Question

The solution set of ${\log }_{|\sin x|}(x^2-8x+23)>\frac{3}{{\log }_2|\sin x|}$ contains

• A. $(3,\pi )$
• B. $(\pi ,\frac{3\pi }{2})$
• C. $(\frac{3\pi }{2},\frac{7\pi }{4}]$
• D. $(\pi ,5)-\left\{\frac{3\pi }{2}\right\}$

Answer 1

Answer: (A), (B), (D)

$(\pi ,5)-\left\{\frac{3\pi }{2}\right\}$

$x^2-8x+23>0$
$\Rightarrow (x-4{)}^2+7>0\text{true}\forall x\in R$
And $|\sin x|\ne 0,1$
$\Rightarrow \sin x\ne 0,\pm 1$
$\Rightarrow x\ne \dots ,-2\pi ,-\pi ,0,\pi ,2\pi ,\dots$
and $x\ne \dots ,\frac{-3\pi }{2},\frac{-\pi }{2},0,\frac{\pi }{2},\frac{3\pi }{2},\dots$

Now, $\frac{{\log }_2(x^2-8x+23)}{{\log }_2|\sin x|}>\frac{3}{{\log }_2|\sin x|}$
$\Rightarrow {\log }_2(x^2-8x+23)<3(\because {\log }_2|\sin x|<0)$
$\Rightarrow x^2-8x+23<2^3\Rightarrow x^2-8x+15<0\Rightarrow (x-5)(x-3)<0\Rightarrow x\in (3,5)$
$\therefore x\in (3,5)-\{\pi ,\frac{3\pi }{2}\}$