= Solution
The <function> is nonnegative and satisfies $\cos^2(\pi-x)=\cos^2x$. Its <derivative> is $-\sin(2x)$, so it decreases from $(0,1)$ to its <minimum> $(\pi/2,0)$ and then increases to $(\pi,1)$. These three points all have horizontal <tangent lines>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1-elementary-graphs.png]
{title=The cosine-squared curve and the two branches of the exponential quotient}
{description=The left panel shows the symmetric minimum of $\cos^2x$. The right panel marks the vertical and horizontal <asymptotes> of $1/(1-e^x)$.}
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