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# The function f(x) is defined in [0, 1], then the domain of definition of the function $$f\left[ {\log \left( {1 - {x^2}} \right)} \right]$$$$\begin{array}{l}A.{\rm{ x}} \in \left\{ 0 \right\}\\B.{\rm{ x}} \in \left[ { - \sqrt {1 + e} , - 1} \right] \cup \left[ {1,\sqrt {1 + e} } \right]\\C.{\rm{ x}} \in \left( { - \infty ,\infty } \right)\\D.{\rm{ None of the above}}\end{array}$$  Hint: To solve the question given above, we will find out what a function is and what the domain of function is. Then, we will assume that, the value of $$\log \left( {1 - {x^2}} \right) = t$$ such that, we will get f(t). Now, t belongs to (0, 1). So, we will write $$0 \le \log \left( {1 - {x^2}} \right) \le 1$$ and from here, we will find the value of x. Then, we will choose the option which matches with our answer.

Before we solve the question, we must know what a function is and what the domain of the function is. A function is a binary relation over two sets that associates every element of the first set, to exactly one element of the second set. The domain of a function is the set of all possible inputs for the function. In other words, the domain is the set of all values of x over which the function exists.
Now, it is given that the domain of f(x) is [0, 1]. Now, we will assume that the value of $$f\left[ {\log \left( {1 - {x^2}} \right)} \right]$$ is y.
Thus, we have:
$$y = f\left[ {\log \left( {1 - {x^2}} \right)} \right]$$
Now, we will assume that $$\left( {1 - {x^2}} \right) = t$$ Thus, we have,
$$y = f\left( t \right)$$
Here, $$t \in \left[ {0,1} \right]$$ so we will get,
$$\begin{array}{l}\log \left( {1 - {x^2}} \right) \in \left[ {0,1} \right]\\ \Rightarrow 0 \le \log \left( {1 - {x^2}} \right) \le 1\end{array}$$
Here, we are assuming that the base of logarithm is e. Thus:
$$\begin{array}{l}0 \le {\log _e}\left( {1 - {x^2}} \right) \le 1\\ \Rightarrow \log 1 \le {\log _e}\left( {1 - {x^2}} \right) \le {\log _e}e\\ \Rightarrow 0 \le 1 - {x^2} \le e\end{array}$$
Now, we will subtract one from all sides. Thus, we will get:
$$0 \le - {x^2} \le e - 1$$
Now, we will multiply the above equation by (-1) and we will change the sign. Thus, we will get:
$$\Rightarrow 0 \ge {x^2} \ge 1 - e{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (i)}}$$
Now, we know that, square of any real number is greater than equal to zero. Thus:$${x^2} \ge 0{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (ii)}}$$
From (i) and (ii):
$$\begin{array}{l}{x^2} = 0\\x = 0\end{array}$$
Hence, option A is correct.

Note: While solving the question, we have chosen 'e' as the base of the logarithm function. We could have chosen any other value of base. It does not matter what we choose as the value of base, the domain will always be $$x \in \left\{ 0 \right\}$$ as long as the value of base is greater than zero and not equal to 1.

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