#### Provide solution for rd sharma maths class 12 chapter Functions exercise 2.4 question 4 sub-question (0)

$\left ( gof \right )^{-1}=f^{-1}og^{-1}$

Given:

$A=\left \{ 1,2,3,4 \right \},B=\left \{ 3,5,7,9 \right \},C=\left \{ 7,23,47,79 \right \}$

$f:A\rightarrow B,g:B\rightarrow Cdefined\: by\: f\left ( x \right )=2x+1,g\left ( x \right )=x^{2}-2.$

Hint:

Clearly $f$ and $g$ are bijections.

Solution:

$f\left ( x \right )=2x+1$

$f=\left \{ \left ( 1,2\left ( 1 \right )+1 \right ),\left ( 2,2\left ( 2 \right ) +1 \right ),\left ( 3,2\left ( 3 \right ) +1\right ),\left ( 4,2\left ( 4 \right ) +1\right )\right \}$

$=\left \{ \left ( 1,3 \right ),\left ( 2,5 \right ),\left ( 3,7 \right ),\left ( 4,9 \right ) \right \}$

$g\left ( x \right )=x^{2}-2$

$g=\left \{ \left ( 3,3^{2} -2\right ),\left ( 5,5^{2}-2 \right ),\left ( 7,7^{2} -2\right ),\left ( 9,9^{2}-2 \right ) \right \}$

$=\left \{ \left ( 3,7 \right ),\left ( 5,23 \right ),\left ( 7,47 \right ),\left ( 9,79 \right ) \right \}$

So,  $f^{-1}=\left \{ \left ( 3,1 \right ),\left ( 5,2 \right ),\left ( 7,3 \right ),\left ( 9,4 \right ) \right \}$

$g^{-1}=\left \{ \left ( 7,3 \right ),\left ( 23,5 \right ),\left ( 47,7 \right ),\left ( 79,9 \right ) \right \}$

Now, $\left ( f^{-1}0g^{-1} \right ):C\rightarrow A$

$\left ( f^{-1}og^{-1} \right )=\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}$

Also $f:A\rightarrow B,g:B\rightarrow C$

$gof:A\rightarrow C,\left ( gof \right )^{-1}:C\rightarrow A$

So,  $f^{-1}og^{-1}\: \: and\: \: \left ( gof \right )^{-1}$ have the same domains.

$\left ( gof \right )\left ( x \right )=g\left ( x \right )=g\left ( 2x+1 \right )=\left ( 2x+1 \right )^{2}-2$

$\left ( gof \right )\left ( x \right )=g\left [ f\left ( x \right ) \right ]$

$\left ( gof \right )\left ( x \right )=4x^{2}+4x+1-2$

$\left ( gof \right )\left ( x \right )=4x^{2}+4x-1$

Then, $\left ( gof \right )\left ( 1 \right )=g\left ( f\left ( 1 \right ) \right )=4+4-1=7$

$\left ( gof \right )\left ( 2 \right )=g\left ( f\left ( 2 \right ) \right )=16+8-1=23$

$\left ( gof \right )\left ( 3 \right )=g\left ( f\left ( 3 \right ) \right )=36+12-1=47$

$\left ( gof \right )\left ( 4 \right )=g\left ( f\left ( 4 \right ) \right )=64+16-1=79$

So,   $\left ( gof \right )=\left \{ \left ( 1,7 \right ),\left ( 2,23 \right ),\left ( 3,47 \right ),\left ( 4,79 \right ) \right \}$

$\left ( gof \right )^{-1}=\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}$

From $\left ( i \right ),\left ( ii \right )$

$\left ( i \right ),\left ( ii\left ( gof \right ) \right )^{-1}=f^{-1}og^{-1}$

$\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}=\left \{ \left ( 7,1 \right ) ,\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right )\right \}$