Question

Let we have a function $f\left( x \right) = \sqrt {\cot \left( {5 + 3x} \right)\left( {\cot \left( 5 \right) + \cot \left( {3x} \right)} \right) - \sqrt {\cot 3x} + 1}$, find the domain of the function.
A). R - \left\\{ {\dfrac{{n\pi }}{3}} \right\\},n \in I
B). $\left( {2n + 1} \right)\dfrac{\pi }{6},n \in I$
C). R - \left\\{ {\dfrac{{n\pi }}{3},\dfrac{{n\pi - 5}}{3}} \right\\},n \in I
D). R - \left\\{ {\dfrac{{n\pi - 5}}{3}} \right\\},n \in I

Answer 1

Here, in the question, we are given a function in the form of $f\left( x \right)$ and asked to find the domain of the function. The domain refers to the set of all possible values of $x$ for which a function is defined. To find the domain, we will first simplify the given function to a possible extent and then check for the possible input values of $x$ to get the desired result.
Formula used:
$\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot B + \cot A}}$

Complete step-by-step solution:
Given, $f\left( x \right) = \sqrt {\cot \left( {5 + 3x} \right)\left( {\cot \left( 5 \right) + \cot \left( {3x} \right)} \right) - \sqrt {\cot 3x} + 1}$
To simplify, we will use trigonometric identity, $\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot B + \cot A}}$
Therefore, $f\left( x \right) = \sqrt {\dfrac{{\cot 5\cot 3x - 1}}{{\cot 3x + \cot 5}}\left( {\cot \left( 5 \right) + \cot \left( {3x} \right)} \right) - \sqrt {\cot 3x} + 1}$
Now, we can cancel the similar terms in numerator and denominator,
$\therefore f\left( x \right) = \sqrt {\cot 5\cot 3x - 1 - \sqrt {\cot 3x} + 1}$
Adding $1$ and $- 1$, we get,
$\therefore f\left( x \right) = \sqrt {\cot 5\cot 3x - \sqrt {\cot 3x} }$
To find the domain, Put $f\left( x \right) = 0$
\therefore \sqrt {\cot 5\cot 3x - \sqrt {\cot 3x} } = 0$$$$$$ Squaring both sides, we get, \cot 5\cot 3x - \sqrt {\cot 3x} = 0$Now, the above equation cannot be simplified further. But it is clearly visible that$x$is the only independent variable and$3x$is the only angle left with its cotangent. Now, we have to find those values for$x$such that$\cot 3x$is defined. Or, in other words, domain will be all the real numbers except for the numbers where$\cot 3x$is undefined. And we know, the cotangent of any angle is undefined at$n\pi ,n \in I$. It means$3x$should not be equal to$n\pi ,n \in I$. Hence, the domain for$f\left( x \right) = \sqrt {\cot \left( {5 + 3x} \right)\left( {\cot \left( 5 \right) + \cot \left( {3x} \right)} \right) - \sqrt {\cot 3x} + 1} $is$R - \left\{ {\dfrac{{n\pi }}{3}} \right\},n \in I$**Hence the correct option is A.$R - \left\{ {\dfrac{{n\pi }}{3}} \right\},n \in I$$ is the correct option.**

Note: Generally, we determine the domain of the function by looking for those values of the independent variable (usually$x$) which we are permitted to use. There are two key-points which should be kept in mind while finding the domain of any function: (i) There should not be negative values under a square root sign, and, (ii) There should not be zero in the denominator of the function.