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 } $