= Solution
Under the same independence assumption, <inclusion-exclusion> gives
$$
\boxed{R(t)=P_1(t)+P_2(t)-P(t)=\begin{cases}
0,&t\leq0,\\s+s^2-s^3,&0<t<T,\\1,&t\geq T.
\end{cases}}
$$
On $0<t<T$,
$$
R'(t)=\frac{1+2s-3s^2}{T}=\frac{(1-s)(3s+1)}T>0,
\qquad R''(t)=\frac{2-6s}{T^2}.
$$
There is one interior <inflection point>, where the <curvature> changes from positive to negative:
$$
\boxed{\left(t,R(t)\right)=\left(\frac T3,\frac{11}{27}\right)}.
$$
At $t=T$ the graph reaches one with zero slope and then stays constant. There is no further change from one sign of <curvature> to the opposite sign there.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2-failure-probabilities.png]
{title=Failure densities and cumulative probabilities, with independent lifetimes for the joint curves}
Without independence, $R=P_1+P_2-P$ still holds, but only $\boxed{s\leq R(t)\leq\min(1,s+s^2)}$ is forced by the marginals. Its inflection behaviour is then not determined. The shared-$U$ construction in (e), for example, gives $R=s$ before the plateau.
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