Verified Solution • 2025–26 Session moderate

QJEEM25Ma2b49: JEE Main 2025 Mathematics - Class 12 JEEM25Ma2b49 Solution: If , then equals

If ∫−π2π296x2cos⁡2x(1+ex)dx=π(απ2+β),α,β∈Z\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} \mathrm{d} x=\pi\left(\alpha \pi^2+\beta\right), \alpha, \beta \in \mathbb{Z} , then (α+β)2(\alpha+\beta)^2 equals

Multiple Choice Options

  • A 196
  • B 100
  • C 64
  • D 144

Step-by-Step Verified Solution

\int_\limits{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} d x \quad (Apply King Property)

amp;∫0π296x2cos⁡2x=48∫0π2x2(1+cos⁡2x)dxamp;48[(x33)0π/2+∫0π2x12cos⁡II2xdx]amp;⇒ On solving π(2π2−12)amp;⇒α=2,β=−12amp;⇒(α+β)2=100\begin{aligned}& \int_0^{\frac{\pi}{2}} 96 x^2 \cos ^2 x=48 \int_0^{\frac{\pi}{2}} x^2(1+\cos 2 x) d x \\& 48\left[\left(\frac{x^3}{3}\right)_0^{\pi / 2}+\int_0^{\frac{\pi}{2}} x_1^2 \cos _{I I} 2 x d x\right] \\& \Rightarrow \text { On solving } \pi\left(2 \pi^2-12\right) \\& \Rightarrow \alpha=2, \beta=-12 \\& \Rightarrow(\alpha+\beta)^2=100\end{aligned}

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