Question

The set of solution in ${\log }_{\frac{1}{2}}\left(x^2-6x+12\right)\ge -2is$

Answer 1

Given ${\log }_{0.5}\left(x^2-6x+12\right)\ge -2$

For log to be defined $x^2-6x+12>0$

$x^2-6x+9+3>0$

$(x-3{)}^2+3>0$

We see that for any value of $x$. This is always true.

Since, base of log lies between $0\text{to}1$. So, given logarithm is a decreasing function.

Then inequality is equivalent to

So, $x^2-6x+12\le {\left(\frac{1}{2}\right)}^{-2}$

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

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

$\Rightarrow (x-2)(x-4)\le 0$

Using wavy curve method,
i.e $x\in [2,4]$

Hence the correct answer is Option B.