The set of real values of x satisfying log12(x2−6x+12)≥−2 is
A
(−∞,2]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
[2,4]
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
[4,+∞]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Noneofthese
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B[2,4] 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].