Q.

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 (-,-115)  
2 (-,-3)  
3 (-,-2)(5,)  
4 (-,-115)(5,)  

Ans.

(1)

  α<1<β

f(1)<0

1+2a+(3a+10)<0

5a+11<0

a<-115

  a(-,-115)