Question

For $x \in R$, let $f\left(x\right)=\left|\sin\,x\right|$ and $g \left(x\right)=\int\limits_{0}^{x} f \left(t\right)dt$. 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 $x \in R$
$f \left(x\right)=\left|\sin \,x\right|$
$g\left(x\right)=\int\limits_{0}^{x}f \left(t\right)dt$, and
$p\left(x\right)=g\left(x\right)-\frac{2}{\pi}x$
$\because P\left(x+\pi\right)=g \left(x+\pi\right)-\frac{2}{\pi}\left(x+\pi\right)$
$=\int\limits_{0}^{x+\pi} \left|\sin \left(t\right)\right|dt -\frac{2}{\pi}x-2$
$=\int\limits_{0}^{\pi}\left|\sin\,t\right|dt+\int\limits_{\pi}^{\pi+x}\left|\sin\,t\right|dt-\frac{2}{\pi}x-2$
$=2\int\limits_{0}^{\pi /2} \sin\,t\,dt+\int\limits_{0}^{x}\left|\sin\,t\right|dt-\frac{2}{\pi}x-2$
$[\because |\sin\,t|$ 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 $x$