JEE Main 2025 (Online) 29th January Evening ShiftQuestion 1
If $24\int_0^{\frac\pi4}\left(\sin\left|4x-\frac\pi{12}\right|+[2\sin x]\right)dx=2\pi+\alpha$, where $[\cdot]$ denotes the greatest integer function, then $\alpha$ is equal to
Definite Integration
Answer$12$
Solution
Concept. Handle the two parts separately: $[2\sin x]$ is $0$ or $1$, and $\sin|u|$ is split at $u=0$. Formulas used. (1) $2\sin x\ge1\iff x\ge\frac\pi6$ on $[0,\frac\pi4]$, and $2\sin\frac\pi4=\sqrt2<2$(2) $\int_0^c\sin u\,du=1-\cos c$ and $\cos\frac{11\pi}{12}=-\cos\frac\pi{12}$Step 1 - Greatest-integer part. By (1), $$\int_0^{\pi/4}[2\sin x]\,dx=\frac\pi4-\frac\pi6=\frac\pi{12}.$$ Step 2 - Modulus part. Put $u=4x-\frac\pi{12}$, $du=4\,dx$, $u:-\frac\pi{12}\to\frac{11\pi}{12}$. By (2), $$\frac14\int_{-\pi/12}^{11\pi/12}\sin|u|\,du=\frac14\Big[\Big(1-\cos\frac\pi{12}\Big)+\Big(1-\cos\frac{11\pi}{12}\Big)\Big]=\frac12.$$ Step 3 - Value. $24\Big(\frac12+\frac\pi{12}\Big)=12+2\pi$, so $\alpha=12$. Answer: 12