Question

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

• A. $x\in (3,\pi )\cup (\pi ,\frac{3\pi }{2})\cup (\frac{3\pi }{2},5)$
• B. $x\in (3,\pi )\cup (\pi ,5)$
• C. $x\in (3,\frac{5\pi }{2})$
• D. None of these

Answer 1

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

$x\ne \frac{(2n+1)\pi }{2},n\pi n\in I$. The given inequality can be written as
$\frac{{\log }_2(x^2-8x+23)}{{\log }_2\left|\sin x\right|}>\frac{3}{{\log }_2\left|\sin x\right|}[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
As ${\log }_2\left|\sin x\right|<0$, we get
${\log }_2(x^2-8x+23)<3\Rightarrow x^2-8x+23<2^3=8$
$\Rightarrow x^2-8x+15<0\Rightarrow (x-5)(x-3)\Rightarrow 3<x<5$
Ans: A