A) \[{{[f(x)]}^{2}}\]
B) \[{{[f(x)]}^{3}}\]
C) \[2f(x)\]
D) \[3f(x)\]
Correct Answer: C
Solution :
\[f(x)=\log (x+\sqrt{{{x}^{2}}+1})\] \[\therefore \,\,\,f\left( \frac{2x}{1+{{x}^{2}}} \right)=\log \,\left[ \frac{1+\frac{2x}{1+{{x}^{2}}}}{1-\frac{2x}{1+{{x}^{2}}}} \right]=\log \,\left[ \frac{{{x}^{2}}+1+2x}{{{x}^{2}}+1-2x} \right]\] \[=\log \,{{\left[ \frac{1+x}{1-x} \right]}^{2}}=2\,\log \,\left[ \frac{1+x}{1-x} \right]=2\,f(x)\].
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