The inverse function of $f(\mathrm{x})=\frac{8^{2 \mathrm{x}}-8^{-2 \mathrm{x}}}{8^{2 \mathrm{x}}+8^{-2 \mathrm{x}}}, \mathrm{x} \in(-1,1),$ is
The inverse function of $f(\mathrm{x})=\frac{8^{2 \mathrm{x}}-8^{-2 \mathrm{x}}}{8^{2 \mathrm{x}}+8^{-2 \mathrm{x}}}, \mathrm{x} \in(-1,1),$ is
- [JEE MAIN 2020]
- A
$\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1-x}{1+x}\right)$
- B
$\frac{1}{4} \log _{e}\left(\frac{1-x}{1+x}\right)$
- C
$\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1+x}{1-x}\right)$
- D
$\frac{1}{4} \log _{e}\left(\frac{1+x}{1-x}\right)$
Similar Questions
If $f\left( x \right) = {\left( {2x - 3\pi } \right)^5} + \frac{4}{3}x + \cos x$ and $g$ is the inverse of $f$, then $g'\left( {2\pi } \right)$ = ?
If $f\left( x \right) = {\left( {2x - 3\pi } \right)^5} + \frac{4}{3}x + \cos x$ and $g$ is the inverse of $f$, then $g'\left( {2\pi } \right)$ = ?
Let $f: X \rightarrow Y$ be an invertible function. Show that the inverse of $f^{-1}$ is $f$, i.e., $\left(f^{-1}\right)^{-1}=f$.
Let $f: X \rightarrow Y$ be an invertible function. Show that the inverse of $f^{-1}$ is $f$, i.e., $\left(f^{-1}\right)^{-1}=f$.
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If $f(x) = 3x - 5$, then ${f^{ - 1}}(x)$
- [IIT 1998]
If $f$ be the greatest integer function and $g$ be the modulus function, then $(gof)\left( { - \frac{5}{3}} \right) - (fog)\left( { - \frac{5}{3}} \right) = $
If $f$ be the greatest integer function and $g$ be the modulus function, then $(gof)\left( { - \frac{5}{3}} \right) - (fog)\left( { - \frac{5}{3}} \right) = $
Which of the following function is invertible
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