TS EAMCET · Maths · Functions
The inverse of the function \(y=\frac{10^x-10^{-x}}{10^x+10^{-x}}+1\) is \(x=\)
- A \(\log \left(\frac{y}{2-y}\right)\)
- B \(\log _{10}\left(\frac{y}{2-y}\right)\)
- C \(\frac{1}{10} \log \left(\frac{y}{1-y}\right)\)
- D \(\frac{1}{2} \log _{10}\left(\frac{y}{2-y}\right)\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2} \log _{10}\left(\frac{y}{2-y}\right)\)
Step-by-step Solution
Detailed explanation
\(y-1=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) \(y-1=\frac{10^{2x}-1}{10^{2x}+1}\) \((y-1)(10^{2x}+1)=10^{2x}-1\) \((y-1)10^{2x}+y-1=10^{2x}-1\) \((y-1)10^{2x}-10^{2x}=-1-(y-1)\) \(10^{2x}(y-1-1)=-y\) \(10^{2x}(y-2)=-y\) \(10^{2x}=\frac{-y}{y-2}=\frac{y}{2-y}\)…
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