The set of real values of x satisfying {log _ | vedclass

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Question

(b) ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$&hellip;..$(i)$

For log to be defined, ${x^2} - 6x + 12 > 0$

$ \Rightarrow $${(x - 3)^2} + 3 >

Vedclass · Accepted Answer

$[2,\,4]$

Vedclass · Answer

(b) ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$&hellip;..$(i)$

For log to be defined, ${x^2} - 6x + 12 > 0$

$ \Rightarrow $${(x - 3)^2} + 3 > 0$, which is true $\forall x \in R$.

From $(i),$ ${x^2} - 6x + 12 \le {\left( {{1 \over 2}} ight)^{ - 2}}$

$ \Rightarrow $${x^2} - 6x + 12 \le 4$ $ \Rightarrow $ ${x^2} - 6x + 8 \le 0$

$ \Rightarrow $ $(x - 2)(x - 4) \le 0$ $ \Rightarrow $ $2 \le x \le 4$;

$	herefore x \in [2,\,4]$.

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

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