Question

Find the domain of the function $f\left(x\right)=\log \left\{{\log }_{\left|\sin x\right|}\right(x^2-8x+23)-\frac{3}{{\log }_2\left|\sin x\right|}\}$

• A. $=(3,\pi )\cup (\pi ,\frac{3\pi }{2})\cup (\frac{3\pi }{2},5)$
• B. $=(3,\pi )\cup (\pi ,\frac{2\pi }{2})\cup (\frac{4\pi }{2},5)$
• C. $=(3,\pi )\cup (\pi ,\frac{3\pi }{2})\cup (\frac{6\pi }{2},5)$
• D. $=(3,\pi )\cup (\pi ,\frac{6\pi }{2})\cup (\frac{3\pi }{2},5)$

Answer 1

The correct option is A $=(3,\pi )\cup (\pi ,\frac{3\pi }{2})\cup (\frac{3\pi }{2},5)$

f(x) is defined if $\left({\log }_{\left|\sin x\right|}\right(x^2-8x+23)-\frac{3}{{\log }_{\left|\sin x\right|}\left|\sin x\right|})>0$
$\Rightarrow {\log }_{\left|\sin x\right|}(\frac{x^2-8x+23)}{8}>0)(as\frac{3}{{\log }_2\left|\sin x\right|}=\frac{{\log }_{2}8}{{\log }_2}\left|\sin x\right|)={\log }_{\left|\sin x\right|}8$
The is true, if
$\left|\sin x\right|\ne 0,1$ and $\frac{x^2-8x+23}{8}<1$
(as $\left|\sin x\right|<1\Rightarrow {\log }_{\left|\sin x\right|}a>0\Rightarrow a<1$)
Now, $\frac{x^2-8x+23}{8}<1$
$\Rightarrow x^2-8x+23<0$
$\Rightarrow x\in (3,5)-\{\pi ,\frac{3\pi }{4}\}$
Hence, domain $=(3,\pi )\cup (\pi ,\frac{3\pi }{2})\cup (\frac{3\pi }{2},5)$