
A function $f(x)$ is defined such that $f(x) = \sqrt {\left( {\left| {\tan x} \right| + \tan x} \right)} /\sqrt 3 x$ . What is the domain of this function?
A. $R$
B. $R - \left\{ {1/3} \right\}$
C. $R - \left\{ {n\pi + \dfrac{\pi }{2}:n \in {I^ + }} \right\}$
D. None of these
Answer
232.8k+ views
Hint: Find out the domains of all the functions used within the given function. The required values for which the function is defined for will be lying within the common values of all the domains, that is, their intersection. Therefore, it is important to note the domains of every function used in a particular function to get its range and domain.
Complete step by step solution:
The given function is:
$f(x) = \sqrt {\left( {\left| {\tan x} \right| + \tan x} \right)} /\sqrt 3 x$
As $\tan x$ function is defined for $x \notin n\pi + \dfrac{\pi }{2},n \in I$ , therefore the domain of the required function is narrowed down to that in the beginning itself.
Now, for the function $f(x)$ to be defined, the following points should be noted:
1. The value within the square root in the numerator of the given function should be greater than or equal to zero.
2. The value of the denominator should not be equal to zero, as then the function becomes meaningless.
Hence, keeping in mind the first point we get that:
$\left( {\left| {\tan x} \right| + \tan x} \right) \geqslant 0$
For $\tan x < 0$ , $\left( {\left| {\tan x} \right| + \tan x} \right) = - \tan x + \tan x = 0$
For $\tan x = 0$, $\left( {\left| {\tan x} \right| + \tan x} \right) = 0$
For $\tan x > 0$ , $\left( {\left| {\tan x} \right| + \tan x} \right) = \tan x + \tan x = 2\tan x$
Similarly, keeping in mind the second point, we get that $x \ne 0$ .
Hence, the domain of the given function will be:
$x \notin \left\{ {n\pi + \dfrac{\pi }{2}:n \in I} \right\} \cup \{ 0\} $
Thus, the given function is defined for $R - \left( {\left\{ {n\pi + \dfrac{\pi }{2}:n \in I} \right\} \cup \{ 0\} } \right)$ .
From the given options, the set $R - \left\{ {n\pi + \dfrac{\pi }{2}:n \in {I^ + }} \right\}$ is a subset of the obtained domain.
Hence, it can also be said that the given function is defined for $R - \left\{ {n\pi + \dfrac{\pi }{2}:n \in {I^ + }} \right\}$ .
Hence the correct option is C.
Note: All the functions have different domains with respect to them. The trigonometric functions are not defined at points where their values reach till infinity, making them meaningless. For example, $\tan x$ function is not defined for values like $x = - \dfrac{{3\pi }}{2}, - \dfrac{\pi }{2}, \dfrac{\pi }{2}, \dfrac{{3\pi }}{2}$ , etc.
Complete step by step solution:
The given function is:
$f(x) = \sqrt {\left( {\left| {\tan x} \right| + \tan x} \right)} /\sqrt 3 x$
As $\tan x$ function is defined for $x \notin n\pi + \dfrac{\pi }{2},n \in I$ , therefore the domain of the required function is narrowed down to that in the beginning itself.
Now, for the function $f(x)$ to be defined, the following points should be noted:
1. The value within the square root in the numerator of the given function should be greater than or equal to zero.
2. The value of the denominator should not be equal to zero, as then the function becomes meaningless.
Hence, keeping in mind the first point we get that:
$\left( {\left| {\tan x} \right| + \tan x} \right) \geqslant 0$
For $\tan x < 0$ , $\left( {\left| {\tan x} \right| + \tan x} \right) = - \tan x + \tan x = 0$
For $\tan x = 0$, $\left( {\left| {\tan x} \right| + \tan x} \right) = 0$
For $\tan x > 0$ , $\left( {\left| {\tan x} \right| + \tan x} \right) = \tan x + \tan x = 2\tan x$
Similarly, keeping in mind the second point, we get that $x \ne 0$ .
Hence, the domain of the given function will be:
$x \notin \left\{ {n\pi + \dfrac{\pi }{2}:n \in I} \right\} \cup \{ 0\} $
Thus, the given function is defined for $R - \left( {\left\{ {n\pi + \dfrac{\pi }{2}:n \in I} \right\} \cup \{ 0\} } \right)$ .
From the given options, the set $R - \left\{ {n\pi + \dfrac{\pi }{2}:n \in {I^ + }} \right\}$ is a subset of the obtained domain.
Hence, it can also be said that the given function is defined for $R - \left\{ {n\pi + \dfrac{\pi }{2}:n \in {I^ + }} \right\}$ .
Hence the correct option is C.
Note: All the functions have different domains with respect to them. The trigonometric functions are not defined at points where their values reach till infinity, making them meaningless. For example, $\tan x$ function is not defined for values like $x = - \dfrac{{3\pi }}{2}, - \dfrac{\pi }{2}, \dfrac{\pi }{2}, \dfrac{{3\pi }}{2}$ , etc.
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