If ∫_ -(π)/(2) ^ (π)/(2) (96 x^2 ^2 x)/((1+e^x)) d x=π(α π^2+β), α, β ∈ ℤ , then (α+β)^2 equals

Correct answer: D. 100 aligned & I=∫_0^ (π)/(2) (96 x^2 ^2 x)/(1+e^x) d x & 2 I=2 ∫_0^ (π)/(2) 96 x^2 ^2 x d x & I=96 ∫_0^ (π)/(2) x^2 ^2 x d x & =48 ∫_0^ (π)/(2) x^2(1+ 2 x) d x & =2 π^2+48(0-0)-48 ∫_0^ (π)/(2) x 2 x d x & =2 π^2-12 π+[0-0]=π(2 π^2-12) & =π(α π^2+β) & (α+β)^2=100 aligned

JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Definite Integration

If ∫_ -(π)/(2) ^ (π)/(2) (96 x^2 ^2 x)/((1+e^x)) d x=π(α π^2+β), α,...

\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

(\alpha+\beta)^2

equals

Options

A64
B196
C144
D100

Correct answer

D. 100

Step-by-step solution

\begin{aligned} & I=\int_0^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{1+e^x} d x \\ & 2 I=2 \int_0^{\frac{\pi}{2}} 96 x^2 \cos ^2 x d x \\ & I=96 \int_0^{\frac{\pi}{2}} x^2 \cos ^2 x d x \\ & =48 \int_0^{\frac{\pi}{2}} x^2(1+\cos 2 x) d x \\ & =2 \pi^2+48(0-0)-48 \int_0^{\frac{\pi}{2}} x \sin 2 x d x \\ & =2 \pi^2-12 \pi+[0-0]=\pi\left(2 \pi^2-12\right) \\ & =\pi\left(\alpha \pi^2+\beta\right) \\ & \Rightarrow(\alpha+\beta)^2=100\end{aligned}

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