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

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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