Question

For the given function \[f\left( x \right) = \sqrt {{{\log }_x}\left( {\cos 2\pi x} \right)} \]. Domain of the function is:
A.\[\left( {0,\dfrac{1}{4}} \right) \cup \left( {\dfrac{3}{4},1} \right) \cup \left( {\dfrac{5}{4},\dfrac{7}{4}} \right) \cup \left\{ {x:x \in N,x \leqslant 2} \right\}\]
B.\[\left( {0,\dfrac{1}{4}} \right) \cup \left( {\dfrac{3}{4},1} \right) \cup \left\{ {x:x \in N,x \leqslant 2} \right\}\]
C.\[\left( {0,\dfrac{1}{4}} \right) \cup \left( {\dfrac{1}{4},1} \right) \cup \left\{ {x:x \in N,x \leqslant 2} \right\}\]
D.\[\left( {0,\dfrac{1}{4}} \right) \cup \left( {\dfrac{3}{4},\dfrac{3}{2}} \right) \cup \left\{ {x:x \in N,x \leqslant 2} \right\}\]

Answer 1

Hint: In this question, we need to determine the domain of the given function. For this, we will use the concept of domain such that it is the value of the function where the function is defined. Domain of a function is the set of all possible inputs for the function.

Complete step-by-step answer:
Given the function \[f\left( x \right) = \sqrt {{{\log }_x}\left( {\cos 2\pi x} \right)} \]
Here base of the function \[a = x\]
Number \[b = \cos 2\pi x\]
We know the logarithmic function is defined when\[b > 1\],\[0 < b < 1\]or \[b = 1\], but since the value of a cosine function lies in between \[ - 1 < \cos 2\pi x < 1\], hence we can say in this function only the case \[b = 1\] is valid and hence its base is defined when\[x \ne 0,1\]
Now since the cosine function is in square root so the square root is defined when \[{\log _x}\left( {\cos 2\pi x} \right) \geqslant 0\], now by taking logarithm inverse we get the function
\[
\Rightarrow \left( {\cos 2\pi x} \right) \geqslant {e^0} \\
\Rightarrow \left( {\cos 2\pi x} \right) \geqslant 1 \;
 \]
We can also write this as
\[2\pi x \geqslant 2n\pi \]
Hence we get \[x \geqslant n\]
So from the above cases we can conclude the base\[x \ne 0,1\], \[{\log _x}\left( {\cos 2\pi x} \right) \geqslant 0\], so we can say the domain of the function\[f\left( x \right) = \sqrt {{{\log }_x}\left( {\cos 2\pi x} \right)} \] will be equal to \[\left( {0,\dfrac{1}{4}} \right) \cup \left( {\dfrac{3}{4},1} \right) \cup \left\{ {x:x \in N,x \leqslant 2} \right\}\]
So, the correct answer is “Option B”.

Note: The domain of the function \[{\log _a}b\] will be positive when it’s both the base and number is \[a > 1,b > 1\]or when the value of the base and the number lies in between \[0 < a < 1,0 < b < 1\]and it will be equal to zero when then number \[b = 1\]and \[a \ne 0,1\].