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#### Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 7 Maths Textbook Solution.

$f^{-1}\left ( x \right )=\sqrt{x-4}$

Given:

$f\left ( x \right )=x^{2}+4,f:R\rightarrow R_{+}\rightarrow \left [ 4,\infty \right ]$.

Hint:

Bijection function should be fulfil the injectivity and surjectivity.

Solution:

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

$\! \! \! \! \! \! \! \! \! \! f\left ( x \right )=f\left ( y \right )\\\\x^{2+4}=y^{2}+4\\\\x^{2}=y^{2}\\\\x=y$

So $f$is one-one.

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

$\! \! \! \! \! \! \! \! f\left ( x \right )=y\\\\x^{2}+4=y\\\\x^{2}=y-4\\\\x=\sqrt{y-4}\epsilon \: R$

$f$ is onto, so $f$ is bijection.

$\! \! \! \! \! \! \! \! f^{-1}\left ( x \right )=y\\\\x=f\left ( y \right )\\\\x=y^{2}+4\\\\x-4=y^{2}y=\sqrt{x-4}\\\\f^{-1}\left ( x \right )=\sqrt{x-4}$                                                                                                                                       … (i)