= Solution
The <cycloid> has limiting endpoints $(0,0)$ and $(2\pi a,0)$, with its highest point $(\pi a,2a)$ at $t=\pi$. The transformation $t\mapsto2\pi-t$ sends $x$ to $2\pi a-x$ and preserves $y$, so the arch is symmetric about $x=\pi a$. Its slope decreases from $+\infty$ through zero to $-\infty$, giving vertical limiting <tangent lines> at the cusps and a horizontal <tangent line> at the top. More explicitly,
$$
\frac{d^2y}{dx^2}=-\frac1{4a\sin^4(t/2)}<0,
$$
so the arch is a <concave function> throughout its interior. The endpoint cusps are limiting points, excluded by the open parameter interval.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-cycloid-arch.png]
{title=One cycloid arch, its limiting cusps, highest point, and the point with slope square root of three}
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