Question

The domain of $f(x)$ is $(0,1)$ , therefore, the domain of $y = f({e^x}) + f(\ln |x|)$ is:
A) $\left( {\dfrac{1}{e},1} \right)$
B) $\left( { - e, - 1} \right)$
C) $\left( { - 1, - \dfrac{1}{e}} \right)$
D) $\left( { - e, - 1} \right) \cup \left( {1,e} \right)$

Answer 1

Hint:
We will first find the individual domain of all the given functions individually, after that we will find the intersection of the domain so that it becomes a common domain for all the functions. Hence, the intersection of the domains would be the final answer.

Complete step by step solution:
We are given that the domain of $f(x)$ is $(0,1)$ , hence
$ \Rightarrow 0 < x < 1$ … (1)
For the function $y = f({e^x}) + f(\ln |x|)$ ,
For ${f_1}(x) = {e^x}$ , we get
$ \Rightarrow 0 < {e^x} < 1$ … (2)
According to the graph ${e^{ - \infty }} = 0$ and ${e^0} = 1$ , hence
$ \Rightarrow - \infty < x < 0$
Hence domain of ${f_1}(x) = {e^x}$ is
$ \Rightarrow {D_1} = ( - \infty ,0)$
Now, for ${f_2}(x) = \ln |x|$ , we get
$ \Rightarrow 0 < \ln |x| < 1$ … (2)
Taking Euler's number in (2), we get
\[ \Rightarrow {e^0} < {e^{\ln |x|}} < {e^1}\]
On substituting ${e^0} = 1$ , ${e^1} = e$ and ${e^{\ln (a)}} = a$ , we get
\[ \Rightarrow 1 < |x| < e\]
Since the domain of \[a < |x| < b\] is
\[D = ( - b, - a) \cup (a,b)\]
Hence domain of ${f_2}(x) = \ln |x|$ is
⇒${D_2} = ( - e, - 1) \cup (1,e)$
Hence, the final domain of the function
$y = {D_1} \cap {D_2}$
On substituting the values of domains, we get
$ \Rightarrow ( - \infty ,0) \cap (( - e, - 1) \cup (1,e))$
According to distributive property, we get
$ \Rightarrow (( - \infty ,0) \cap ( - e, - 1)) \cup (( - \infty ,0) \cap (1,e))$
Since nothing is common between $( - \infty ,0)$ and $(1,e)$ , hence
$ \Rightarrow ( - \infty ,0) \cap (1,e) = 0$
Therefore
$ \Rightarrow ( - \infty ,0) \cap ( - e, - 1)$
Since the whole set $( - e, - 1)$ is present in $( - \infty ,0)$ , hence
$ \Rightarrow ( - \infty ,0) \cap ( - e, - 1) = ( - e, - 1)$
Hence, the final domain is $( - e, - 1)$

Therefore, the final answer is B.

Note:
We can also find the intersection of domains by the number line method, consider a number line where numbers are present from $ - \infty $ to $ + \infty $ , Only marks the places where all the domains are present, hence make sets of the required answer to get the final answer.