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JEE Main202621 January 2026Evening ShiftMathematicsQuadratic EquationActual

Let and be the roots of the equation x²+2 a x+(3 a+10)=0 such that <1< . Then the set of all possible values of a is :

Options

  1. A(- , -11 5 )
  2. B(- ,-2) (5, )
  3. C(- ,-3)
  4. D(- , -11 5 ) (5, )

Correct answer

A. (- , -11 5 )

Step-by-step solution

Let f(x) = x^2 + 2ax + (3a + 10) . For f(1) = 1 + 2a + 3a + 10 = 5a + 11 a When f(1) a (- , - 11 5 ) .

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