The correct option is C x∈(−1,0)∪(1,3) if −2<x<0 and 1<x<∞
For log to be defined,
x+3>0, x+3≠1 and x2−x>0
⇒x>−3, x≠−2 and x∈(−∞,0)∪(1,∞)
⇒x∈(−3,−2)∪(−2,0)∪(1,∞)
Now, log(x+3)(x2−x)<1
If x+3>1
i.e., x>−2
i.e., x∈(−2,0)∪(1,∞) ⋯(1)
then x2−x<x+3
⇒x2−2x−3<0⇒(x−3)(x+1)<0
⇒x∈(−1,3) ⋯(2)
Hence, from (1) and (2),
x∈(−1,0)∪(1,3)
If 0<x+3<1
i.e., x∈(−3,−2) ⋯(3)
then x2−x>x+3
⇒x2−2x−3>0⇒(x−3)(x+1)>0⇒x∈(−∞,−1)∪(3,∞) ⋯(4)
Hence, from (3) and (4),
x∈(−3,−2)