JEE Main 2025 Jan 29 Shift 2, Mathematics Q1: Quadratic Equations

JEE Main2025Jan 29, Shift 2Mathematics

If the set of all $a \in R$ , for which the equation $2 x^{2} + (a - 5) x + 15 = 3 a$ has no real root, is the interval $(α, β)$ , and $X = {x \in Z : α < x < β}$ , then $x \in X \sum x^{2}$ is equal to

A
$2109$
B
$2129$
C
$2139$
D
$2119$

Solution

Rewrite the equation as $2 x^{2} + (a - 5) x + (15 - 3 a) = 0$ . No real root demands a negative discriminant:

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

$a^{2} - 10 a + 25 - 120 + 24 a < 0 \Rightarrow a^{2} + 14 a - 95 < 0$

$\Rightarrow (a + 19) (a - 5) < 0 \Rightarrow a \in (- 19, 5)$ , so $α = - 19$ , $β = 5$ .

Integers in $(- 19, 5)$ are $- 18, - 17, \dots, - 1, 0, 1, 2, 3, 4$ .

$x \in X \sum x^{2} = (1^{2} + 2^{2} + \dots + 1 8^{2}) + (1^{2} + 2^{2} + 3^{2} + 4^{2})$

$= \frac{18 \cdot 19 \cdot 37}{6} + \frac{4 \cdot 5 \cdot 9}{6} = 2109 + 30 = 2139$ .

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q1: Quadratic Equations

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