Find the Value of \( k \)
If
\[ e^{f(x)}=\frac{10+x}{10-x}, \qquad x\in(-10,10) \]
and
\[ f(x)=k\,f\left(\frac{200x}{100+x^2}\right), \]
then \(k=\)
(a) \(0.5\)
(b) \(0.6\)
(c) \(0.7\)
(d) \(0.8\)
Given,
\[ e^{f(x)}=\frac{10+x}{10-x} \]
Therefore,
\[ f(x)=\log\left(\frac{10+x}{10-x}\right) \]
Now,
\[ f\left(\frac{200x}{100+x^2}\right) = \log\left( \frac{ 10+\frac{200x}{100+x^2} }{ 10-\frac{200x}{100+x^2} } \right) \]
\[ = \log\left( \frac{(10+x)^2}{(10-x)^2} \right) \]
\[ = 2\log\left( \frac{10+x}{10-x} \right) \]
\[ =2f(x) \]
Given,
\[ f(x)=k\cdot 2f(x) \]
\[ 1=2k \]
\[ k=\frac12=0.5 \]
\[ \boxed{\text{Correct Answer: (a)}} \]