The correct option is C (12,1)
For logx(4x+56−5x)<−1 to be defined,
4x+56−5x>0, x>0, x≠1⇒x∈(−54,65), x>0, x≠1∴x∈(0,65)−{1}
Case 1: When x∈(0,1)
logx4x+56−5x<−1⇒4x+56−5x>x−1⇒4x+56−5x−1x>0⇒4x2+10x−6x(6−5x)>0⇒2(2x−1)(x+3)x(5x−6)<0⇒x∈(−3,0)∪(12,65)
⇒x∈(12,1) ⋯(1) as x∈(0,1)
Case 2: When x∈(1,65)
logx4x+56−5x<−1
⇒4x+56−5x<x−1
⇒4x+56−5x−1x<0
⇒4x2+10x−6x(6−5x)<0
⇒2(2x−1)(x+3)x(5x−6)>0
⇒x∈(−∞,−3)∪(0,12)∪(65,∞)
⇒x∈ϕ ⋯(2) as x∈(1,65)
From (1) and (2),
x∈(12,1)