Q43 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
Let $\alpha, \beta$ be the roots of the quadratic equation $12 x^{2}-20 x+3 \lambda=0, \lambda \in \mathbf{Z}$. If $\frac{1}{2} \leqslant|\beta-\alpha| \leqslant \frac{3}{2}$, then the sum of all possible values of $\lambda$ is :
- A1
- B6
- Correct answerC3
- D4
Solution
For $12x^2-20x+3\lambda=0$: $\alpha+\beta=\frac{5}{3}$, $\alpha\beta=\frac{\lambda}{4}$.
$|\beta-\alpha|=\frac{\sqrt{25-9\lambda}}{3}$
Given $\frac{1}{2}\leq\frac{\sqrt{25-9\lambda}}{3}\leq\frac{3}{2}$
$\frac{9}{4}\leq 25-9\lambda\leq\frac{81}{4}$
From left: $\lambda\leq\frac{91}{36}\approx 2.53$
From right: $\lambda\geq\frac{19}{36}\approx 0.53$
Since $\lambda\in\mathbb{Z}$: $\lambda\in\{1,2\}$. Sum $=3$.
$|\beta-\alpha|=\frac{\sqrt{25-9\lambda}}{3}$
Given $\frac{1}{2}\leq\frac{\sqrt{25-9\lambda}}{3}\leq\frac{3}{2}$
$\frac{9}{4}\leq 25-9\lambda\leq\frac{81}{4}$
From left: $\lambda\leq\frac{91}{36}\approx 2.53$
From right: $\lambda\geq\frac{19}{36}\approx 0.53$
Since $\lambda\in\mathbb{Z}$: $\lambda\in\{1,2\}$. Sum $=3$.
Hence, the correct answer is option: 'C'.