#### Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 12 Maths Textbook Solution.

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

Given:

$f:Q\rightarrow Q,g:Q\rightarrow Q$ defined by $f\left ( x \right )=2x,g\left ( x \right )=x+2$.

Hint:

Bijective function should be fulfil the injective and surjective function.

Solution:

$x,y$ be two elements of domain$\left ( Q \right )$, such

$\! \! \! \! \! \! \! \! \! f\left (x \right )=f\left ( y \right )\\\\2x=2y\\\\x=y$

$f$ is one-one.

Let $y$ be in the co-domain$\left ( Q \right )$, such that

$f\left ( x \right )=y$

$2x =y$

$x=\frac{y}{2}\epsilon \: Q$.

$f$ is onto.

So, $f$ is a bijection and hence, it’s invertible
Let          $f^{-1}\left ( x \right )=y$                                                                                                                                           …(i)

$\\\\x=f\left ( y \right )\\\\y=\frac{x}{2}\\\\f^{-1}\left ( x \right )=\frac{x}{2}$                                                                                                                                                                                                             Injectivity of $g$

Let  $x,y$ be two elements of domain$\left ( Q \right )$.

$\! \! \! \! \! \! \! \! g\left ( x \right )=g\left ( y \right )\\\\x+2=y+2\\\\x=y$

So, $g$ is one-one.

Subjectivity of $g$:

Let $y$ be in the co-domain$\left ( Q \right )$, such that $g\left ( x \right )=y$.

$x+2=y$

$x=2-y\: \epsilon \:$ (domain)

$g$ is onto.

So, $g$ is a bijection, hence it’s invertible.

Let          $g^{-1}\left ( x \right )=y$

$x=g\left ( y \right )$                                                                                                                                                         …(ii)

$x=y+2$

$y=x-2$

So,          $g^{-1}\left ( x \right )=x-2$                                                                                                                                   [from (ii)]

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

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

$\left ( f^{-1} og^{-1}\right )\left ( x \right )=f^{-1}\left ( x-2 \right )$                                                   $\left [ \because \left ( f^{-1} og^{-1}\right )\left ( x \right )=f^{-1}\left ( g^{-1}\left ( x \right ) \right ) \right ]$

$\left ( f^{-1}og^{-1} \right )\left ( x \right )=f^{-1}\left ( \frac{x-2}{2} \right )$                                                                                                              …(iii)

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

$=g\left (2 x \right )$

$=2x+1$

Let          $\left ( gof \right )\left ( x \right )=y$                                                                                                                                 …(vi)

$x=\left ( gof \right )\left ( y \right )$

$x=2y+2$

$y=\frac{x-2}{2}$

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

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