The correct option is C 0
(x2−x−1)(x2−x−7)<−5
Let x2−x=t
Then, (t−1)(t−7)<−5
⇒t2−8t+12<0
⇒(t−2)(t−6)<0
⇒t∈(2,6)
⇒2<x2−x<6
Case I:x2−x>2
∴x2−x−2>0
⇒(x+1)(x−2)>0
∴x∈(−∞,−1)∪(2,∞) ⋯(1)
Case II:x2−x<6
∴x2−x−6<0
⇒(x+2)(x−3)<0
∴x∈(−2,3) ⋯(2)
From (1)∩(2),
x∈(−2,−1)∪(2,3)
There is no integral value of x in the solution set.