Question

If $f\left(x\right)=\frac{\cos mx}{x^{{}^2}+a^2}$ and $\left|{\int }_0^{\infty }\frac{\cos mxdx}{x^2+a^2}\right|\le A$ then $A$ equals ?

• A. $\pi a^{{}^2}$
• B. $\frac{\pi }{2a}$
• C. $\frac{\pi a^{{}^2}}{4}$
• D. None of these

Answer 1

The correct option is B $\frac{\pi }{2a}$

$-1\le {\cos }^{-1}\theta \le 1\Rightarrow -1\le \cos mx\le 1$
$\Rightarrow \frac{-1}{x^2+a^2}\le \frac{\cos mx}{x^2+a^2}\le \frac{1}{x^2+a^2}$
$\Rightarrow -{\int }_0^{\infty }\frac{1}{x^2+a^2}dx\le {\int }_0^{\infty }\frac{\cos mx}{x^2+a^2}dx\le \frac{1}{x^2+a^2}dx$
$\Rightarrow \left|{\int }_0^{\infty }\frac{\cos mx}{x^2+a^2}\right|\le \left|{\int }_0^{\infty }\frac{1}{x^2+a^2}\right|$
$\Rightarrow \left|{\int }_0^{\infty }\frac{\cos mx}{x^2+a^2}\right|\le {\left[\frac{1}{a}{\tan }^{-1}\frac{x}{a}\right]}_0^{\infty }=\frac{1}{a}\frac{\pi }{2}=\frac{\pi }{2a}$