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

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

Answers (1)

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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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