Question

The domain of ${\sin ^{ - 1}}\left( {\ln x} \right)$ is
A. $\left[ {\dfrac{1}{e},e} \right]$
B. $\left( {\dfrac{1}{e},2} \right]$
C. $\left( {0,\infty } \right)$
D. $\left( { - \infty ,0} \right]$

Answer 1

Hint: We first determine the range of $\sin x$, hence finding the domain of ${\sin ^{ - 1}}\left( x \right)$. Next, find the values of $x$ which satisfy the domain of ${\sin ^{ - 1}}\left( x \right)$. From those values of $x$, find the values that will satisfy the domain of $\ln x$, as $x > 0$ for $\ln x$.

Complete step by step answer:

First of all, determine the range of $\sin x$.
As $\sin x$ can take values from $ - 1$ to 1 only, therefore, the range of $\sin x$ is $ - 1 \leqslant x \leqslant 1$.
Therefore, the domain of ${\sin ^{ - 1}}\left( x \right)$ is $ - 1 \leqslant x \leqslant 1$.
In this question, we are given ${\sin ^{ - 1}}\left( {\ln x} \right)$, then the value of $\ln x$ will be from $ - 1$ to 1, that implies,
$ - 1 \leqslant \ln x \leqslant 1$
Solve for $x$ in the above inequality.
As we know, \[\ln {a^m} = b \Rightarrow {e^b} = a\]
Therefore, $ - 1 \leqslant \ln x \leqslant 1$ can be written as, ${e^{ - 1}} \leqslant x \leqslant e$ which is equivalent to $\dfrac{1}{e} \leqslant x \leqslant e$
Now, eliminate those values of \[x\] which are negative because $\ln x$is only defined for values $x > 0$.
In $\dfrac{1}{e} \leqslant x \leqslant e$, all values of \[x\] are positive. Hence, the domain of ${\sin ^{ - 1}}\left( {\ln x} \right)$ is $\dfrac{1}{e} \leqslant x \leqslant e$ which can be written as $\left[ {\dfrac{1}{e},e} \right]$
Therefore, option A is correct.

Note: The domain of ${\sin ^{ - 1}}\left( x \right)$ is $\left[ { - 1,1} \right]$ and the range of ${\sin ^{ - 1}}\left( x \right)$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$. The domain of $\ln x$is $\left( {0,\infty } \right)$ and the range of $\ln x$ is $\left( { - \infty ,\infty } \right)$. Therefore., after calculating the values of $\ln x$, take only those values of $x$ which belongs to the domain of $\ln x$.