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If $f(x)=\large\frac{A}{\pi}\frac{1}{16+x^{2'}}$$-\infty<x<\infty$\[\]is a p.d.f of a continuous random variable $X,$ then the value of $A$ is

\[\begin{array}{1 1}(1)16&(2)8\\(3)4&(4)1\end {array}\]

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Given that $f(c) =\large\frac{A}{\pi}. \frac{1}{16^2+x^2}$$ - \infty < x < \infty $ is a p.d.f
$\therefore \int \limits_{-\infty}^{\infty} $$f(x) dx=1$
$\int \limits_{-\infty}^{\infty} \large\frac{A}{\pi} \frac{1}{16+x^2}$$dx=1$
$\large\frac{A}{\pi} \int \limits_{-\infty}^{\infty} \large\frac{dx}{4^2+x^2}$$dx=1$
$\large\frac{2A}{\pi} \int \limits_0^{\infty} \frac{ dx}{4^2+x^2} $$=1$
Since $\large\frac{1}{4^2+x^2}$ is an even function.
$\large\frac{2A}{\pi} .\frac{1}{4} . \tan ^{-1} (\frac{x}{4} ) \bigg]_0^{\infty}$
$\large\frac{A}{2 \pi} [ \frac{\pi}{2} -0 ]$$=1$
$\large\frac{A}{2}$$=1 => A=4$
Hence 3 is the correct answer.
answered May 22, 2014 by meena.p

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