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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Quadratic Equation

Let α and β be the roots of the equation x^ 2 +2 a x+(3 a+10)=0...

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

Options

  1. A(−∞,−115)\left(-\infty, \frac{-11}{5}\right)
  2. B(−∞,−2)∪(5,∞)(-\infty,-2) \cup(5, \infty)
  3. C(−∞,−3)(-\infty,-3)
  4. D(−∞,−115)∪(5,∞)\left(-\infty, \frac{-11}{5}\right) \cup(5, \infty)

Correct answer

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

Step-by-step solution

Let f(x)=x2+2ax+(3a+10)f(x) = x^2 + 2ax + (3a + 10). For α<1<β\alpha < 1 < \beta, since the leading coefficient is positive, we need f(1)<0f(1) < 0.
f(1)=1+2a+3a+10=5a+11<0f(1) = 1 + 2a + 3a + 10 = 5a + 11 < 0
a<−115a < -\frac{11}{5}
When f(1)<0f(1) < 0 with positive leading coefficient, the parabola is negative at x=1x = 1, so 1 lies strictly between the two real roots. Discriminant is automatically positive.
a∈(−∞,−115)a \in \left(-\infty, -\frac{11}{5}\right).
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