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

2

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3

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4

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none of these

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Solution

The correct option is C none of these
${log}_{1 / 2} (x^{2} - 6 x + 12) \geq 2$
$\Rightarrow x^{2} - 6 x + 12 \leq (1 / 2)^{2} = 1 / 4$
$\Rightarrow 4 x^{2} - 24 x + 47 \leq 0$
Discriminant of above quadratic is, $D = \sqrt{24^{2} - 16 \times 47} = - 176 < 0$
Hence given inequality is not possible for any $x \in R$

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