Question

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

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

Answer 1

The correct option is B

[2, 4]

${\log }_{0.5}\left(x^2-6x+12\right)\ge -2$
--------(1)

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 in equation (1) lies between 0 to 1. So, given logarithm is a decreasing function.

Then inequality is equivalent to

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

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

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

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

So, x should be greater than 2 and less than 4

x ∈ [2,4]