Let f: [0, 2] → ℝ be the function defined by f(x) = (3 - (2 )) ( -...

Let $\pix - \dfrac{\pi}{4} = \theta ⇒ \pix = \theta + \dfrac{\pi}{4}$

$∵ f\left(x\right) \ge 0 ⇒ \left[3 - \sin \left(2\theta + \dfrac{\pi}{2}\right)\right]sin\theta - \sin \left(3\theta + \pi\right) \ge 0$

$⇒ \left(3 - \cos 2\theta\right)sin\theta + \sin 3\theta \ge 0$

$⇒ \left(2 + 2\sin^{2}\theta\right)sin\theta + 3sin\theta - 4\sin^{3}\theta \ge 0$

$⇒ sin\theta\left[5 - 2\sin^{2}\theta\right] \ge 0$

$⇒ sin\theta \ge 0 ⇒ \theta \in \left[0, \pi\right]$

(Since, $x \in \left[0, 2\right] ⇒ \pi x \in \left[0, 2\pi\right] ⇒ \theta \in \left[- \dfrac{\pi}{4}, \dfrac{7\pi}{4}\right]$)

$⇒ sin\theta \ge 0 ⇒ \theta \in \left[0, \pi\right] ⇒ \pix \in \left[\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right]$

$⇒ x \in \left[\dfrac{1}{4}, \dfrac{5}{4}\right] ⇒ \alpha = \dfrac{1}{4}, \beta = \dfrac{5}{4}$

Thus, $\beta - \alpha = \dfrac{5}{4} - \dfrac{1}{4} = 1$.