Question

Let $f:[0,2] \rightarrow R$ 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 ______.

Answer 1

Let $\pi x=\theta$
$f(x)=(3-\sin 2 \theta) \sin \left(\theta-\frac{\pi}{4}\right)-\sin \left(3 \theta+\frac{\pi}{4}\right)$
$=(3-\sin 2 \theta) \frac{(\sin \theta-\cos \theta)}{\sqrt{2}}-\left(\frac{\sin 3 \theta}{\sqrt{2}}+\frac{\cos 3 \theta}{\sqrt{2}}\right)$
$=\frac{1}{\sqrt{2}}\left[(3-\sin 2 \theta)(\sin \theta-\cos \theta)-\left(3 \sin \theta-3 \cos \theta-4 \sin ^{3} \theta+4 \cos ^{3} \theta\right)\right]$
$=\frac{1}{\sqrt{2}}[(3-\sin 2 \theta)(\sin \theta-\cos \theta)-\{3(\sin \theta-\cos \theta)-4(\sin \theta-\cos \theta)$
$\left.\left.\left(1+\frac{\sin 2 \theta}{2}\right)\right\}\right]$
$=\frac{(\sin \theta-\cos \theta)}{\sqrt{2}}(3-\sin 2 \theta+1+2 \sin 2 \theta)$
$\therefore $ For $f(x) \geq 0$

$\therefore [\alpha, \beta]=\left[\frac{1}{4}, \frac{5}{4}\right]$
$\Rightarrow [\beta-\alpha]=\frac{5}{4}-\frac{1}{4}=1$