If we are given a logarithmic function ${{\log }_{{{e}^{2}}x}}\left( \dfrac{2\ln x+2}{-x} \right)$ and $g(x)=\left\{ x \right\}$ then the range of g(x) for the existence of f(g(x)) is
A. $\left( 0,\dfrac{1}{e} \right)-\left\{ \dfrac{1}{{{e}^{2}}} \right\}$
B. $\left( 0,\dfrac{2}{e} \right)-\left\{ \dfrac{1}{{{e}^{2}}} \right\}$
C. $\left( 0,\dfrac{3}{e} \right)-\left\{ \dfrac{1}{{{e}^{2}}} \right\}$
D. None of these

Hint: To find the range of g(x) for the existence of f(g(x)), we will find the domain of f(x). From the rule of logarithmic functions, we will get ${{e}^{2}}x\ne 1\Rightarrow x\ne \dfrac{1}{{{e}^{2}}}$ . Also we know that $x>0$ . Hence, we can write $\left( \dfrac{2\ln x+2}{-x} \right)>0\Rightarrow \left( \dfrac{\ln x+1}{x} \right)<0$ . Now, we can find the critical points and they are $x=0$ and $x=\dfrac{1}{e}$ . From this, we can find the domain of the function f(x) that will be the range of g(x).

Complete step-by-step solution
We have to find the range of g(x) for the existence of f(g(x)).
This means that if g(x) is defined from $A\to B$ and f(x) is defined by \[B\to C\] then f(g(x)) is defined such that the range of g(x) will be the domain of f(x).
Now let us find the domain of f(x).
We know that the base of logarithmic functions will always be positive and not equal to 1.
Hence, we can write the base of given log functions as
$\begin{align}
  & {{e}^{2}}x\ne 1 \\
 & \Rightarrow x\ne \dfrac{1}{{{e}^{2}}} \\
\end{align}$
And we also know that $x>0$
Hence, we can write
$\left( \dfrac{2\ln x+2}{-x} \right)> 0$
Let us take 2 outside, we will get
$2\left( \dfrac{\ln x+1}{-x} \right)> 0$
This can be written as
$\left( \dfrac{\ln x+1}{x} \right)< 0$
Now, let us find the critical points.
We know that the denominator will be 0 when $x=0$ making the condition invalid.
And also, when $x=\dfrac{1}{e}$ ,
\[\left( \dfrac{\ln \left( \dfrac{1}{e} \right)+1}{\dfrac{1}{e}} \right)< 0\]
When we simplify this, we get
\[\begin{align}
  & \Rightarrow \left( -1+1 \right)e< 0 \\
 & \Rightarrow 0<0 \\
\end{align}\]
We know that the above condition is invalid.
Hence $x=\dfrac{1}{e}$ is another critical point.
Hence, we have 3 conditions.
$x\ne \dfrac{1}{{{e}^{2}}}$ and the critical points $x=0$ and $x=\dfrac{1}{e}$ .
We can write this as $x\in \left( 0,\dfrac{1}{e} \right)-\left\{ \dfrac{1}{{{e}^{2}}} \right\}$
It is given that $g(x)=\left\{ x \right\}$ , that is the fractional part of x.
So, obviously the range of g(x) will be
$\left( 0,\dfrac{1}{e} \right)-\left\{ \dfrac{1}{{{e}^{2}}} \right\}$
Hence, the correct option is A.

Note: The basic rules of log functions must be thorough, for example, the base of a log function is never negative and 1. One may make a mistake in this, leading to the wrong solution. When moving on to $\left( \dfrac{\ln x+1}{x} \right)< 0$ from the previous step $2\left( \dfrac{\ln x+1}{-x} \right)> 0$ , be cautious about the change of greater than sign due to the negative sign in the denominator. The student must know the value of basic log functions like $\log \left( \dfrac{1}{e} \right)$ .