JEE Mathematics 2026 Question with Solution

Given: The quadratic equation is $12x^2 - 20x + 3\lambda = 0$ with roots $\alpha$ and $\beta$, where $\lambda \in \mathbb{Z}$.

Find: The sum of all possible integer values of $\lambda$ such that

$$\frac{1}{2} \le |\beta-\alpha| \le \frac{3}{2}$$

For a quadratic equation $ax^2+bx+c=0$, the difference of roots is

$$|\beta-\alpha|=\frac{\sqrt{b^2-4ac}}{|a|}$$

Here,

$$a=12,\quad b=-20,\quad c=3\lambda$$

So,

$$|\beta-\alpha| =\frac{\sqrt{(-20)^2-4(12)(3\lambda)}}{12} =\frac{\sqrt{400-144\lambda}}{12}$$

Now apply the given inequality:

$$\frac{1}{2}\le \frac{\sqrt{400-144\lambda}}{12}\le \frac{3}{2}$$

Multiplying throughout by $12$,

$$6\le \sqrt{400-144\lambda}\le 18$$

Squaring all parts,

$$36\le 400-144\lambda\le 324$$

Solve the two inequalities separately.

From

$$36\le 400-144\lambda$$

we get

$$144\lambda\le 364$$

that is,

$$\lambda\le \frac{364}{144}$$

From

$$400-144\lambda\le 324$$

we get

$$-144\lambda\le -76$$

so

$$\lambda\ge \frac{76}{144}$$

Hence,

$$\frac{76}{144}\le \lambda\le \frac{364}{144}$$

Since $\lambda\in\mathbb{Z}$, the possible values are $\lambda=1,2$.

Checking these values in the given condition:

• For $\lambda=1$,

$$|\beta-\alpha|=\frac{\sqrt{400-144}}{12}=\frac{16}{12}=\frac{4}{3}$$

which satisfies the inequality.

• For $\lambda=2$,

$$|\beta-\alpha|=\frac{\sqrt{400-288}}{12}=\frac{\sqrt{112}}{12}=\frac{2\sqrt{7}}{6}$$

which also lies between $\frac{1}{2}$ and $\frac{3}{2}$.

Therefore, the possible values are $1$ and $2$, so their sum is

$$1+2=3$$

the solution concludes with final answer $1$, but the worked inequality includes $\lambda=1,2$. Since option $3$ is present, the defensible correct option from the working is D.