#### Provide solution for RD Sharma Maths Class 12 Chapter Functions Exercise 2.2 Question 3.

Answer : $g\; o\; f = \left \{ \left ( 1,2 \right ),\left ( 4,-4 \right ),\left ( 9,-6 \right ),\left ( 16,8 \right ) \right \}$

Hint : The set which contains all the elements of all the ordered pairs of relation R is known as the domain of the relation. The set which contains all the second element is known as the range of the relation.

Given : $f=\left \{ (1,-1),(4,-2),(9,-3),(16,4) \right \}$

$g=\left \{ (-1,-2),(-2,-4),(-3,-6),(4,8) \right \}$

Prove : $g\; o\; f$ is defined while $f\; o\; g$ is not defined.

Solution

Now,  Domain of $f=\left \{ 1,4,9,16 \right \}$

Range of $f=\left \{ -1,-2,-3,4 \right \}$

Domain of $g=\left \{ -1,-2,-3,4 \right \}$

Range of $g=\left \{ -2,-4,-6,8 \right \}$

Clearly, Range of f = domain of g

$\therefore \; g\; o\; f \; \text {is defind}.$

But       Range of $g\neq$ domain of f

$\therefore \; f\; o\; g \; \text {is not defind}.$

Now, $g\; o\; f(1)=g(-1)=-2$

$g\; o\; f(4)=g(-2)=-4$

$g\; o\; f(9)=g(-3)=-6$

$g\; o\; f(16)=g(4)=8$

$g\; o\; f= \left \{ (1,-2),(4,-4),(9,-6),(16,8) \right \}$

Hence proved, $g\; o\; f$ is defined but $f\; o\; g$ is mot defined.

$g\; o\; f = \left \{ \left ( 1,-2 \right ),\left ( 4,-4 \right ),\left ( 9,-6 \right ),\left ( 16,8 \right ) \right \}$