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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

                 

              

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infoexpert27

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