#### Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 10 Maths Textbook Solution.

$f^{-1}\left ( x \right )=\sqrt[3]{x+3},f^{-1}\left ( 24 \right )=3,f^{-1}\left ( 5 \right )=2$

Given:

If  $f:R\rightarrow R$ be defined by $f\left ( x \right )=x^{3}-3$.

Hint:

Bijection function should be fulfil the injectivity, surjectivity condition.

Solution:

Let $x,y$ be two elements in domain$\left ( R \right )$, such that $x^{3}-3=y^{3}-3$.

$\! \! \! \! \! \! \! \! x^{3}=y^{3}\\\\x=y$

So, $f$ is one-one.

Surjectivity of $f$.

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

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

$f$ is onto.

So, $f$is a bijection and hence it’s convertible.

$\! \! \! \! \! \! \! \! \!Let, f^{-1}\left (x \right )=y\\\\x=f\left ( y \right )\\\\x=y^{3}-3\\\\x+3=y^{3}\\\\y=\sqrt[3]{x+3}=f^{-1}\left ( x \right )$                                                                                                                                       …(i)

From (i)

$\! \! \! \! \! \! \! \! f^{-1}\left ( x \right )=\sqrt[3]{x+3}\\\\f^{-1}\left ( 24+3 \right )=\sqrt[3]{27}=\sqrt[3]{3^{3}}=3\\\\f^{-1}\left ( 5 \right )=\sqrt[3]{5+3}=\sqrt[3]{8}=2$