Verified Solution • 2025–26 Session moderate QJEEM25Ma6f94: JEE Main 2025 Mathematics - Class 12 JEEM25Ma6f94 Solution: If , where denotes the greatest integer function, then is equal … If 24∫0π4[sin∣4x−π12∣+[2sinx]]dx=2π+α24 \int\limits_0^{\frac{\pi}{4}} \bigg[\sin \left| 4x - \frac{\pi}{12} \right| + [2 \sin x] \bigg] dx = 2\pi + \alpha240∫4π[sin4x−12π+[2sinx]]dx=2π+α , where [⋅][\cdot][⋅] denotes the greatest integer function, then α\alphaα is equal to ________. Step-by-Step Verified Solution =amp;24∫0π48−sin(4x−π12)+∫π/48π/4sin(4x−π12)amp;+∫0π6[0]dx+∫π/6π/4[2sinx]dx\begin{aligned}= & 24 \int_0^{\frac{\pi}{48}}-\sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)+\int_{\pi / 48}^{\pi / 4} \sin \left(4 \mathrm{x}-\frac{\pi}{12}\right) \\& +\int_0^{\frac{\pi}{6}}[0] \mathrm{dx}+\int_{\pi / 6}^{\pi / 4}[2 \sin \mathrm{x}] \mathrm{dx}\end{aligned}=amp;24∫048π−sin(4x−12π)+∫π/48π/4sin(4x−12π)amp;+∫06π[0]dx+∫π/6π/4[2sinx]dx amp;=24[(1−cosπ12)4−(−cosπ12−1)4]+π4−π6amp;=24(12+π12)=2π+12amp;α=12\begin{aligned}& =24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right]+\frac{\pi}{4}-\frac{\pi}{6} \\& =24\left(\frac{1}{2}+\frac{\pi}{12}\right)=2 \pi+12 \\& \alpha=12\end{aligned}amp;=24[4(1−cos12π)−4(−cos12π−1)]+4π−6πamp;=24(21+12π)=2π+12amp;α=12 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring