Question

Find domain of the function $f\left( x \right)=\sqrt{{{\log }_{x}}\left( \cos 2\pi x \right)}$?

Answer 1

We start solving the problem by applying the property that the function $\sqrt{g\left( x \right)}$ is valid if and only if the values of $g\left( x \right)\ge 0$ for the function ${{\log }_{x}}\left( \cos 2\pi x \right)$. We then recall the properties of logarithmic function {{\log }_{a}}b=\left\\{ \begin{array}{*{35}{l}} +ve,\text{ if }a > 1,b > 1 \\\ +ve,\text{ if }0 < a < 1,0 < b < 1 \\\ 0,\text{ if }b=1,a > 0\text{ }and\text{ }a\ne 1 \\\ -ve,\text{ if }a > 1,0 < b < 1 \\\ -ve,\text{ if }0 < a < 1,b > 1 \\\ \end{array} \right. and apply those which are feasible for ${{\log }_{x}}\left( \cos 2\pi x \right)\ge 0$. We then make the necessary calculations to find the domain for the feasible properties which in turn gives us the domain of the function $f\left( x \right)$.

Complete step-by-step answer :
According to the problem, we need to find the domain of the function $f\left( x \right)=\sqrt{{{\log }_{x}}\left( \cos 2\pi x \right)}$.
We know that the function $\sqrt{g\left( x \right)}$ is valid if and only if the values of $g\left( x \right)\ge 0$.
So, we have ${{\log }_{x}}\left( \cos 2\pi x \right)\ge 0$ ---(1).
We know that the properties of the logarithmic function ${{\log }_{a}}b$ is defined as follows:

+ve,\text{ if }a > 1,b > 1 \\\ +ve,\text{ if }0 < a < 1,0 < b < 1 \\\ 0,\text{ if }b=1,a > 0\text{ }and\text{ }a\ne 1 \\\ -ve,\text{ if }a > 1,0 < b < 1 \\\ -ve,\text{ if }0 < a < 1,b > 1 \\\ \end{array} \right.$$. So, the feasible conditions for equation (1) is (i) $x > 1\text{ and }\cos \left( 2\pi x \right) > 1$. (ii) $0 < x < 1\text{ and }0 < \cos \left( 2\pi x \right) < 1$. (iii) $x > 0\text{ and }x\ne 1\text{ and }\cos \left( 2\pi x \right)=1$. Let us first check the feasible condition (i), $x > 1\text{ and }\cos \left( 2\pi x \right) > 1$. We know that the range of the cosine function is $\left[ -1,1 \right]$. So, $\cos \left( 2\pi x \right) > 1$ is not possible for any value of ‘x’. Now, let us check the feasible condition (ii), $0 < x < 1\text{ and }0 < \cos \left( 2\pi x \right) < 1$. We know that if $0 < \cos x < 1$, then $\left( 4n-1 \right)\dfrac{\pi }{2} < x < \left( 4n+1 \right)\dfrac{\pi }{2},n\in I$. So, we get $$\left( 4n-1 \right)\dfrac{\pi }{2} < 2\pi x < \left( 4n+1 \right)\dfrac{\pi }{2},n\in I$$. $\Rightarrow \dfrac{4n-1}{4} < x < \dfrac{4n+1}{4},n\in I$. Let us substitute the value of integers. $\Rightarrow .....\cup \left( \dfrac{-1}{4},\dfrac{1}{4} \right)\cup \left( \dfrac{3}{4},\dfrac{5}{4} \right)\cup \left( \dfrac{7}{4},\dfrac{9}{4} \right)\cup .......$ ---(1). But we already have another condition for ‘x’. i.e., $0 < x < 1$. So, we neglect all the negative values and the values that were greater than or equal to 1 of ‘x’ in equation (1). So, the domain we get for this condition is $x\in \left( 0,\dfrac{1}{4} \right)\cup \left( \dfrac{3}{4},1 \right)$ ---(2). Now, let us check the feasible condition (iii), $x > 0\text{ and }x\ne 1\text{ and }\cos \left( 2\pi x \right)=1$. We know that if $\cos x=1$, then $2n\pi ,n\in I$. So, we get $2\pi x=2n\pi ,n\in I$. $\Rightarrow x=n,n\in I$. But we already have the condition $x > 0\text{ and }x\ne 1$. This gives us the domain for ‘x’ as $\left\\{ 2,3,4,...... \right\\}$ ---(3). From (2) and (3), we have found the domain of $f\left( x \right)$ as $x\in \left\\{ 2,3,4,5,..... \right\\}\cup \left( 0,\dfrac{1}{4} \right)\cup \left( \dfrac{3}{4},1 \right)$. We have found the domain of $f\left( x \right)$ as $x\in \left\\{ 2,3,4,5,..... \right\\}\cup \left( 0,\dfrac{1}{4} \right)\cup \left( \dfrac{3}{4},1 \right)$. **Note** : We should not include 0, 1 in the domain of $f\left( x \right)$ as the logarithmic function is not valid at that value of ‘x’. We should know that the both the conditions of $a$ and $b$ of ${{\log }_{a}}b$ should be valid while checking the feasibility conditions of $${{\log }_{x}}\left( \cos 2\pi x \right)\ge 0$$. We can see that the given problem contains a lot of calculation, so we need to perform each step carefully. Similarly, we can expect the problems to find the domain of $$f\left( x \right)=\sqrt{{{\log }_{x}}\left( \cos \left| 2\pi x \right| \right)}$$ and $$f\left( x \right)=\sqrt{{{\log }_{x}}\left( \cos \left[ 2\pi x \right] \right)}$$.