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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)

best_answer

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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