Let $f : [0, 2] \to \mathbf{R}$ be the function defined by $f(x) = (3 - \sin(2\pi x))\sin\left(\pi x - \frac{\pi}{4}\right) - \sin\left(3\pi x + \frac{\pi}{4}\right)$. If $\alpha, \beta \in [0, 2]$ are such that $\{x \in [0, 2] : f(x) \geq 0\} = [\alpha, \beta]$, then the value of $\beta - \alpha$ is

$\text{Given: } f:[0,2]\to\mathbb{R}$

$f(x)=(3-\sin 2\pi x)\sin\left(\pi x-\frac{\pi}{4}\right)-\sin\left(3\pi x+\frac{\pi}{4}\right)$

$\text{To find: } \beta-\alpha,\ \text{where } \{x\in[0,2]:f(x)\geq 0\}=[\alpha,\beta]$

$\text{Let } u=\pi x:\quad x\in[0,2]\ \Rightarrow\ u\in[0,2\pi]$

$f=(3-\sin 2u)\sin\left(u-\frac{\pi}{4}\right)-\sin\left(3u+\frac{\pi}{4}\right)$

$\sqrt{2}\sin\left(u-\frac{\pi}{4}\right)=\sin u-\cos u$

$\sqrt{2}\sin\left(3u+\frac{\pi}{4}\right)=\sin 3u+\cos 3u$

$\sqrt{2}\,f=(3-\sin 2u)(\sin u-\cos u)-(\sin 3u+\cos 3u)$

$s=\sin u,\quad c=\cos u,\qquad \sin 2u=2sc$

$\sin 3u+\cos 3u=(3s-4s^{3})+(4c^{3}-3c)$

$=3(s-c)-4(s-c)(s^{2}+sc+c^{2})$

$=(s-c)\left[3-4(1+sc)\right]=-(s-c)(1+4sc)$

$\sqrt{2}\,f=(s-c)(3-2sc)+(s-c)(1+4sc)$

$\sqrt{2}\,f=(s-c)(4+2sc)=2(s-c)(2+sc)$

$f=\sqrt{2}\,(\sin u-\cos u)(2+\sin u\cos u)$

$2+\sin u\cos u=2+\tfrac12\sin 2u \ge \tfrac32 > 0$

$\operatorname{sign} f=\operatorname{sign}(\sin u-\cos u)$

$\sin u-\cos u=\sqrt{2}\sin\left(u-\frac{\pi}{4}\right)$

$\sin\left(u-\frac{\pi}{4}\right)\geq 0 \Rightarrow u-\frac{\pi}{4}\in[0,\pi]$

$\Rightarrow u\in\left[\frac{\pi}{4},\frac{5\pi}{4}\right]$

$x=\frac{u}{\pi}\in\left[\frac14,\frac54\right]$

$\alpha=\frac14,\quad \beta=\frac54$

$\beta-\alpha=\frac54-\frac14=1$