= Solution
{c}
If $K$ has one sign, the weighted mean-value theorem applied to the continuous <derivative> $u^{(n+1)}$ gives
$$
\int_0^1K(t)u^{(n+1)}(t)\,dt
=u^{(n+1)}(\eta)\int_0^1K(t)\,dt
$$
for some $\eta\in[0,1]$. One may prove this by bounding the weighted integral between the minimum and maximum <derivative> values times the total weight, then using the <intermediate value theorem>. If the total weight is zero, the fixed-sign kernel is zero almost everywhere and the conclusion is immediate.
Apply the <Peano kernel theorem> to $u(x)=x^{n+1}$ to get $X[x^{n+1}]=(n+1)!\int_0^1K$. Consequently
$$
\boxed{X[u]=\frac{u^{(n+1)}(\eta)}{(n+1)!}X[x^{n+1}].}
$$
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