JEE Mathematics 2025 Question with Solution

Given: The equation is

$$2x^2 + (a - 5)x + 15 = 3a$$

So, in standard form,

$$2x^2 + (a - 5)x + (15 - 3a) = 0$$

Find: If the set of all $a$ for which this quadratic has no real root is $(\alpha, \beta)$, then find

$$\sum_{x \in X} x^2$$

where

$$X = \{ x \in \mathbb{Z} : \alpha < x < \beta \}$$

For a quadratic equation in $x$ to have no real roots, its discriminant must be negative:

$$D = b^2 - 4ac < 0$$

Here,

$$a_1 = 2, \quad b_1 = a - 5, \quad c_1 = 15 - 3a$$

Therefore,

$$(a - 5)^2 - 4(2)(15 - 3a) < 0$$ $$a^2 - 10a + 25 - 8(15 - 3a) < 0$$ $$a^2 - 10a + 25 - 120 + 24a < 0$$ $$a^2 + 14a - 95 < 0$$

Factorizing,

$$(a + 19)(a - 5) < 0$$

Hence,

$$-19 < a < 5$$

So,

$$(\alpha, \beta) = (-19, 5)$$

Now,

$$X = \{ x \in \mathbb{Z} : -19 < x < 5 \} = \{-18, -17, \ldots, -1, 0, 1, 2, 3, 4\}$$

Thus,

$$\sum_{x \in X} x^2 = (1^2 + 2^2 + \cdots + 18^2) + (1^2 + 2^2 + 3^2 + 4^2)$$

Using

$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$

we get

$$\sum_{k=1}^{18} k^2 = \frac{18 \times 19 \times 37}{6} = 2109$$

and

$$\sum_{k=1}^{4} k^2 = \frac{4 \times 5 \times 9}{6} = 30$$

Therefore,

$$\sum_{x \in X} x^2 = 2109 + 30 = 2139$$

So the correct option is C.

Given:

$$2x^2 + (a - 5)x + 15 = 3a$$

Find: The value of

$$\sum_{x \in X} x^2$$

after determining the interval of all real values of $a$ for which the equation has no real root.

First rewrite the equation as

$$2x^2 + (a - 5)x + (15 - 3a) = 0$$

This is a quadratic in $x$. For no real root,

$$D < 0$$

So,

$$(a - 5)^2 - 8(15 - 3a) < 0$$

Expanding,

$$a^2 + 25 - 10a - 120 + 24a < 0$$ $$a^2 + 14a - 95 < 0$$

Now solve the corresponding quadratic equation:

$$a^2 + 14a - 95 = 0$$ $$a = \frac{-14 \pm \sqrt{14^2 - 4(1)(-95)}}{2}$$ $$a = \frac{-14 \pm \sqrt{196 + 380}}{2}$$ $$a = \frac{-14 \pm \sqrt{576}}{2}$$ $$a = \frac{-14 \pm 24}{2}$$

So the roots are

$$a = -19 \quad \text{and} \quad a = 5$$

Since the coefficient of $a^2$ is positive, the inequality

$$a^2 + 14a - 95 < 0$$

holds between the roots. Hence,

$$-19 < a < 5$$

Therefore,

$$(\alpha, \beta) = (-19, 5)$$

Now the set

$$X = \{ x \in \mathbb{Z} : -19 < x < 5 \}$$

becomes

$$X = \{-18, -17, -16, \dots, 3, 4\}$$

Then,

$$\sum_{x \in X} x^2 = (-18)^2 + (-17)^2 + \dots + 4^2$$

The squares of negative integers are positive, so this is

$$\sum_{k=1}^{18} k^2 + \sum_{k=1}^{4} k^2$$

Now,

$$\sum_{k=1}^{18} k^2 = \frac{18 \cdot 19 \cdot 37}{6} = 2109$$

And,

$$\sum_{k=1}^{4} k^2 = 1 + 4 + 9 + 16 = 30$$

Hence,

$$\sum_{x \in X} x^2 = 2109 + 30 = 2139$$

Therefore, the required value is $2139$, so the correct option is C.