 Solution

For the linear test equation $f(y)=\lambda y$ put $z=h\lambda$. The second stage solves $(1-az)k_2=\lambda y[1+(1-a)z]$, and substitution into the update gives the stability functionA-stability requires no poles and $|R(z)|\le 1$ throughout the closed left half-plane. If $a\ne 1/2$, the quadratic numerator gives unbounded magnitude as $z\to -\infty$; when $a=0$ the denominator is constant and the same conclusion holds. Thus $a=1/2$ is necessary. At this value $R(z)=(1+z/2)/(1-z/2)$, whose only pole is at positive $z=2$. Moreoveron the left half-plane. Hence the method is A-stable exactly for $\boxed{a=1/2}$. Its limit at negative infinity is $-1$, so it is not L-stable.