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MathsJEE Main 21 Jan 2026, Shift 2·Q9·MCQ·Easy–moderate

Quadratic Equations · Location of roots

Let α\alpha and β\beta be the roots of equation x2+2ax+(3a+10)=0x^2 + 2ax + (3a + 10) = 0 such that α<1<β\alpha < 1 < \beta. Then the set of all possible values of aa is:
  1. (1)(−∞,−115)∪(5,∞)\left(-\infty, \dfrac{-11}{5}\right) \cup (5, \infty)
  2. (2)(−∞,−2)∪(5,∞)(-\infty, -2) \cup (5, \infty)
  3. (3)(−∞,−3)(-\infty, -3)
  4. (4)(−∞,−115)\left(-\infty, \dfrac{-11}{5}\right)

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