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If the set of all \(\mathrm{a} \in \mathbf{R}\), for which the equation \(2 x^2+(a-5) x+15=3a\) has no real root, is the…

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

[JEE Main 2025, 29 Jan (Shift 2)]

a

2129

b

2119

c

2109

d

2139

✓ Correct answer: d)

2139

Explanation

Given quadratic equation does not have real roots.

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

so \(D<0\)

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

\(a^2+14 a+25-120<0\)

\(a^2+14 a-95<0\)

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

\(a \in(-19,5)\)

\(\therefore-19<x<5\)

\( \therefore \sum_{x \in X} x^2= \left(1^2+2^2+\ldots+4^2\right) +\left(1^2+2^2+\ldots+18^2\right)\)

\(=\frac{4 \times 5 \times 9}{6}+\frac{18 \times 19 \times 37}{6}\)

\(=2139\)

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