= Solution
{c}
Take $\boxed{p(x)=x(L-x)}$. It vanishes at both endpoints. Its <odd periodic extension> is continuously differentiable across the joins, whereas its <even extension> has corners because $p'(0)=L$ and $p'(L)=-L$ are nonzero.
Two <integrations by parts>, with $q=n\pi/L$, give
$$
\int_0^Lp(x)\sin(qx)\,dx=\frac{2[1-(-1)^n]}{q^3},\qquad
\int_0^Lp(x)\cos(qx)\,dx=-\frac{L[1+(-1)^n]}{q^2}.
$$
The <Fourier coefficients> are therefore
$$
\boxed{b_n=\frac{4L^2[1-(-1)^n]}{n^3\pi^3},\qquad
a_0=\frac{L^2}{3},\qquad
a_n=-\frac{2L^2[1+(-1)^n]}{n^2\pi^2}.}
$$
The <Fourier sine series> has a uniform absolute tail $O(N^{-2})$, and each of its <partial sums> satisfies both zero <boundary values> exactly. The <Fourier cosine series> has an absolute tail bound $O(N^{-1})$; at $x=0$ its omitted even-index <Fourier coefficients> have the same sign and yield an error of order $N^{-1}$. This verifies a strictly faster uniform error bound for the sine approximation in this example.
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