Question

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

• A. $(\frac{1}{2},2)$
• B. $(1,\frac{5}{2})$
• C. $[\frac{1}{2},1)\cup (2,\frac{5}{2}]$
• D. None of these

Answer 1

The correct option is C $[\frac{1}{2},1)\cup (2,\frac{5}{2}]$

We get
$(x^2-3x+2)\le (\frac{\sqrt{3}}{2}{)}^2$
Or
$x^2-3x+2\le \frac{3}{4}$
Or
$x^2-3x+\frac{5}{4}\le 0$
Or
$(x-\frac{3}{2}{)}^2\le 1$
Or
$-1\le x-\frac{3}{2}\le 1$
$\frac{1}{2}\le x\le \frac{5}{2}$...(i)
Now
$x^2-3x+2>0$ for the domain of the function.
Or
$(x-1)(x-2)>0$
$x<1$ or $x>2$...(ii)
Hence from i and ii
$xϵ[\frac{1}{2},1)\cup (2,\frac{5}{2}]$.