The correct option is C (1,52)
Given equation
x2−(a2−5a+5)x+(2a2−3a−4)=0
Let the roots of the equation be m,n.
So, m+n<1,mn<1
m+n<1
⇒a2−5a+5<1⇒a2−5a+4<0⇒(a−4)(a−1)<0⇒a∈(1,4) ⋯(1)
mn<1
⇒2a2−3a−4<1⇒2a2−3a−5<0⇒(a+1)(2a−5)<0⇒a∈(−1,52) ⋯(2)
From (1) and (2),
a∈(1,52)