JEE Main 2025 Jan 29 Shift 2, Mathematics Q21: Some Properties of Definite Integrals

Split the integral into the modulus part $I_1$ and the floor part $I_2$.

$I_1=24{\int }_0^{\pi /4}\left|\sin (4x-\pi /12)\right|dx$. The argument $4x-\pi /12$ runs from $-\pi /12$ (at $x=0$) to $11\pi /12$ (at $x=\pi /4$) and crosses zero at $x=\pi /48$.

On $[0,\pi /48]$, $\sin (\cdot )$ is negative; on $[\pi /48,\pi /4]$ it is positive.

${\int }_0^{\pi /48}-\sin (4x-\pi /12)dx=\frac{1}{4}(1-\cos (\pi /12))$.

${\int }_{\pi /48}^{\pi /4}\sin (4x-\pi /12)dx=\frac{1}{4}(1+\cos (\pi /12))$.

Sum $=\frac{1}{2}$, so $I_1=24\cdot \frac{1}{2}=12$.

$I_2=24{\int }_0^{\pi /4}[2\sin x]dx$. On $[0,\pi /4]$, $2\sin x$ rises from $0$ to $\sqrt{2}\approx 1.414$. So $[2\sin x]=0$ when $2\sin x<1$ (i.e. $x<\pi /6$) and $=1$ when $\pi /6\le x\le \pi /4$.

${\int }_{\pi /6}^{\pi /4}1dx=\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}$, so $I_2=24\cdot \frac{\pi }{12}=2\pi$.

Total $=12+2\pi$. Matching with $2\pi +\alpha$ gives $\alpha =12$.