The area of the region enclosed by the curve f(x) = max x, x , - π...

Plotting the required diagram we get,

Now from diagram we can say that,

$\sin x$ is maximum for $x \in$\left(- \pi, \dfrac{- 3\pi}{4}\right) \cup $\left(\dfrac{\pi}{4}, \pi\right)$ and $\cos x$ is maximum for $x \in \left(\dfrac{- 3\pi}{4}, \dfrac{- \pi}{2}\right) \cup \left(\dfrac{- \pi}{2}, \dfrac{\pi}{4}\right)$

So, area of the region is given by,

$A = \left|{∫}_{- \pi}^{\dfrac{- 3\pi}{4}}sinxdx\right| + \left|{∫}_{\dfrac{\pi}{4}}^{\pi}sinxdx\right| + \left|{∫}_{\dfrac{- 3\pi}{4}}^{\dfrac{- \pi}{2}}cosxdx\right| + \left|{∫}_{\dfrac{- \pi}{2}}^{\dfrac{\pi}{4}}cosxdx\right|$

$⇒ A = \left|\dfrac{1}{\sqrt{2}} - 1\right| + \left|1 + \dfrac{1}{\sqrt{2}}\right| + \left|- 1 + \dfrac{1}{\sqrt{2}}\right| + \left|\dfrac{1}{\sqrt{2}} + 1\right|$

$⇒ A = 2 + \dfrac{2}{\sqrt{2}} + 2 - \dfrac{2}{\sqrt{2}} = 4$

Hence this is the required option.