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)

JEE Main 21 Jan 2026 Shift 2 Paper Solved Paper

Section: Mathematics

Question : 2 of 75

Marks: +1, -0

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 :

[21 Jan 2026 Shift 2]

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