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)

$\frac{1}{\pi }$

Answer 2

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