The correct option is A (2,3)∪(4,∞)
(x−2)x2−6x+8>1
For x>2
⇒(x2−6x+8)log10(x−2)>0
⇒(x−2)(x−4)log10(x−2)>0
Case 1:
(x−2)(x−4)>0 and log10(x−2)>0
⇒x∈((−∞,2)∪(4,∞))∩(3,∞)
⇒x∈(4,∞) ⋯(1)
Case 2:
⇒(x−2)(x−4)<0 and log10(x−2)<0
⇒x∈(2,4)∩(2,3)
⇒x∈(2,3) ⋯(2)
Now, (1)∪(2)
⇒x∈(2,3)∪(4,∞)