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Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 17 Maths Textbook Solution.

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Answer:

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

Given:

                f\left ( x \right )=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}   is invertible

Hint:

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

Solution:

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

                f\left ( y\right )=x

                \frac{10^{y}-10^{-y}}{10^{y}+10^{-y}}=x

                \frac{10^{-y}\left ( 10^{2y}-1 \right )}{10^{-y}\left ( 10^{2y}+1 \right )}=x

                10^{2y}-1=x10^{2y}+x

                10^{2y}\left ( 1-x \right )=1+x

                10^{2y}=\frac{1+x}{1-x}

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

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

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

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