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JEE Main Mathematics Quadratic Equation 2026 JEE Main 2026 (21 January Shift 2)

JEE Main Mathematics Question (2026) — Solution

Question

Let and be the roots of the equation x^ 2 +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, )

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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