Question

If $e^{f\left(x\right)}=\frac{10+x}{10-x}$, x ∈ (−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

(a) 0.5

$e^{f\left(x\right)}=\frac{10+x}{10-x}$
$\Rightarrow f\left(x\right)=\log {}_e\left(\frac{10+x}{10-x}\right)$ ...(1)
$f\left(x\right)=kf\left(\frac{200x}{100+x^2}\right)$
$$\begin{gathered} \Rightarrow \log {}_e\left(\frac{10+x}{10-x}\right)=k{\log }_e\left(\frac{10+{\displaystyle \frac{200x}{100+x^2}}}{10-\frac{200x}{100+x^2}}\right)\{\mathrm{from}(1\left)\right\} \\ \Rightarrow \log {}_e\left(\frac{10+x}{10-x}\right)=kl{\mathrm{og}}_e\left(\frac{1000+10x^2+200x}{1000+10x^2-200x}\right) \\ \Rightarrow \log {}_e\left(\frac{10+x}{10-x}\right)=kl{\mathrm{og}}_e\left(\frac{{\left(x+10\right)}^2}{{\left(x-10\right)}^2}\right) \\ \Rightarrow \log {}_e\left(\frac{10+x}{10-x}\right)=2kl{\mathrm{og}}_e\frac{\left(x+10\right)}{\left(x-10\right)} \\ \Rightarrow 1=2k \\ \Rightarrow k=1/2=0.5 \end{gathered}$$