Question

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

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

Answer 1

Answer: (A)

Given, for $x\in R$
$f\left(x\right)=\left|\sin x\right|$
$g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$, and
$p\left(x\right)=g\left(x\right)-\frac{2}{\pi }x$
$\because P\left(x+\pi \right)=g\left(x+\pi \right)-\frac{2}{\pi }\left(x+\pi \right)$
$=\underset{0}{\overset{x+\pi }{\int }}\left|\sin \left(t\right)\right|dt-\frac{2}{\pi }x-2$
$=\underset{0}{\overset{\pi }{\int }}\left|\sin t\right|dt+\underset{\pi }{\overset{\pi +x}{\int }}\left|\sin t\right|dt-\frac{2}{\pi }x-2$
$=2\underset{0}{\overset{\pi /2}{\int }}\sin tdt+\underset{0}{\overset{x}{\int }}\left|\sin t\right|dt-\frac{2}{\pi }x-2$
$[\because |\sin t|$ is periodic function having period $\pi ]$
$=2{\left[-\cos t\right]}_0^{\pi 2}+g\left(x\right)-\frac{2}{\pi }x-2$
$=2\left(0-\left(-1\right)\right)+g\left(x\right)-\frac{2}{\pi }x-2$
$=2\left(0-\left(-1\right)\right)+g\left(x\right)-\frac{2}{\pi }x-2$
$=2+g\left(x\right)-\frac{2}{\pi }x-2=g\left(x\right)-\frac{2}{\pi }$
$x=p\left(x\right)$
$\therefore p\left(x+\pi \right)=P\left(x\right)$ for all $x$