Question

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

• A. $(-\infty ,2]$
• B. $[2,4]$
• C. $[4,+\infty )$
• D. none of these

Answer 1

Answer: (B)

$[2,4]$

${\log }_{\frac{1}{2}}(x^2-6x+12)\ge -2$

$x^2-6x+12>0$

$\Rightarrow x^2-6x+12\le {\left(\frac{1}{2}\right)}^{-2}$, since base of log is less than $1$.

$\Rightarrow x^2-6x+12\le 4$
$\Rightarrow x^2-6x+8\le 0$

$\Rightarrow (x-4)(x-2)\le 0$
$\Rightarrow x\in [2,4]$
Ans: B