Consider $f\left(x\right)=\{\begin{array}{cc}\cos x& 0\le x<\frac{\pi }{2}\\ {(\frac{\pi }{2}-x)}^2& \frac{\pi }{2}\le x<\pi \end{array}$  such that $f$  is periodic with period $\pi$ , then:

Given,  $f\left(x\right)=\{\begin{array}{cc}\cos x& 0\le x<\frac{\pi }{2}\\ {(\frac{\pi }{2}-x)}^2& \frac{\pi }{2}\le x<\pi \end{array}$ and $f$  is periodic with period  $\pi$.
  Let us draw the graph of  $y=f\left(x\right)$
From the graph, the range of the function  $[0,\frac{{\pi }^2}{4})\Rightarrow \left(a\right)$
It is discontinuous at $x=n\pi ,n\in I$ .
It is not differentiable at $x=\frac{n\pi }{2},n\in I$ .
 
Area bounded by $y=f\left(x\right)$  and the  $X$-axis from $-n\pi$  to $n\pi$  for  $n\in N$
  $$\begin{gathered} =2n\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx=2n[\underset{0}{\overset{\pi /2}{\int }}\cos xdx+\underset{\pi /2}{\overset{\pi }{\int }}{(\frac{\pi }{2}-x)}^{2}dx] \\ =2n(1+\frac{{\pi }^3}{24}) \end{gathered}$$