If

(a) 0.5

$e^{f(x)}=\frac{10+x}{10-x}$

$\Rightarrow f(x)=\log _{e}\left(\frac{10+x}{10-x}\right) \ldots(1)$

$f(x)=k f\left(\frac{200 x}{100+x^{2}}\right)$

$\Rightarrow \log _{e}\left(\frac{10+x}{10-x}\right)=k \log _{e}\left(\frac{10+\frac{200 x}{100+x^{2}}}{10-\frac{200 x}{100+x^{2}}}\right) \quad\{$ from $(1)\}$

$\Rightarrow \log _{e}\left(\frac{10+x}{10-x}\right)=k l \operatorname{og}_{e}\left(\frac{1000+10 x^{2}+200 x}{1000+10 x^{2}-200 x}\right)$

$\Rightarrow \log _{e}\left(\frac{10+x}{10-x}\right)=k l \log _{e}\left(\frac{(x+10)^{2}}{(x-10)^{2}}\right)$

$\Rightarrow \log _{e}\left(\frac{10+x}{10-x}\right)=2 k \log _{e} \frac{(x+10)}{(x-10)}$

$\Rightarrow 1=2 k$

$\Rightarrow k=1 / 2=0.5$