The correct option is C (−∞,4)
f(x)=x2−2kx+k2+k−5
both roots are less than 5, then
(i) Discriminant ≥0
(ii)f(5)>0(iii)Sum of roots2<5
Hencei) (4k2−4(k2+k−5)≥04k2−4k2−4k+20≥04k≤20⇒k≤5
(ii) f(5)>0;25−10k+k2+k−5>0or k2−9k+20>0or k(k−4)−5(k−4)>0or(k−5)(k−4)>0⇒kϵ(−∞,4)⋃(5,∞)(iii) Sum of roots2=−b2a=2k2<5
The intersection of (i), (ii) & (iii) gives k ϵ(−∞,4).