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प्रश्न
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
उत्तर
(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)
\[f\left( x \right) = kf\left( \frac{200 x}{100 + x^2} \right)\]
\[\Rightarrow \log {}_e \left( \frac{10 + x}{10 - x} \right) = k \log_e \left( \frac{10 + \frac{200x}{100 + x^2}}{10 - \frac{200x}{100 + x^2}} \right) {\text{ from } (1)}\]
\[ \Rightarrow \log {}_e \left( \frac{10 + x}{10 - x} \right) = \text{ k } l {og}_e \left( \frac{1000 + 10 x^2 + 200x}{1000 + 10 x^2 - 200x} \right)\]
\[ \Rightarrow \log {}_e \left( \frac{10 + x}{10 - x} \right) =\text{ k} l {og}_e \left( \frac{\left( x + 10 \right)^2}{\left( x - 10 \right)^2} \right)\]
\[ \Rightarrow \log {}_e \left( \frac{10 + x}{10 - x} \right) = 2\text{ k } l {og}_e \frac{\left( x + 10 \right)}{\left( x - 10 \right)}\]
\[ \Rightarrow 1 = 2k\]
\[ \Rightarrow k = 1/2 = 0 . 5\]
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