Question 1

Mathematics Function and Relation Hard

Inverse of the function \(f(x)=\frac{10^x-10^{-x}}{10^{x}+10^{-x}}\) is

(A) \(\log_{10}(2-x)\)
(B) \(\frac{1}{2}\log_{10}\Bigg{(}\frac{1+x}{1-x}\Bigg{)}\)
(C) \(\frac{1}{2}\log_{10}(2x-1)\)
(D) \(\frac{1}{4}\log_{10}\Bigg{(}\frac{2x}{2-x}\Bigg{)}\)
View Dynamic Solution & Explanation
Correct Solution: Option B

Step-by-step Solution:

The given function is: \( f(x) = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} \). To find the inverse, assume \( f(x) = y \) and solve the equation for \( x \): \[ y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} \] Multiply through by \( 10^x + 10^{-x} \): \[ y(10^x + 10^{-x}) = 10^x - 10^{-x} \] Rearranging terms: \[ 10^x (1 + y) = 10^{-x} (1 - y) \] Multiply through by \( 10^x \): \[ 10^{2x} (1 + y) = 1 - y \] Rearranging further: \[ 10^{2x} = \frac{1 - y}{1 + y} \] Taking the logarithm: \[ 2x = \log\left(\frac{1 + y}{1 - y}\right) \] Divide by 2: \[ x = \frac{1}{2} \log\left(\frac{1 + y}{1 - y}\right) \] Finally, replace \( y \) with \( x \) and \( x \) with \( f^{-1}(x) \): \[ f^{-1}(x) = \frac{1}{2} \log\left(\frac{1 + x}{1 - x}\right) \] This is the inverse function.