$\int \limits_0^{2 \pi} | \sin \large\frac{x}{4} $$- \cos \large\frac{x}{4}| $$dx$
=>$ \sin \large\frac{x}{4} - $$\cos \large\frac{x}{4}$$ =0$
=> $\tan \frac{x}{4} =1, \large\frac{x}{4} =\frac{\pi}{4} ,$$ x=\pi$
$\int \limits_0^{\pi} -\bigg( \sin \large\frac{x}{4} $$- \cos\large\frac{x}{4} )$$ dx + \int \limits _{\pi}^{2 \pi} \bigg( \sin \large\frac{x}{4} -$$\cos \large\frac{x}{4} \bigg ) $$dx$
=> $ \bigg[4 \bigg[ \cos \large\frac{x}{4} $$+ \sin \large\frac{x}{4} \bigg] \bigg]_0^{\pi} $$+4 \bigg[ \bigg[ \cos \large\frac{x}{4} \bigg] -$$\sin \large\frac{x}{4}\bigg]_x^{2 \pi}$
=> $ 4 \bigg[ \cos \large\frac{\pi}{4} $$ +\sin \large \frac{\pi}{4} $$- \cos 0 -\sin 0] $$+4 \bigg [\cos \large\frac{\pi}{2}$$ -\sin \large\frac{\pi}{2}-$$ \cos \large\frac{\pi}{4} +$$ \sin \large\frac{\pi}{4}\bigg]$
$4 [\sqrt 2-1]+4 [\sqrt 2 -1]= 8 [\sqrt 2 -1]$
Hence a is the correct answer.