If we have functions ${e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}},x \in \left( { - 10,10} \right)$ and $f\left( x \right) = k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$ then k =
$
  {\text{A}}{\text{. 0}}{\text{.5}} \\
  {\text{B}}{\text{. 0}}{\text{.6}} \\
  {\text{C}}{\text{. 0}}{\text{.7}} \\
  {\text{D}}{\text{. 0}}{\text{.8}} \\
$

Hint: -To solve this question first we have to find f (x ) taking log on both sides and then we have to find $f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$ and thereafter further calculation to get answer.

Complete step-by-step solution -
We have
${e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}}$
Now taking log on both side with base e we get,
${\log _e}{e^{f\left( x \right)}} = {\log _e}\dfrac{{10 + x}}{{10 - x}}$
Now as we know the property of log $\left( {{{\log }_a}{a^m} = m{{\log }_a}a = m} \right)$
So now,
$f\left( x \right) = {\log _e}\dfrac{{10 + x}}{{10 - x}}$$ \ldots \ldots \left( i \right)$
Now we have to find $f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$ so we will replace x by $\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$
On doing this we get,
$f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{10 + \dfrac{{200x}}{{100 + {x^2}}}}}{{10 - \dfrac{{200x}}{{100 + {x^2}}}}}} \right)$
Now on further calculation we get,
$f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{\dfrac{{1000 + 10{x^2} + 200x}}{{100 + {x^2}}}}}{{\dfrac{{1000 + 10{x^2} - 200x}}{{100 + {x^2}}}}}} \right)$
On cancel out we get,
$f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{1000 + 10{x^2} + 200x}}{{1000 + 10{x^2} - 200x}}} \right)$
Now we can write it as
$f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{100 + {x^2} + 20x}}{{100 + {x^2} - 20x}}} \right) = {\log _e}\left( {\dfrac{{{{\left( {10 + x} \right)}^2}}}{{{{\left( {10 - x} \right)}^2}}}} \right)$
$f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}{\left( {\dfrac{{\left( {10 + x} \right)}}{{\left( {10 - x} \right)}}} \right)^2} = 2{\log _e}\left( {\dfrac{{10 + x}}{{10 - x}}} \right)$
Now we have following result
$f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = 2{\log _e}\left( {\dfrac{{10 + x}}{{10 - x}}} \right)$$ \ldots \ldots \left( {ii} \right)$
And we have given in the question
$f\left( x \right) = k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)$
Using results obtained from equation (i) and equation ( ii ) we get,
${\log _e}\dfrac{{10 + x}}{{10 - x}} = k.2\log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)$
On cancel out we get,
1 = 2k
$\therefore k = \dfrac{1}{2} = 0.5$
Hence option A is the correct option.

Note:- Whenever we get this type of question the key concept of solving is we have knowledge of function as well as good understanding of function and relation. This solution is a standard method of solving this type of question so keep in mind this method for further this type of question.