= Solution
{c}
Put $x=1-t$. For $0<k\leq1$,
$$
(1-t)^k=1-kt+\frac{k(k-1)}2t^2+O(t^3),
$$
whereas the constrained <rational interpolant> expands as
$$
\frac{a(1-t)}{1+b-bt}
=\frac a{1+b}\left[1-\frac{t}{1+b}+O(t^2)\right].
$$
Matching the constant term gives $a=1+b$, and matching the coefficient of $t$ gives $1/(1+b)=k$. Thus $a=1/k$, $b=(1-k)/k$, and
$$
\boxed{r(x)=\frac{x}{k+(1-k)x}.}
$$
This is an <endpoint-constrained rational approximation to a power>: it matches value and derivative at $x=1$ and also enforces $r(0)=0$ for $k>0$.
There is a terminology qualification. A standard degree-$(1,1)$ <Padé approximant> about $t=0$ matches through $t^2$, but the displayed constrained expression has quadratic coefficient $-k(1-k)$ rather than $-k(1-k)/2$. The unconstrained Padé approximant is instead
$$
R(x)=\frac{(1-k)+(1+k)x}{(1+k)+(1-k)x}.
$$
The source's requested formula is obtained by the two stated endpoint matching conditions within its restricted family, rather than by full Padé matching. At $k=1$ both reduce to $x$. At $k=0$, the requested expression is one for $x>0$ with a removable value at zero; finite $a,b$ in the original restricted form do not represent that constant function. The endpoint is understood by this limit.
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