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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