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(C) Step $1$: Find the value of $k$ using the property $\int_{-\infty}^{\infty} f(x) dx = 1$. Since $f(x) = 0$ for $x \leq 0$,we have $\int_{0}^{\inft

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$\frac{2}{\pi} 	an^{-1} x$

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(C) Step $1$: Find the value of $k$ using the property $\int_{-\infty}^{\infty} f(x) dx = 1$. Since $f(x) = 0$ for $x \leq 0$,we have $\int_{0}^{\infty} \frac{k}{x^2+1} dx = 1$. Step $2$: Evaluate the integral: $k [	an^{-1} x]_{0}^{\infty} = 1$. $k (\frac{\pi}{2} - 0) = 1 \implies k = \frac{2}{\pi}$. Step $3$: The c.d.f. $F(x)$ is defined as $P(X \leq x) = \int_{0}^{x} f(t) dt$ for $x > 0$. $F(x) = \int_{0}^{x} \frac{2/\pi}{t^2+1} dt = \frac{2}{\pi} [	an^{-1} t]_{0}^{x} = \frac{2}{\pi} 	an^{-1} x$.

If a random variable $X$ has the p.d.f. $f(x) = \begin{cases} \frac{k}{x^2+1} & , \text{if } 0 < x < \infty \\ 0 & , \text{otherwise} \end{cases}$,then the c.d.f. of $X$ is:

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