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  4. If ef(x) =(10 +x/10 -x), x∈(-10, 10) and f(x) =kf((200x/100 +x2)) then k =

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Q. If $e^{f\left(x\right)} =\frac{10 +x}{10 -x}, x\in\left(-10, 10\right)$ and $f\left(x\right) =kf\left(\frac{200x}{100 +x^{2}}\right)$ then $k =$

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Solution:

$e^{f\left(x\right)} =\frac{10 +x}{10 -x}, x\in\left(-10, 10\right)$
$\Rightarrow f\left(x\right) =log\left(\frac{10 +x}{10 -x}\right)$
$\Rightarrow f\left(\frac{200x}{100 +x^{2}}\right) = log \left[\frac{10+\frac{200x}{100 +x^{2}}}{10-\frac{200x}{100 +x^{2}}}\right]$
$=log \left[\frac{10\left(10 +x\right)^{2}}{10\left(10 -x\right)^{2}}\right] = 2$ log$\left(\frac{10 +x}{10-x}\right) = 2f\left(x\right)$
$\therefore f\left(x\right) =\frac{1}{2}f\left(\frac{200x}{100 +x^{2}}\right)$
$\Rightarrow k=\frac{1}{2} = 0.5$