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Let α and β be the roots of the equation x^2+2 a x+(3 a+10)=0 such that α<1<β. Then the set of all possible values of a is :

JEE Main 2026 — Mathematics Algebra

2026mcqmedium

Let α\alphaα and β\betaβ be the roots of the equation x2+2ax+(3a+10)=0x^{2}+2 a x+(3 a+10)=0x2+2ax+(3a+10)=0 such that α<1<β\alpha<1<\betaα<1<β. Then the set of all possible values of aaa is :

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

  1. A

    (−∞,−115)\left(-\infty, \frac{-11}{5}\right)(−∞,5−11​)

  2. B

    (−∞,−2)∪(5,∞)(-\infty,-2) \cup(5, \infty)(−∞,−2)∪(5,∞)

  3. C

    (−∞,−3)(-\infty,-3)(−∞,−3)

  4. D

    (−∞,−115)∪(5,∞)\left(-\infty, \frac{-11}{5}\right) \cup(5, \infty)(−∞,5−11​)∪(5,∞)

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