Question

Let $f(x)=\frac{k}{1+x^2},-\infty <x<\infty$ be the probability density of a random variable. Then, $k$ equals to

• A. $\pi$
• B. $-\pi$
• C. $\frac{1}{\pi }$
• D. $1$

Answer 1

Answer: (C)

$\because$ ${\int }_{-\infty }^{\infty }f(x)dx=1$
$\therefore$ ${\int }_{-\infty }^{\infty }\frac{k}{1+x^2}dx=1$
$\Rightarrow$ ${\int }_0^{\infty }\frac{k}{1+x^2}dx=1$
$\Rightarrow$ $2{\int }_0^{\infty }\frac{k}{1+x^2}dx=1$
$\Rightarrow$ $2k({\tan }^{-1}x{)}_0^{\infty }=1$
$\Rightarrow$ $2k({\tan }^{-1}\infty -{\tan }^{-1}0)=1$
$\Rightarrow$ $2k\left(\frac{\pi }{2}-0\right)=1$
$\Rightarrow$ $k=\frac{1}{\pi }$