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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Definite Integration

If 24 ∫_0^ (π)/(4) ( |4 x-(π)/(12)|+[2 x]) d x=2 π+α , where [·]...

If 24∫0π4(sin⁡∣4x−π12∣+[2sin⁡x])dx=2π+α24 \int_0^{\frac{\pi}{4}}\left(\sin \left|4 x-\frac{\pi}{12}\right|+[2 \sin x]\right) \mathrm{d} x=2 \pi+\alpha, where [⋅][\cdot] denotes the greatest integer function, then α\alpha is equal to _______.

Correct answer

0

Step-by-step solution

Let I=24∫0π2(sin⁡∣4x−π2∣+[2sin⁡x])dxI=24 \int_0^{\frac{\pi}{2}}\left(\sin \left|4 x-\frac{\pi}{2}\right|+[2 \sin x]\right) d x ...(i) Now ∣4x−π12∣=\left|4 x-\frac{\pi}{12}\right|=\begin{cases}-4 x+\frac{\pi}{12} & ; x < \frac{\pi}{48} \\ 4 x-\frac{\pi}{12} & ; \quad x \geq \frac{\pi}{48}\end{cases}∴\therefore from (i)(\mathrm{i})I=24∫0π48−sin⁡(4x−π12)dx+∫π48π4sin⁡(4x−π12)+∫0π6[2sin⁡x]dx+∫π6π4[2sin⁡x]dxI=24[(1−cos⁡π12)4−(−cos⁡π12−1)4)+π4−π6I=24(12)+π4−π6I=2π+12=2π+α (from above) ∴α=12\begin{aligned} & I=24 \int_0^{\frac{\pi}{48}}-\sin \left(4 x-\frac{\pi}{12}\right) d x+\int_{\frac{\pi}{48}}^{\frac{\pi}{4}} \sin \left(4 x-\frac{\pi}{12}\right) \\ & +\int_0^{\frac{\pi}{6}}[2 \sin x] d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}[2 \sin x] d x \\ & I=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} \\ & I=24\left(\frac{1}{2}\right)+\frac{\pi}{4}-\frac{\pi}{6} \\ & I=2 \pi+12=2 \pi+\alpha \text{ (from above) } \\ & \therefore \quad \alpha=12\end{aligned}$$
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