If {e^{f(x)}} = frac{{10 + x}}{{10 - x}},;x i | vedclass

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Question

(A) Given ${e^{f(x)}} = \frac{{10 + x}}{{10 - x}}$,taking the natural logarithm on both sides,we get $f(x) = \ln \left( \frac{10 + x}{10 - x} \right)$

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) Given ${e^{f(x)}} = \frac{{10 + x}}{{10 - x}}$,taking the natural logarithm on both sides,we get $f(x) = \ln \left( \frac{10 + x}{10 - x} ight)$.Now,evaluate $f\left( \frac{200x}{100 + x^2} ight)$:$f\left( \frac{200x}{100 + x^2} ight) = \ln \left( \frac{10 + \frac{200x}{100 + x^2}}{10 - \frac{200x}{100 + x^2}} ight)$$= \ln \left( \frac{10(100 + x^2) + 200x}{10(100 + x^2) - 200x} ight)$$= \ln \left( \frac{1000 + 10x^2 + 200x}{1000 + 10x^2 - 200x} ight)$$= \ln \left( \frac{10(x^2 + 20x + 100)}{10(x^2 - 20x + 100)} ight)$$= \ln \left( \frac{(x + 10)^2}{(10 - x)^2} ight)$$= 2 \ln \left( \frac{10 + x}{10 - x} ight) = 2f(x)$.Given $f(x) = k f\left( \frac{200x}{100 + x^2} ight)$,we substitute the result:$f(x) = k \cdot 2f(x)$.Since $f(x) 
eq 0$ for $x 
eq 0$,we have $1 = 2k$,which implies $k = 0.5$.

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

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