Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 11 Maths Textbook Solution.

Bijection,  $f^{-1}\left ( 3 \right )=-1$.

Given:

$f:R\rightarrow R$ is defined as $f\left ( x \right )=x^{3}+4$.

Hint:

Bijection function should be fulfil the injection and surjection condition.

Solution:

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

$\! \! \! \! \! \! \! f\left ( x \right )=f\left ( y \right )\\\\x^{3}+4=y^{3}+4\\\\x^{3}=y^{3}\\\\x=y$

$f$is one-one.

Let $y$ be in the co-domain$\left ( R \right )$, such that $f\left ( x \right )=y$

$x^{2}+4=y\\\\x^{3}=y-4\\\\x=\sqrt[3]{y-4}\: \epsilon \: R \: domain$

$f$ is onto.

So, $f$ is a bijection, hence is invertible.

$f^{-1}\left ( x \right )=y$                                                                                                                                           …(i)

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

$\! \! \! \! \! \! \! x=y^{3}+4\\\\x-4=y^{3}\\\\y=\sqrt[3]{x-4}$

$\! \! \! \! So\: \: \: \: \: \: f^{-1}\left ( x \right )=\sqrt[3]{x-4}\\\\f^{-1}\left ( 3 \right )=\sqrt[3]{3-4}=\sqrt[3]{-1}=-1$                                                                                                                   [from (i)]