 Solution

A highest-order method in this family has order $s+1$. The Second Dahlquist barrier says that an irreducible A-stable multistep method has order at most two, so $s+1\le 2$ and hence $s=1$. Part b then leaves only the trapezoidal rule. Its amplification factor isand $|R(z)|\le 1$ whenever $\mathrm{Re}z\le 0$. Thus the trapezoidal rule is the unique highest-order A-stable method of the stated form, apart from representations containing removable common factors.