The correct options are
A x∈(−3,−2)
C x∈(1,3)
D x∈(−1,0)
logx+3(x2−x)<1
For the inequality to be defined,
x2−x>0 and x+3>0
⇒x(x−1)>0 and x>−3
⇒x∈(−∞,0)∪(1,∞) and x∈(−3,∞)
∴x∈(−3,0)∪(1,∞)
Case 1: When x+3>1 i.e., x∈(−2,0)∪(1,∞)
x2−x<x+3⇒x2−2x−3<0⇒(x−3)(x+1)<0⇒x∈(−1,3)
∴x∈(−1,0)∪(1,3) ⋯(1)
Case 2: When 0<x+3<1 i.e., x∈(−3,−2)
x2−x>x+3
⇒x2−2x−3>0⇒(x−3)(x+1)>0
⇒x∈(−∞,−1)∪(3,∞)
∴x∈(−3,−2) ⋯(2)
Hence, from (1) and (2),
x∈(−3,−2)∪(−1,0)∪(1,3)