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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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The value of K for which function $ f(x) =\left\{ \begin{array}{l l} k\: e^{-3x} & \quad \text{ $x$ >0}\\ 0 & \quad \text{ elsewhere} \end{array} \right.$ is a Probability Distribution Function.

$\begin {array} {1 1} (A)\;3 & \quad (B)\;\large\frac{1}{3} \\ (C)\;2 & \quad (D)\;-3 \end {array}$


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We know that $ f(x)$ is a PDF, if
$ \int_{-\infty}^{\infty}\: f(x)dx=1$
or $ \int_{-\infty}^{0} 0\: dx+k \int_{0}^{\infty} e^{-3x}dx=1$
or $ \large\frac{k}{-3} [e^{-3x}]^{\infty}_{0}$
$ \Rightarrow \large\frac{k}{3}=1$
$\therefore k = 3$
Hence answer is (a)
answered Jan 15, 2014 by thanvigandhi_1
edited Mar 26, 2014 by rvidyagovindarajan_1

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