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

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

Answers (1)

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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