The set of real values of x satisfying log1/2(x2−6x+12)≥−2 is
A
(−∞,2]
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B
[2,4]
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C
[4,∞)
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D
none of these
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Solution
The correct option is B[2,4] As x2−6x+12>0 for all x. Therefore, log1/2(x2−6x+12) is defined for all x∈R Now, log1/2(x2−6x+12)≥−2 ⇒x2−6x+12≤(12)−2 ⇒x2−6x+8≤0⇒(x−2)(x−4)≤0⇒x∈[2,4]