solution:

Let A = {a,b}

A x A = {a,b} x {a,b}

R = {(a,a),(a,b),(b,a),(b,b)}

Number of elements are 4.

Hence, the number of binary operations on the set will be 24 = 16

Hence, B is the correct option.

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