Question

$\text{For}x\in R,\text{let}f\left(x\right)=|\sin x|\text{and}g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt.$
$\text{Let}p\left(x\right)=g\left(x\right)-\frac{2}{\pi }x.$ Then

• A. $p(x+\pi )=p\left(x\right)$ for all $x$
• B. $p(x+\pi )\ne p\left(x\right)$ for at least one but finitely many $x$
• C. $p(x+\pi )\ne p\left(x\right)$ for infinitely many $x$
• D. $p$ is a one-one function

Answer 1

The correct option is A $p(x+\pi )=p\left(x\right)$ for all $x$

$g(x+\pi )=\underset{0}{\overset{x+\pi }{\int }}|\sin x|dx=\underset{0}{\overset{\pi }{\int }}|\sin x|dx+\underset{\pi }{\overset{x+\pi }{\int }}|\sin x|dx$

We know that
$\underset{T}{\overset{x+T}{\int }}f\left(x\right)dx=\underset{0}{\overset{x}{\int }}f\left(x\right)dx$, where $T$ is the period of the $f\left(x\right)$

Using the above property,
$g(x+\pi )=\underset{0}{\overset{\pi }{\int }}|\sin x|dx+\underset{0}{\overset{x}{\int }}|\sin x|dx(\because \pi \text{is period of}|\sin x|)=2+g\left(x\right)$

Now,
$p(x+\pi )=g(x+\pi )-\frac{2}{\pi }(x+\pi )=2+g\left(x\right)-\frac{2}{\pi }x-2=g\left(x\right)-\frac{2}{\pi }x=p\left(x\right)$

$\text{So,}p(x+\pi )=p\left(x\right)\text{for all}x$