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

Chapter 1 of NCERT Class 12 Mathematics Part I focuses on Relations and Functions, building upon the foundational concepts introduced in earlier classes. This chapter delves deeper into various types of relations and functions, exploring their properties, compositions, and applications. Students will learn to analyze and manipulate functions, understand inverse functions, and solve problems involving different function types.

The content of this article has been merged with Chapter 1 Relations And Functions - Miscellaneous Exercise as per the revised syllabus of NCERT.

Question 11: Let S = {a, b, c} and T = {1, 2, 3}. Find F^–1 of the following functions F from S to T, if it exists.

(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 ∗

Question 13. Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A^–1 = A. 

(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. 

Question 14. Define a binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as 

a*b= \begin{cases} a+b, \hspace{0.2cm}a+b<6\\ a+b-6,\hspace{0.2cm}a+b\geq6 \end{cases}

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

i.e.,\begin{cases} a+b=0=b+a, \hspace{0.2cm}a+b<6\\ a+b-6=0=b+a-6,\hspace{0.2cm}a+b\geq6 \end{cases}

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

Question 15. Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined by f(x) = x² – x, x ∈ A and g(x) = 2|x-\frac{1}{2}|-1     x ∈ A. Are f and g equal? Justify your answer. 

(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) = 2|x-\frac{1}{2}|-1    x ∈ A

At x = -1

f(0) = (-1)² - (-1) = 2

g(0) = 2|-1-\frac{1}{2}|-1   = 2

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

At x = 0

f(0) = 0² - 0 = 0

g(0) = 2|0-\frac{1}{2}|-1   = 0

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

At x = 1

f(1) = 1² - 1 = 0

g(1) = 2|1-\frac{1}{2}|-1   = 0

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

At x = 2

f(1) = 2² - 2 = 2

g(1) = 2|2-\frac{1}{2}|-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.

Question 16. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

(A) 1 

(B) 2 

(C) 3 

(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.

Question 18. Let f : R → R be the Signum Function defined as

f(x)= \begin{cases} 1, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}x=0\\ -1\hspace{0.2cm}x>0 \end{cases}

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

g(x)= \begin{cases} 1, \hspace{0.2cm}x=1\\ 0,\hspace{0.2cm}0<x<1 \end{cases}

Now, fog(x)=f(g(x)) = f([x])

f([x])= \begin{cases} f(1), \hspace{0.2cm}x=1\\ f(0),\hspace{0.2cm}0<x<1 \end{cases} =   \begin{cases} 1, \hspace{0.2cm}x=1\\ 1,\hspace{0.2cm}0<x<1 \end{cases}

And, Now gof(x) = g(f(x))

g(1) = [1] = 1

g(0) = [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].

Question 19. Number of binary operations on the set {a, b} are

(A) 10 

(B) 16 

(C) 20 

(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.

Summary

This chapter equips students with advanced tools for analyzing relations and functions, crucial for higher mathematics and real-world applications. It covers topics such as function composition, invertibility, domain and range analysis, and special types of functions. The exercises reinforce these concepts, preparing students for more complex mathematical reasoning in future chapters.