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Given,

$\LARGE f(x)=\frac{e^{x}}{1+e^{x}}$

Differentiating with respect to x,

$\large f'(x)=\frac{(1+e^{x}).\frac{\mathrm{d} }{\mathrm{d} x}e^{x}-e^{x}\frac{\mathrm{d} }{\mathrm{d} x}(1+e^{x})}{(1+e^{x})^{2}}$

$\large f'(x)=\frac{1+e^{2x}-e^{2x}}{(1+e^{x})^{2}}$

$\large f'(x)=\frac{1}{(1+e^{x})^{2}}$

$\large f'(x)$ is always greater than 0 , as denominator $\large (1+e^{x})^{2}$ >0 and numerator is 1 so ,

$\large f'(x)>0$ in $\large \left [ -\infty,+\infty \right ]$.

It indicates that if we draw a tangent in the graph for $\large f(x)=\frac{e^{x}}{1+e^{x}}$at any point in $\large \left [ -\infty,+\infty \right ]$ the slop of the tangent for the graph is always positive which indicates that the graph is monotonically increasing.

The graph will look like this ,

so correct answer is option (a).