Question

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

• A. 0.5
• B. 0.6
• C. 0.7
• D. 0.8

Answer 1

The correct option is A

0.5

$f\left(x\right)=\frac{10+x}{10-x}\Rightarrow f\left(x\right)=lo{g}_e\left(\frac{10+x}{10-x}\right)\dots \dots \left(1\right)f\left(x\right)=kf\left(\frac{200x}{100+x^2}\right)$
$\Rightarrow lo{g}_e\left(\frac{10+x}{100-x}\right)=klo{g}_e\left(\frac{10+\frac{200x}{100+x^2}}{10-\frac{200x}{100+x^2}}\right)$ [from (i)]
$\Rightarrow lo{g}_e\left(\frac{10+x}{100-x}\right)=klo{g}_e\left(\frac{1000+10x^2+200x}{1000+10x^2-200x}\right)\Rightarrow lo{g}_e\left(\frac{10+x}{10-x}\right)=klo{g}_e\left(\frac{(x+10{)}^2}{(x-10{)}^2}\right)\Rightarrow lo{g}_e\left(\frac{10+x}{10-x}\right)=2klo{g}_e\left(\frac{x+10}{x-10}\right)\Rightarrow 1=2k\Rightarrow k=\frac{1}{2}=0.5$