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

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

Answers (1)

Answer:

                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, fis 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

                 

              

Posted by

infoexpert27

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