Hard MCQ +4 / -1 PYQ · JEE Mains 2025

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:

  1. A <p>2139</p> Correct answer
  2. B <p>2119</p>
  3. C <p>2109</p>
  4. D <p>2129</p>

Solution

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

About this question

Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane

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