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Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 18 Maths Textbook Solution.

Answers (1)

Answer:

                f^{-1}\left ( x \right )= \frac{1}{2}\log _{e}\left ( \frac{x}{2-x} \right )

Given:

                f\left ( x \right )= \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}+1  is invertible.

Hint:

                f^{-1}\left ( x \right )=y\Rightarrow f\left ( y \right )=x

Solution:

Let          f^{-1}\left ( x \right )=y

                f\left ( y \right )=x

                \frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}+1=x

                \frac{e^{-y}\left ( e^{2y}-1 \right )}{e^{-y}\left ( e^{2y}+1 \right )}=x-1

                e^{2y}-1=xe^{2y}-e^{2y}+x-1

                e^{2y}\left ( 2-x \right )=x

                e^{2y}=\frac{x}{2-x}

                2y=\log_{e}\frac{x}{2-x}

                y=\frac{1}{2}\log_{e}\left (\frac{x}{2-x} \right )= f^{-1}\left ( x \right )

      So,             f^{-1}\left ( x \right )=\frac{1}{2}\log_{e}\left (\frac{x}{2-x} \right )

      

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infoexpert27

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