Question
The set of real values of x satisfying log12(x2−6x+12)≥−2 is
Answer: Option B
:
B
log12(x2−6x+12)≥−2 ...(i)
For log to be defined, x2−6x+12>0
⇒(x−3)2+3>0, which is true ∀xϵR.
From (i), x2−6x+12≤(12)−2
⇒x2−6x+12≤4⇒x2−6x+8≤0⇒(x−2)(x−4)≤0⇒2≤x≤4;∴xϵ[2,4].
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:
B
log12(x2−6x+12)≥−2 ...(i)
For log to be defined, x2−6x+12>0
⇒(x−3)2+3>0, which is true ∀xϵR.
From (i), x2−6x+12≤(12)−2
⇒x2−6x+12≤4⇒x2−6x+8≤0⇒(x−2)(x−4)≤0⇒2≤x≤4;∴xϵ[2,4].
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