b
(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 \((i),\) \({x^2} - 6x + 12 \le {\left( {{1 \over 2}} \right)^{ - 2}}\)

\( \Rightarrow \)\({x^2} - 6x + 12 \le 4\) \( \Rightarrow \) \({x^2} - 6x + 8 \le 0\)

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

\(\therefore x \in [2,\,4]\).