NCERT Solutions for CBSE Class 12 Maths, Chapter 1: Relations & Functions are available in this article. Here, you will get solutions to the questions of exercise 1.3 from question number 8 to question number 14. These questions are related to composition of functions and invertible function.

*NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions (Exercise 1.3) are given below*

**Question8:**

**Solution 8:**

**NCERT Exemplar Class 12 Mathematics – Chapter 1 Relations and Functions**

**Question 9:**

**Solution 9:**

**NCERT Solutions for CBSE Class 12 Physics: All Chapters**

**Question 10:** Let *f *: X → Y be an invertible function. Show that f has unique inverse. (Hint: suppose *g*_{1} and *g*_{2} are two inverses of *f*. Then for all *y* ∈ Y, *f*o*g*_{1}(*y*) = *I _{Y}*(y) =

*f*o

*g*

_{2}(

*y*). Use one-one ness of

*f*).

**Solution 10:**

Let* Y* is an invertible function

Suppose *g*_{1} & *g*_{2} are two inverses of *f *

Now, for all *y* ϵ Y,

*f*o*g*_{1} (*y*) = *f*o*g*_{2} (*y*)

*f*[*g*_{1}(*y*)] = *f*[*g*_{2}(*y*)]

*g*_{1}(*y*) = *g*_{2}(*y*) {f is one-one}

*g*_{1} = *g*_{2} (g is one-one)

**Question 11: **Consider *f* : {1, 2, 3} → {*a*, *b*, *c*} given by *f *(1) = a, *f* (2) = b and *f *(3) = *c*. Find *f*^{ –1} and show that (*f* ^{–1})^{–1} = *f*.

**Solution 11:**

*f*: {1, 2, 3} → {*a*, *b*, *c*}

*f*(1) = *a*

*f*(2) = *b*

*f*(3) = *c*

Let *g*: {*a*, *b*, *c*} → {1, 2, 3}

*g*(*a*) = 1

*g*(*b*) = 2

*g*(*c*) = 3

Now,

(*f*o*g*) (*a*) = *f* [*g*(*a*)] = *f* (1) = *a*

(fog) (*b*) = *f* [*g*(*b*)] = *f*(2) = *b*

(*f*o*g*) (*c*) = *f* [*g*(*c*)] = *f *(3) = *c*

(*g*o*f*) (1) = *g* [*f*(1)] = *g*(*a*) =1

(*g*o*f*) (2) = *g* [*f*(2)] = *g*(*b*) =2

(*g*o*f*) (3) = *g* [*f*(3)] = *g*(*c*) =3

Thus, *gof* = *fog*

*gof* = *I*_{X}

fog = *I*_{Y}

Where *X* = {1, 2, 3} and *Y*= {*a*, *b*, *c*}

Thus, *f*^{−1} **=** *g*

Hence, *f*^{−1}: {*a*, *b*, *c*} → {1, 2, 3}

*f*^{−1}(*a*) = 1

*f*^{−1}(*b*) = 2

*f*^{-1}(*c*) = 3

Let us calculate the inverse of *f*^{−1}

*h*: {1, 2, 3} → {*a*, *b*, *c*}

*h*(1) = *a*, *h*(2) = *b*, *h*(3) = *c*

Then,

(*g*o*h*) (1) = *g*[*h*(1)] = *g*(*a*) = 1

(*g*o*h*) (2) = *g*[*h*(2)] = *g*(*b*) = 2

(*g*o*h*) (3) = *g*[*h*(3)] = *g*(*c*) = 3

(*h*o*g*)(*a*) = *h* [*g*(*a*)]= *h* (1) = *a*

(*h*o*g*)(*b*) = *h* [*g*(*b*)]= *h* (2) = *b*

(*h*o*g*)(*c*) = *h* [*g*(*c*)]= *h* (3) = *c*

*g*o*h* = *I*_{X}

*h*o*g* = *I*_{Y}

Now,

*g*^{−1} = *h*

(*f*^{−1})^{−1} = *h*

We can observe that

*h* = *f*

Hence, (*f*^{‒1})^{‒1} = *f.*

**Question 12:** Let* f* : X → Y be an invertible function. Show that the inverse of *f*^{–1 }is *f*, i.e., (*f*^{ –1})^{–1} = *f*.

**Solution 12:**

**Question 13:** If *f* : R → R be given by* f* (*x*) = (3 − x^{3} )^{1/3} , then *f*o*f* (*x*) is

(A) *x*^{3}

(B) *x*^{3 }

(C) *x*

(D) (3 – *x*^{3}).

**Solution 13:**

**Question 14:**

**Solution 14:**

**Download NCERT Solutions for Class 12 Maths: Chapter 1 Relations and Functions in PDF format**