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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Quadratic Equation

Let α, β be the roots of the quadratic equation 12 x^ 2 -20 x+3...

Let α,β\alpha, \beta be the roots of the quadratic equation 12x2−20x+3λ=0,λ∈Z12 x^{2}-20 x+3 \lambda=0, \lambda \in \mathbf{Z}. If 12⩽∣β−α∣⩽32\frac{1}{2} \leqslant|\beta-\alpha| \leqslant \frac{3}{2}, then the sum of all possible values of λ\lambda is :

Options

  1. A1
  2. B6
  3. C3
  4. D4

Correct answer

C. 3

Step-by-step solution

For 12x2−20x+3λ=012x^2-20x+3\lambda=0: α+β=53\alpha+\beta=\frac{5}{3}, αβ=λ4\alpha\beta=\frac{\lambda}{4}.
∣β−α∣=25−9λ3|\beta-\alpha|=\frac{\sqrt{25-9\lambda}}{3}
Given 12≤25−9λ3≤32\frac{1}{2}\leq\frac{\sqrt{25-9\lambda}}{3}\leq\frac{3}{2}
94≤25−9λ≤814\frac{9}{4}\leq 25-9\lambda\leq\frac{81}{4}
From left: λ≤9136≈2.53\lambda\leq\frac{91}{36}\approx 2.53
From right: λ≥1936≈0.53\lambda\geq\frac{19}{36}\approx 0.53
Since λ∈Z\lambda\in\mathbb{Z}: λ∈{1,2}\lambda\in\{1,2\}. Sum =3=3.
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