JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 7)
If the set of all $a \in \mathbf{R}$, for which the equation $2 x^2+(a-5) x+15=3 a$ has no real root, is the interval ( $\alpha, \beta$ ), and $X=|x \in Z ; \alpha < x < \beta|$, then $\sum\limits_{x \in X} x^2$ is equal to:
2139
2119
2109
2129
Explanation
$$\begin{aligned}
& (a-5)^2-8(15-3 a)<0 \\
& a^2+14 a+25-120<0 \\
& a^2+14 a-95<0 \\
& (a+19)(a-5)<0 \\
& a \in(-19,5) \\
& \therefore-19< x<5 \\
& \therefore \sum_{x \in X} x^2=\left(1^2+2^2+\ldots .+4^2\right)+\left(1^2+2^2+\ldots+18^2\right) \\
& =\frac{4 \times 5 \times 9}{6}+\frac{18 \times 19 \times 37}{6} \\
& =30+2109 \\
& =2139
\end{aligned}$$
Comments (0)
