A) 0.5
B) 0.6
C) 0.7
D) 0.8
Correct Answer: A
Solution :
\[{{e}^{f(x)}}=\frac{10+x}{10-x},\,x\in (-10,\,10)\] Þ \[f(x)=\log \left( \frac{10+x}{10-x} \right)\] Þ \[f\left( \frac{200x}{100+{{x}^{2}}} \right)=\log \left[ \frac{10+\frac{200x}{100+{{x}^{2}}}}{10-\frac{200x}{100+{{x}^{2}}}} \right]=\log {{\left[ \frac{10(10+x)}{10(10-x)} \right]}^{2}}\] \[=2\log \left( \frac{10+x}{10-x} \right)=2f(x)\] \ \[f(x)=\frac{1}{2}f\left( \frac{200x}{100+{{x}^{2}}} \right)\Rightarrow k=\frac{1}{2}=0.5.\]
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