If the set of all a ∈ 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 ∈ Z: α x β\ , then ∑_ x ∈ X x^2 is equal to :

Correct answer: D. 2139 aligned & (a-5)^2-8(15-3 a) < 0 & a^2+14 a+25-120 < 0 & a^2+14 a-95 < 0 & (a+19)(a-5) < 0 & a ∈(-19,5) & -19 < x < 5 & ∑_ x ∈ X x^2=(1^2+2^2+ .+4^2)+(1^2+2^2+ +18^2) & =(4 × 5 × 9)/(6)+(18 × 19 × 37)/(6) & =30+2109 & =2139 aligned

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Quadratic Equation

If the set of all a ∈ R , for which the equation 2 x^2+(a-5) x+15=3...

If the set of all

\mathrm{a} \in \mathbf{R}

, for which the equation

2 x^2+(a-5) x+15=3 \mathrm{a}

has no real root, is the interval

(\alpha, \beta)

, and

X=\{x \in Z: \alpha \lt x \lt \beta\}

, then

\sum_{x \in X} x^2

is equal to :

Options

A2109
B2129
C2119
D2139

Correct answer

D. 2139

Step-by-step solution

\begin{aligned} & (a-5)^2-8(15-3 a) < 0 \\ & a^2+14 a+25-120 < 0 \\ & a^2+14 a-95 < 0 \\ & (a+19)(a-5) < 0 \\ & a \in(-19,5) \\ & \therefore-19 < x < 5 \\ & \therefore \sum_{x \in X} x^2=\left(1^2+2^2+\ldots .+4^2\right)+\left(1^2+2^2+\ldots+18^2\right) \\ & =\frac{4 \times 5 \times 9}{6}+\frac{18 \times 19 \times 37}{6} \\ & =30+2109 \\ & =2139\end{aligned}

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