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JEE Main 29-Jan-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 29-Jan-2025 Paper - Shift 24 Marks
Question
If $24 \int_0^{\frac{\pi}{4}}\left(\sin \left|4 x-\frac{\pi}{12}\right|+[2 \sin x]\right) d x=2 \pi+\alpha$, where [.] denotes the greatest integer function, then $\alpha$ is equal to _________
✓

Answer

(12)
$\begin{array}{l}Sol. = 24 \int_0^{\frac{\pi}{48}}-\sin \left(4 x-\frac{\pi}{12}\right)+\int_{\pi / 48}^{\pi / 4} \sin \left(4 x-\frac{\pi}{12}\right) \\ +\int_0^{\frac{\pi}{6}}[0] dx +\int_{\pi / 6}^{\pi / 4}[2 \sin x] dx \\ = 24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right]+\frac{\pi}{4}-\frac{\pi}{6} \\ = 24\left(\frac{1}{2}+\frac{\pi}{12}\right)=2 \pi+12 \\ \alpha=12\end{array}$

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