Question

The set of real values of x satisfying $lo{g}_{\frac{1}{2}}(x^2-6x+12)\ge -2$ is

• A. $(-\infty ,2]$
• B. $[2,4]$
• C. $[4,+\infty ]$
• D. $Noneofthese$

Answer 1

The correct option is B $[2,4]$

$lo{g}_{\frac{1}{2}}(x^2-6x+12)\ge -2$ ...(i)
For log to be defined, $x^2-6x+12>0$
$\Rightarrow (x-3{)}^2+3>0$, which is true $\forall xϵR.$
From (i), $x^2-6x+12\le (\frac{1}{2}{)}^{-2}$
$\Rightarrow x^2-6x+12\le 4\Rightarrow x^2-6x+8\le 0\Rightarrow (x-2)(x-4)\le 0\Rightarrow 2\le x\le 4;\therefore xϵ[2,4].$