Question

The domain of $f\left(x\right)=\sqrt{{\log }_x\left(\cos 2\pi x\right)},$ is

• A. $(0,\frac{1}{4})\cup (\frac{3}{4},1)\cup N-\left\{1\right\}$
• B. $(0,\frac{1}{4})\cup (\frac{3}{4},1)$
• C. $(0,\frac{1}{4})\cup (\frac{3}{4},1)\cup N$
• D. $(0,1)\cup N-\left\{1\right\}$

Answer 1

The correct option is A $(0,\frac{1}{4})\cup (\frac{3}{4},1)\cup N-\left\{1\right\}$

For,
$f\left(x\right)=\sqrt{{\log }_x\left(\cos 2\pi x\right)}$ to be defined,
$\left(i\right){\log }_x\left(\cos 2\pi x\right)\ge 0$
$\left(ii\right)x>0,$ $x\ne 1,\cos 2\pi x>0$

For $x\in (0,1)\Rightarrow 2\pi x\in (0,2\pi )$
and ${\log }_x\left(\cos 2\pi x\right)\ge 0$
$\Rightarrow \cos 2\pi x\le 1\therefore 0<\cos 2\pi x\le 1\Rightarrow 2\pi x\in (0,\frac{\pi }{2})\cup (\frac{3\pi }{2},2\pi )\Rightarrow x\in (0,\frac{1}{4})\cup (\frac{3}{4},1)$

For $x\in (1,\infty )\Rightarrow 2\pi x\in (2\pi ,\infty )$
$\Rightarrow {\log }_x\left(\cos 2\pi x\right)\ge 0$
$\Rightarrow \cos 2\pi x\ge 1$ but $\cos 2\pi x\le 1$
$\Rightarrow \cos \left(2\pi x\right)=1\Rightarrow 2\pi x\in \{2n\pi ,n>1,n\in N\}\Rightarrow x\in N-\left\{1\right\}$