Let \alpha and \beta be the roots of the equation x^2+2 a x+(3 a+10)=0 such that \alpha< 1< \beta. Then the set of all possible values of a is : [21 Jan 2026 Shift 2]

Correct Answer is : (−∞,−115)

(-\infty, \frac{-11}{5}) \cup(5, \infty)

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