# NCERT Solutions for Miscellaneous Exercise Chapter 1 Class 12 - Relations and Functions

NCERT solutions for class 12 maths chapter 1 miscellaneous exercise discusses concepts related to inverse, injective, onto functions. Also in Class 12 maths chapter 1 miscellaneous exercise some of the topics of binary operations signum function, inverse functions. All in all it can be said that in miscellaneous exercises, diverse questions are asked unlike in earlier exercises of this chapter. Hence class 12 maths chapter 1 miscellaneous exercise is a good source to cover the syllabus in a comprehensive manner. There are a total of 5 exercises in the NCERT solutions for class 12 maths chapter 1 including miscellaneous exercise which is present in NCERT Class 12th book. Also with NCERT questions, students should practice NCERT exemplar and class 12 CBSE previous year questions for better understanding.

**Also Read**

NCERT solutions for class 12 maths chapter 1 exercise 1.1

NCERT solutions for class 12 maths chapter 1 exercise 1.2

NCERT solutions for class 12 maths chapter 1 exercise 1.3

NCERT solutions for class 12 maths chapter 1 exercise 1.4

**NCERT solutions for Class 12 Maths Chapter 1 Relations and Functions: Miscellaneous Exercise**

** ****Question:1 ** Let be defined as . Find the function such that .

** Answer: **

and

For one-one:

Thus, f is one-one.

For onto:

For ,

Thus, for , there exists such that

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Let as

and

Hence, defined as

** Question:2 ** Let be defined as , if n is odd and , if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

** Answer: **

if n is odd

if n is even.

For one-one:

Taking x as odd number and y as even number.

Now, Taking y as odd number and x as even number.

This is also impossible.

** If both x and y are odd ** :

** If both x and y are even : **

f is one-one.

Onto:

Any odd number 2r+1 in codomain of N is an image of 2r in domain N and any even number 2r in codomain N is the image of 2r+1 in domain N.

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Sice, f is invertible.

Let as if m is even and if m is odd.

When x is odd.

When x is even

Similarly, m is odd

m is even ,

and

Hence, f is invertible and the inverse of f is g i.e. , which is the same as f.

Hence, inverse of f is f itself.

** Question:3 ** If f : R → R is defined by f(x) = x ^{ 2 } – 3x + 2, find f (f (x)).

** Answer: **

This can be solved as following

f : R → R

** Question:4 ** Show that the function defined by is one one and onto function.

** Answer: **

The function defined by

,

One- one:

Let ,

It is observed that if x is positive and y is negative.

Since x is positive and y is negative.

but 2xy is negative.

Thus, the case of x is positive and y is negative is removed.

Same happens in the case of y is positive and x is negative so this case is also removed.

** When x and y both are positive: **

** When x and y both are negative ** :

f is one-one.

Onto:

Let such that

If y is negative, then

If y is positive, then

Thus, f is onto.

Hence, f is one-one and onto.

** Question:5 ** Show that the function given by is injective.

** Answer: **

One-one:

Let

We need to prove .So,

Let then there cubes will not be equal i.e. .

It will contradict given condition of cubes being equal.

Hence, and it is one -one which means it is injective

** Question:6 ** Give examples of two functions and such that is injective but g is not injective. (Hint : Consider and ).

** Answer: **

One - one:

Since

As we can see but so is not one-one.

Thus , g(x) is not injective.

Let

Since, so x and y are both positive.

Hence, gof is injective. ** **

** Question:7 ** Give examples of two functions and such that is onto but is not onto.

(Hint : Consider and

** Answer: **

and

and

Onto :

Consider element in codomain N . It is clear that this element is not an image of any of element in domain N .

f is not onto.

Now, it is clear that , there exists such that .

Hence, is onto.

** Question:8 ** Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if . Is R an equivalence relation on P(X)? Justify your answer.

** Answer: **

Given a non empty set X, consider P(X) which is the set of all subsets of X.

Since, every set is subset of itself , ARA for all

R is reflexive.

Let

This is not same as

If and

then we cannot say that B is related to A.

R is not symmetric.

If

this implies

R is transitive.

Thus, R is not an equivalence relation because it is not symmetric.

** Question:9 ** Given a non-empty set X, consider the binary operation given by , where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

** Answer: **

Given is defined as .

As we know that

Hence, X is the identity element of binary operation *.

Now, an element is invertible if there exists a ,

such that (X is identity element)

i.e.

This is possible only if .

Hence, X is only invertible element in with respect to operation *

** Question:10 ** Find the number of all onto functions from the set to itself.

** Answer: **

The number of all onto functions from the set to itself is permutations on n symbols 1,2,3,4,5...............n.

