Question

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

• A. $0.5$
• B. $0.6$
• C. $0.7$
• D. $0.8$

Answer 1

Answer: (A)

$0.5$

Given $f\left(x\right)=k.f\left(\frac{200x}{100+x^2}\right)$ .....(1)
Also given $e^{f\left(x\right)}=\frac{10+x}{10-x}$

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

$\Rightarrow f\left(\frac{200x}{100+x^2}\right)=\log \left(\frac{10+\frac{200x}{100+x^2}}{10-\frac{200x}{100+x^2}}\right)$

$=\log \frac{(10+x{)}^2}{(10-x{)}^2}$

$=2\log \frac{10+x}{10-x}$

$=2f\left(x\right)$

$\therefore f\left(x\right)=\frac{1}{2}f\left(\frac{200x}{100+x^2}\right)$

On comparing with (1), we get

$k=\frac{1}{2}=0.5$