Verified Solution • 2025–26 Session moderate

QJEEM25Maf29a: JEE Main 2025 Mathematics - Class 11 JEEM25Maf29a Solution: If the set of all , for which the equation has no real root, is …

If the set of all a∈Ra \in \mathbf{R} , for which the equation 2x2+(a−5)x+15=3a2 x^2+(a-5) x+15=3 a has no real root, is the interval ( α,β\alpha, \beta ), and X=|x \in Z ; \alpha < x < \beta| , then ∑x∈Xx2\sum\limits_{x \in X} x^2 is equal to:

Multiple Choice Options

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

Step-by-Step Verified Solution

amp;(a−5)2−8(15−3a)lt;0amp;a2+14a+25−120lt;0amp;a2+14a−95lt;0amp;(a+19)(a−5)lt;0amp;a∈(−19,5)amp;∴−19lt;xlt;5amp;∴∑x∈Xx2=(12+22+….+42)+(12+22+…+182)amp;=4×5×96+18×19×376amp;=30+2109amp;=2139\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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