Question

Consider $f\left(x\right)=\{\begin{array}{c}\cos x0\le x<\frac{\pi }{2}\\ {\left(\frac{\pi }{2}-x\right)}^2\frac{\pi }{2}\le x<\pi \end{array}$
such that f is periodic with period $\pi$ , then

• A. The range of f is $\left(0,\frac{{\pi }^2}{4}\right)$
• B. f is continuous for all real X,but not differentiable for some real X
• C. f is continuous for all real X
• D. The are bounded by $y=f\left(x\right)$ and the X-axis from
  $x=-n\pi$ to $x=n\pi$ is $2n\left(1+\frac{{\pi }^3}{24}\right)$ for a given $n\in N$

Answer 1

The correct option is D The are bounded by $y=f\left(x\right)$ and the X-axis from

$x=-n\pi$ to $x=n\pi$ is $2n\left(1+\frac{{\pi }^3}{24}\right)$ for a given $n\in N$
if $\begin{array}{c}x=\frac{\pi }{2},f\left(x\right)=0\\ x=\pi ,f\left(x\right)=\frac{{\pi }^2}{4}\end{array}$
so range=$\left[0,\frac{{\pi }^2}{4}\right)$
$A-\underset{-\pi }{\overset{\pi }{\int }}f\left(x\right)dx=\underset{-\pi }{\overset{0}{\int }}+\underset{0}{\overset{\pi }{\int }}=2\underset{0}{\overset{\pi }{\int }}=2\left(1+\frac{{\pi }^3}{24}\right)$
$A=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos x+\underset{\frac{\pi }{2n}}{\overset{\pi }{\int }}{\left(\frac{\pi }{2}-x\right)}^2=2n\left(1+\frac{{\pi }^3}{24}\right)$