Question

The solution set of the inequality
${\log }_{\sin (\pi /3)}(x^2-3x+2)\ge 2$ is

• A. $\left(\frac{1}{2},2\right)$
• B. $\left[\frac{1}{2},2\right]$
• C. $[\frac{1}{2},1)\cup (2,\frac{5}{2}]$
• D. $(\frac{1}{2},\frac{5}{2})$

Answer 1

Answer: (C)

We have,
${\log }_{\left(\sin \frac{\pi }{3}\right)}\left(x^2-3x+2\right)\ge 2$
$\Rightarrow {\log }_{\frac{\sqrt{3}}{2}}\left(x^2-3x+2\right)\ge 2$
$\Rightarrow \left(x^2-3x+2\right)\le {\left(\frac{\sqrt{3}}{2}\right)}^2$
and $x^2-3x+2>0$
$\Rightarrow x^2-3x+2\le \frac{3}{4}$ and $x^2-2x-x+2>0$
$\Rightarrow 4x^2-12x+5\le 0$ and $(x-2)(x-1)>0$
$\Rightarrow (2x-5)(2x-1)\le 0$ and $x\in (-\infty ,1)\cup (2,\infty )$
$x\in \left[\frac{1}{2},\frac{5}{2}\right]$ and $x\in (-\infty ,1)\cup (2,\infty )$
$\therefore x\in \left[\frac{1}{2},1\right)\cup \left(2,\frac{5}{2}\right]$