# NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 1: Relations and Functions (Part IV)

Download NCERT Solutions for CBSE Class 12 Maths - Chapter 1: Relations and Functions. Here, you will find solutions of exercise 1.3 from question number 1 to question number 7. Solutions of other exercises are available in further parts. These questions are important for CBSE Class 12 Maths board exam.

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 1 to question number 7. Most of the questions given in this exercise are related to composition of functions and invertible function. These questions are important CBSE Class 12 Maths board exam.

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

**Question1:** Let *f* : {1, 3, 4} → {1, 2, 5} and *g* : {1, 2, 5} → {1, 3} be given by *f* = {(1, 2), (3, 5), (4, 1)} and *g* = {(1, 3), (2, 3), (5, 1)}. Write down* gof*.

**Solution 1:**

The functions *f*: {1, 3, 4} → {1, 2, 5} and *g*: {1, 2, 5} → {1, 3} are defined as

*f *= {(1, 2), (3, 5), (4, 1)} and *g *= {(1, 3), (2, 3), (5, 1)}.

*g*o*f* (1) = *g* [*f* (1)] = *g* (2) = 3

*g*o*f* (3) = *g* [*f* (3)] = *g* (5) = 1

*g*o*f* (4) = *g* [*f* (4)] = *g* (1) = 3

Thus, *g*o*f* = {(1,3), (3,1),(4,3)}

**Question 2: **Let *f*, *g* and *h* be functions from R to R. Show that

(*f* + *g*)o*h* = *f*o*h* + *g*o*h*

(*f*.*g*)o*h* = (*f*o*h*) . (*g*o*h*).

**Solution 2:**

To prove: (*f *+ *g*) o *h* = *f*o*h* + *g*o*h*

Proof:

[(*f* + *g*) o *h*] (*x*)

= (*f* + *g*) [*h* (*x*)]

= *f* [{*h*(*x*)} + *g*{*h*(*a*)}]

= (*f*o*h*) (*x*) + (*g*o*h*) (*x*)

= {*f*o*h* + *g*o*h*} *x*

Hence, (*f* + *g*)o*h* = *f*o*h* + *g*o*h*

To prove: (*f*.*g*)o*h* = (*f*o*h*) (*g*o*h*)

Proof:

[(*f.g*)o*h*] (x) = (*f.g*) [*h*(*x*)] = *f* [*h*(*x*)].*g *[*h*(*x*)] = (*f*o*h*) (*x*). (*g*o*h*) (*x*) = [(*f*o*h*).(*g*o*h*)] (*x*)

Hence, (*f*.*g*)o*h* = (*f*o*h*) (*g*o*h*)

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

**Question 3:** Find *g*o*f* and *f*o*g*, if

(*a*) *f* (*x*) = | *x* | and g(*x*) = | 5*x* – 2 |

(*b*)* f *(*x*) = 8*x*^{3} and *g*(*x*) = *x*^{1/3}.

**Solution 3:**

(*a*) *f* (*x*) = |*x*| & *g* (*x*) = |5*x* ‒ 2|

(*g*o*f* ) (*x*) = *g* [*f* (*x*)] = *g* (|*x*|) = |5 |*x*| ‒ 2|

(*f*o*g*) (*x*) = *f* [*g* (*x*)] = *f* (|5*x* ‒ 2|) = ||5*x* ‒ 2|| = |5*x* ‒ 2|.

(*b*) *f* (*x*) = 8 *x*^{3} & *g* (*x*) = *x*^{1/3}

(*g*o*f*) (*x*) = *g* [*f* (*x*)] = *g* (8*x*^{3})^{1/3} = 2*x*

(*f*o*g*) (*x*) = *f* [*g *(*x*)] = f (x^{1/3}) = 8(*x*^{1/3})^{3} = 8*x*.

**Question 4:** If *f* (*x*) = (4*x* + 3)/ (6*x* ‒ 4), *x* ≠ 2/3, show that *fof* (*x*) = *x*, for all *x* ≠ 2/3. What is the inverse of *f*?

**Solution 4:**

*f* (*x*) = (4*x* + 3)/(6*x* ‒ 4), *x* ≠ 2/3

(*f*o*f*) (*x*) = *f* [*f* (*x*)]

Hence, the given function *f* is invertible and the inverse of *f* is *f* itself.

**Question 5:** State with reason whether following functions have inverse

**( i)**

*f*: {1, 2, 3, 4} → {10} with

*f*= {(1, 10), (2, 10), (3, 10), (4, 10)}

**( ii)**

*g*: {5, 6, 7, 8} → {1, 2, 3, 4} with

*g*= {(5, 4), (6, 3), (7, 4), (8, 2)}

**( iii)**

*h*: {2, 3, 4, 5} → {7, 9, 11, 13} with

*h*= {(2, 7), (3, 9), (4, 11), (5, 13)}

**Solution 5:**

** **

**(ii)**

**(iii)**

**Question 6:** Show that *f *: [–1, 1] → *R*, given by *f* (*x*) = *x*/(*x* + 2) is one-one. Find the inverse of the function of the function *f* : [–1, 1] → Range* f*.

[Hint: For *y* ∈ Range *f*, *y* = *f* (*x*) = 2/(*x* + 2), for some x in [–1, 1], i.e., x = 2*y*/(1 ‒ *y*)]

**Solution 6:**

**Question 7:** Consider *f *: R → R given by *f *(*x*) = 4*x* + 3. Show that *f* is invertible. Find the inverse of *f*.

**Solution 7:**

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

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