Answer A. III Only (i.e. $\sqrt{x^3 + 1}$ )
Explanation
Decreasing/Increasing nature of a function can be determined by observing the first derivative of equations in given domain.
If the derivative is positive in given domain, It is increasing, else a negative value indicates it is decreasing.
Now testing one by one
I. $e^{-x}$
Here, $\frac{\mathrm{d} }{\mathrm{d} x}e^{-x} = -e^{-x}$
This will remain negative in entire domain [0,1] hence decreasing.
II. $x^{2} - \sin(x)$
Here, $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2} - \sin(x)) = 2x - \cos(x)$
Here the switch happens! Observe $2x - \cos(x)$ is negative for $x=0$ (i.e. = -1) and positive at $x=1$ (i.e. = ~1.4596977). Though we can find the exact point where it switches but that's not required here. We can confirm that till some point in the domain [0,1], this function decreases and then increases.
III. $\sqrt{x^3 + 1}$
Here, (Though this one is intuitive), $\frac{\mathrm{d} }{\mathrm{d} x}\sqrt{x^3 + 1} = \frac{3x^{2}}{2\sqrt{x^3+1}}$ This will remain positive throughout [0,1] and hence will be increasing.