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If $\quad e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10) \quad$ and $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$, then $k$ is equal to
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The correct answer is:
0.5
We have,
$e^{f(x)}=\frac{10+x}{10-x}$
$f(x)=\log \frac{10+x}{10-x}$
Given that $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$
$\Rightarrow \quad \log \frac{10+x}{10-x}=k \cdot \log \left\{\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right\}$
$=k \log \left\{\frac{1000+10 x^2+200 x}{1000+10 x^2-200 x}\right\}$
$=k \log \left\{\frac{x^2+100+20 x}{x^2+100-20 x}\right\}$
$=k \log \left\{\frac{(x+10)^2}{(10-x)^2}\right\}$
$=2 k \log \frac{10+x}{10-x}$
$\Rightarrow \quad \log \frac{10+x}{10-x}=2 k \log \frac{10+x}{10-x}$
$\therefore \quad 2 k=1 \Rightarrow k=\frac{1}{2}=0.5$
$e^{f(x)}=\frac{10+x}{10-x}$
$f(x)=\log \frac{10+x}{10-x}$
Given that $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$
$\Rightarrow \quad \log \frac{10+x}{10-x}=k \cdot \log \left\{\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right\}$
$=k \log \left\{\frac{1000+10 x^2+200 x}{1000+10 x^2-200 x}\right\}$
$=k \log \left\{\frac{x^2+100+20 x}{x^2+100-20 x}\right\}$
$=k \log \left\{\frac{(x+10)^2}{(10-x)^2}\right\}$
$=2 k \log \frac{10+x}{10-x}$
$\Rightarrow \quad \log \frac{10+x}{10-x}=2 k \log \frac{10+x}{10-x}$
$\therefore \quad 2 k=1 \Rightarrow k=\frac{1}{2}=0.5$
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