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

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(B) Given the inequality: $\log _{1/2}(x^2 - 6x + 12) \ge -2$ $(i)$For the logarithm to be defined,the argument must be positive: $x^2 - 6x + 12 > 0$.

Vedclass · Accepted Answer

$[ 2 , 4 ]$

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(B) Given the inequality: $\log _{1/2}(x^2 - 6x + 12) \ge -2$ $(i)$For the logarithm to be defined,the argument must be positive: $x^2 - 6x + 12 > 0$.Since the discriminant $D = (-6)^2 - 4(1)(12) = 36 - 48 = -12  0$ for all $x \in \mathbb{R}$.Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequality sign reverses when we remove the logarithm:$x^2 - 6x + 12 \le (1/2)^{-2}$$x^2 - 6x + 12 \le 4$$x^2 - 6x + 8 \le 0$$(x - 2)(x - 4) \le 0$Solving the quadratic inequality,we get $2 \le x \le 4$.Thus,the solution set is $x \in [2, 4]$.

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

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