Q. If $e^{f(x)} = \frac{10 + x}{10 - x},\ x \in ( - 10,\ 10)$ and $f(x) = kf\left( \frac{200x}{100 + x^{2}} \right)$, then $k$ is equal to
A) 0.5
B) 0.6
C) 0.7
D) 0.8
Explanation
(a)
Here, $f(x) = \log\frac{10 + x}{10 - x}$
Given that,$\ \ f(x) = kf{\left( \frac{200x}{100 + x^{2}} \right)\ \ }^{\ }$
$\Rightarrow \log\frac{10 + x}{10 - x} = k.\log\left\{ \frac{10 + \frac{200x}{100 + x^{2}}}{10 - \frac{200x}{100 + x^{2}}} \right\}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = k{\log\left( \frac{10 + x}{10 - x} \right)}^{2}\ \$
$\Rightarrow \ \ \log{\frac{10 + x}{10 - x} = 2k\log\frac{10 + x}{10 - x}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k = 0.5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.