JEE Mains · Maths · STD 12 - 1. relation and function
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
- 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 \(\frac{1}{4} \log _{e}\left(\frac{1+x}{1-x}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{4}\left(\log _{8} e\right) \log _{e}\left(\frac{1+x}{1-x}\right)\)
Step-by-step Solution
Detailed explanation
\(f(\mathrm{x})=\mathrm{y}=\frac{8^{4 \mathrm{x}}-1}{8^{4 \mathrm{x}}+1}=1-\frac{2}{8^{4 \mathrm{x}}+1}\) so, \(8^{4 x}+1=\frac{2}{1-y} \Rightarrow 8^{4 x}=\frac{1+y}{1-y}\)…
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