= Solution
Put $N=6-2e^{-2}$. The <function graph> is symmetric about the vertical axis, zero outside $[-2,2]$, and has its unique maximum at $x=0$ with height $2/N$. On $0<x<2$, it decreases as $(1+e^{-x})/N$; the negative side is its reflection. The one-sided interior endpoint values are $(1+e^{-2})/N$, so the <probability density function> jumps to zero at both ends of its support. It also has a corner at its central maximum because of the <absolute value> in the exponent.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-q16-density.png]
{title=The symmetric density in Question 16, including its central corner and endpoint jumps}
Values assigned exactly at $x=\pm2$ do not affect any <probability>.
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