Do you really know power function?

The pair $a=0.25$ , $b=0.5$ is a counterexample for each of the first three inequalities, so the only possible answer is the last option, $(1-a)^a>(1-b)^b$ .

To show that this is always true, consider the function $y=(1-x)^x$ on $x \in (0,1)$ . Note that $y$ is always positive on this interval.

We have $\log y=x \log{(1-x)}$ . Differentiating,

$\frac{y'}{y}=\log{(1-x)}-\frac{x}{1-x}$

The right-hand side is clearly negative on the interval; since $y$ is always positive, $y'$ is negative and $y$ is a strictly decreasing function on $(0,1)$ . Therefore $(1-a)^a>(1-b)^b$ when $0<a<b<1$ .