The range of \[f\left( x \right)={{\log }_{2}}\left( {{\log }_{\dfrac{1}{2}}}\left( 1-{{x}^{2}} \right) \right)\] is:
A) $\left( -\infty ,\infty \right)$
B) $\left( -1,1 \right)$
C) $\left( 0,\infty \right)$
D) $\left( -\infty ,0 \right)$

Hint: First, start with the condition check for the log i.e., the value in the log cannot be negative. After that again check the condition, the value in the log cannot be less than 0. After that perform the operations to find the roots of the equation. Then, apply the condition for which the equation satisfies.

Complete step-by-step answer:
Given:- \[f\left( x \right)={{\log }_{2}}\left( {{\log }_{\dfrac{1}{2}}}\left( 1-{{x}^{2}} \right) \right)\]
As we know that the value in the log cannot be less than or equal to 0. Because the negative number in the log is not possible. Then,
\[{{\log }_{\dfrac{1}{2}}}\left( 1-{{x}^{2}} \right)>0\]
Again the value in the log cannot be less than or equal to 0. Because the negative number in the log is not possible. Then,
$1-{{x}^{2}}>0$
Move the variable part on the right side,
${{x}^{2}}<1$
When \[\left( x-a \right)\left( x-b \right)<0\] then x lies between a and b. Hence one term will be positive and one term negative depending on whether a is greater or b is. x belongs to (a, b) or (b, a).
So,
Thus, the range of the function $f\left( x \right)$ is $\left( -1,1 \right)$.

Hence, option (2) is the correct answer.

Note: A function is a relation for which each value from the set of the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.
The domain of a function is the set of all possible inputs for the function.
Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.
In other words, a logarithm is essentially an exponent that is written in a particular manner.