A) \[\frac{1}{3}{{\log }_{10}}\frac{1+x}{1-x}\]
B) \[\frac{1}{2}{{\log }_{10}}\frac{2+3x}{2-3x}\]
C) \[\frac{1}{3}{{\log }_{10}}\frac{2+3x}{2-3x}\]
D) \[\frac{1}{6}{{\log }_{10}}\frac{2-3x}{2+3x}\]
Correct Answer: B
Solution :
[b] if \[y=\frac{2}{3}\frac{{{10}^{x}}-{{10}^{-x}}}{{{10}^{x}}+{{10}^{-x}}},{{10}^{2x}}=\frac{3y+2}{2-3y}\] or \[x=\frac{1}{2}{{\log }_{10}}\frac{2+3y}{2-3y}\] \[\therefore {{f}^{-1}}(x)=\frac{1}{2}{{\log }_{10}}\frac{2+3x}{2-3x}.\]
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