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Solution:

We have,

$\mathrm{f}(\mathrm{x})=\frac{1}{1+\mathrm{x}^{2}}$

Case 1

When $x \in[0, \infty)$

Let $\mathrm{x}_{1}>\mathrm{x}_{2}$

$\Rightarrow \mathrm{x}_{1}^{2}>\mathrm{x}_{2}^{2}$

$\Rightarrow 1+\mathrm{x}_{1}^{2}>1+\mathrm{x}_{2}^{2}$

$\Rightarrow \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}}$

$\Rightarrow f\left(x_{1}\right)

$\Rightarrow \therefore f(x)$ is decreasing on $[0, \infty)$

Case 2

When $x \in(-\infty, 0]$

Let $\mathrm{x}_{1}>\mathrm{x}_{2}$

$\Rightarrow \mathrm{x}_{1}^{2}<\mathrm{x}_{2}^{2}$

$\Rightarrow 1+\mathrm{x}_{1}^{2}<1+\mathrm{x}_{2}^{2}$

$\Rightarrow \frac{1}{1+\mathrm{x}_{1}^{2}}>\frac{1}{1+\mathrm{x}_{2}^{2}}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)>\mathrm{f}\left(\mathrm{x}_{2}\right)$

$\therefore f(x)$ is increasing on $(-\infty, 0] .$

Thus, $f(x)$ is neither increasing nor decreasing on $R$.