Hence, permutations on n symbols 1,2,3,4,5...............n = n

Thus, total number of all onto maps from the set to itself is same as permutations on n symbols 1,2,3,4,5...............n which is n. ** **

** Question:11(i) ** Let and . Find of the following functions F from S to T, if it exists.

(i)

** Answer: **

is defined as

is given by

** **

** Question:11(ii) ** Let and . Find of the following functions F from S to T, if it exists.

(ii)

** Answer: **

is defined as

, F is not one-one.

So inverse of F does not exists.

Hence, F is not invertible i.e. does not exists. ** **

** Question:12 ** Consider the binary operations and defined as and . Show that ∗ is commutative but not associative, is associative but not commutative. Further, show that , . [If it is so, we say that the operation ∗ distributes over the operation ]. Does distribute over ∗? Justify your answer.

** Answer: **

Given and is defined as

and

For , we have

the operation is commutative.

where

the operation is not associative

Let . Then we have :

Hence,

Now,

for

Hence, operation o does not distribute over operation *. ** **

** 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).

** Answer: **

It is given that

be defined as

Now, let .

Then,

And

Therefore,

Therefore, we can say that is the identity element for the given operation *.

Now, an element A P(X) will be invertible if there exists B P(X) such that

Now, We can see that

such that

Therefore, by this we can say that all the element A of P(X) are invertible with

** Question:14 ** Define a binary operation ∗ on the set as Show that zero is the identity for this operation and each element of the set is invertible with being the inverse of .

** Answer: **

X = as

An element is identity element for operation *, if

For ,

Hence, 0 is identity element of operation *.

An element is invertible if there exists ,

such that i.e.

means or

But since we have X = and . Then .

is inverse of a for .

Hence, inverse of element , is 6-a i.e. ,

** Question:15 ** Let , and be functions defined by and . Are and equal? Justify your answer. (Hint: One may note that two functions and such that , are called equal functions).

** Answer: **

Given :

,

are defined by and .

It can be observed that

Hence, f and g are equal functions.

** Question:16 ** Let . Then number of relations containing and which are reflexive and symmetric but not transitive is

(A) 1

(B) 2

(C) 3

(D) 4

** Answer: **

The smallest relations containing and which are

reflexive and symmetric but not transitive is given by

, so relation R is reflexive.

and , so relation R is symmetric.

but , so realation R is not transitive.

Now, if we add any two pairs and to relation R, then relation R will become transitive.

Hence, the total number of the desired relation is one.

Thus, option A is correct.

** Question:17 ** Let . Then number of equivalence relations containing is

(A) 1

(B) 2

(C) 3

(D) 4

** Answer: **

The number of equivalence relations containing is given by

We are left with four pairs , , .

, so relation R is reflexive.

and , so relation R is not symmetric.

, so realation R is not transitive.

Hence , equivalence relation is bigger than R is universal relation.

Thus the total number of equivalence relations cotaining is two.

Thus, option B is correct.

** Question:18 ** Let be the Signum Function defined as and be the Greatest Integer Function given by , where is greatest integer less than or equal to . Then, does and coincide in ?

** Answer: **

is defined as

is defined as

Let

Then we have , if x=1 and

Hence,for , and .

Hence , gof and fog do not coincide with . ** **

** Question:19 ** Number of binary operations on the set are(A) 10(B) 16(C) 20(D ) 8

** Answer: **

Binary operations on the set are is a function from

i.e. * is a function from

Hence, the total number of binary operations on set is

Hence, option B is correct

**More About NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise**

The NCERT Class 12 Maths chapter Relations and Functions has a total of 5 exercises including miscellaneous exercise. NCERT solutions for **Class 12 Maths chapter 1 miscellaneous exercise** covers solutions to 19 main questions and their sub-questions. This exercise covers more NCERT syllabus topics than preceding exercises and holds more weightage for exams also. Hence NCERT solutions for Class 12 Maths chapter 1 miscellaneous exercise is recommended for students to strengthen their conceptual understanding.

Also Read| NCERT Notes For Class 12 Mathematics Chapter 1

**Benefits of NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercises**

By using NCERT book miscellaneous exercise chapter 1 Class 12 students will be able to get an holistic understanding of the topics mentioned in this chapter.

Many questions mentioned in Class 12 Maths chapter 1 miscellaneous solutions are useful for the CBSE class 12 board exam and also for exams like JEE main, VITEEE, etc.

Revision can also be facilitated by using NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercises

This exercise can also be used in similar physics topics also.

**Also see-**

NCERT exemplar solutions class 12 maths chapter 1

NCERT solutions for class 12 maths chapter 1

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