Question

If ${e^{f(x)}} = \dfrac{{10 + x}}{{10 - x}}$, $x \in ( - 10,10)$ and $f(x) = k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$ then $k = $
A. 0.5
B. 0.6
C. 0.7
D. 0.8

Answer 1

Hint: Calculate $f(x)$ from ${e^{f(x)}} = \dfrac{{10 + x}}{{10 - x}}$ and then equate it to $f(x) = k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$. Simplify $k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$ by substituting $\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$ in place of $x$ in \[\log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\].

Formula used:${a^2} + {b^2} + 2ab = {(a + b)^2}$, ${a^2} + {b^2} - 2ab = {(a - b)^2}$

Complete step-by-step solution:
It is given that ${e^{f(x)}} = \dfrac{{10 + x}}{{10 - x}}$.
Taking log on both sides, \[f(x) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]
It is also given that $f(x) = k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$
\[k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]
Solving further, we get
\[k.\log \left( {\dfrac{{10 + \dfrac{{200x}}{{100 + {x^2}}}}}{{10 - \dfrac{{200x}}{{100 + {x^2}}}}}} \right) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]
\[k.\log \left( {\dfrac{{\dfrac{{1000 + 10{x^2} + 200x}}{{100 + {x^2}}}}}{{\dfrac{{1000 + 10{x^2} - 200x}}{{100 + {x^2}}}}}} \right) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]
\[k.\log \left( {\dfrac{{10\left( {100 + {x^2} + 20x} \right)}}{{10\left( {100 + {x^2} - 20x} \right)}}} \right) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]
Simplifying further, we get
\[k.\log \left( {\dfrac{{{{\left( {10 + x} \right)}^2}}}{{{{\left( {10 - x} \right)}^2}}}} \right) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]
\[2k.\log \left( {\dfrac{{10 + x}}{{10 - x}}} \right) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]
$2k = 1$
$k = 0.5$
Therefore, the answer is A. 0.5

Note: Proof that if ${e^n} = m$, then $n = \log m$:
Taking $\log $ on both sides of the equation, we get
$\log {e^n} = \log m$
$n\log e = \log m$
$n = \log m$ as $\log e = 1$
This is why \[f(x) = \log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)\]