Question

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

 

पर्याय

• (a) 0.5

• (b) 0.6

• (c) 0.7

• (d) 0.8

Answer 1

(a) 0.5

\[e^{f\left( x \right)} = \frac{10 + x}{10 - x}\]

\[\Rightarrow f(x) = \log {}_e \left( \frac{10 + x}{10 - x} \right)\]    ...(1)

\[f\left( x \right) = kf\left( \frac{200 x}{100 + x^2} \right)\]

\[\Rightarrow \log {}_e \left( \frac{10 + x}{10 - x} \right) = k \log_e \left( \frac{10 + \frac{200x}{100 + x^2}}{10 - \frac{200x}{100 + x^2}} \right) {\text{ from }  (1)}\]

\[ \Rightarrow \log {}_e \left( \frac{10 + x}{10 - x} \right) = \text{ k } l {og}_e \left( \frac{1000 + 10 x^2 + 200x}{1000 + 10 x^2 - 200x} \right)\]

\[ \Rightarrow \log {}_e \left( \frac{10 + x}{10 - x} \right) =\text{  k}  l {og}_e \left( \frac{\left( x + 10 \right)^2}{\left( x - 10 \right)^2} \right)\]

\[ \Rightarrow \log {}_e \left( \frac{10 + x}{10 - x} \right) = 2\text{ k }  l {og}_e \frac{\left( x + 10 \right)}{\left( x - 10 \right)}\]

\[ \Rightarrow 1 = 2k\]

\[ \Rightarrow k = 1/2 = 0 . 5\]