Let α and β be the roots of the equation x2+2ax+(3a+10)=0 such that α<1<β. Then the set of all possible values of a is : [2026]
(1)
∵ α<1<β
f(1)<0
⇒1+2a+(3a+10)<0
⇒5a+11<0
a<-115
∴ a∈(-∞,-115)