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

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Answer:

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

fis 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)]

 

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