(b) ${\log _{1/2}}({x^2} – 6x + 12) \ge – 2$…..$(i)$

For log to be defined, ${x^2} – 6x + 12 > 0$

$ \Rightarrow $${(x – 3)^2} + 3 > 0$, which is true $\forall x \in R$.

From $(i),$ ${x^2} – 6x + 12 \le {\left( {{1 \over 2}} \right)^{ – 2}}$

$ \Rightarrow $${x^2} – 6x + 12 \le 4$ $ \Rightarrow $ ${x^2} – 6x + 8 \le 0$

$ \Rightarrow $ $(x – 2)(x – 4) \le 0$ $ \Rightarrow $ $2 \le x \le 4$;

$\therefore x \in [2,\,4]$.