Question

Find the domain of $f\left(x\right)={\displaystyle \frac{1}{\sqrt{\ln \left({\cot }^{-1}x\right)}}}$?
(a) $(\cot 1,\mathrm{\infty })$
(b) $R-\cot 1$
(c) $(-\mathrm{\infty },0)\cup (0,\cot 1)$
(d) $(-\mathrm{\infty },\cot 1)$

Answer 1

Hint: To find the domain of the function, find the set of all possible values of x for which the function is defined. For the log function to be defined consider its argument greater than 0 and for the square root in the denominator to be defined consider the term inside the square root greater than 0. Use the fact that the inverse cotangent function is a decreasing function to form the set of values of x.

Complete step by step answer:
Here we have been provided with the function $f\left(x\right)={\displaystyle \frac{1}{\sqrt{\ln \left({\cot }^{-1}x\right)}}}$ and we are asked to find the domain. First we need to understand the term domain.
Now, in mathematics the term domain is the set of values of x for which the function is defined. As we can see that we have a logarithmic term inside the square root symbol and which is present in the denominator. For a function to be defined its denominator must not be 0 and the term inside the square root must be greater than or equal to 0.
Since the function $\ln \left({\cot }^{-1}x\right)$ is in the denominator and also inside the square root so it must not be 0 or less than 0, so we have,
$\Rightarrow \ln \left({\cot }^{-1}x\right)>0$
We can 0 as $\ln 1$ and as the base of log on both the sides is e (natural log) so when we will remove the log function the direction of the inequality will remain same, so we get,
$$\begin{array}{rl}& \Rightarrow \ln \left({\cot }^{-1}x\right)>\ln 1\\ & \Rightarrow {\cot }^{-1}x>1\end{array}$$
1 in the R.H.S means 1 radian which is in the range $(0,\pi )$ that is the range of ${\cot }^{-1}x$ so we can write $1={\cot }^{-1}(\cot x)$, therefore we have,
$$\Rightarrow {\cot }^{-1}x>{\cot }^{-1}(\cot 1)$$
We know that the inverse cotangent function is a decreasing function so when we will remove that function from both the sides then the direction of the inequality will get reversed, so we get,
$$\Rightarrow x<\cot 1$$ ……… (1)
Now, the argument of the log function must be greater than 0 to be defined, so we get,
$\Rightarrow {\cot }^{-1}x>0$
We know that the range of ${\cot }^{-1}x$ is $(0,\pi )$ which is greater than 0 for all real values of x, so we get,
$\Rightarrow -\mathrm{\infty }<x<\mathrm{\infty }$ ……… (2)
We have to take those values of x such that both the relations (1) and (2) gets satisfied that means we have to take the intersection of the two sets, so we get,
$\begin{array}{rl}& \Rightarrow x\in (-\mathrm{\infty },\mathrm{\infty })\cap (-\mathrm{\infty },\cot 1)\\ & \therefore x\in (-\mathrm{\infty },\cot 1)\end{array}$

So, the correct answer is “Option d”.

Note: You must remember the range and domain values of all the trigonometric, inverse trigonometric, logarithmic, exponential functions. Note that if the base value of log is between 0 and 1 then the direction of inequality gets changed while removing the function. Remember the increasing and decreasing nature of common functions which determine the change of inequality sign while removing them.