勘根，a 取 2
\(\begin{align}
  & f\left( \frac{1}{2} \right)=1-\frac{\sqrt{2}}{2}>0 \\
& f\left( \frac{\sqrt{2}}{2} \right)=\frac{1}{2}-\frac{1}{{{2}^{\frac{\sqrt{2}}{2}}}}<0 \\
& f\left( 1 \right)=-\frac{1}{2}<0 \\
& f\left( \sqrt{2} \right)=\frac{1}{2}-\frac{1}{{{2}^{\sqrt{2}}}}>0 \\
\end{align}\)

忘了寫，選擇題，把 a 當 2 就好，若 \(a=\frac{1}{2}\) 也是一樣的函數
不然就 \(f\left( \frac{1}{a} \right),f\left( \frac{1}{\sqrt{a}} \right),f\left( 1 \right),f\left( \sqrt{a} \right)\)吧