= Solution
{c}
The <Sobolev space> $H^1(0,1)$ consists of square-integrable <functions> whose weak first <derivative> is square-integrable, with <norm> $\|v\|_{H^1}^2=\int_0^1(v^2+v'^2)$. Its one-dimensional representatives have well-defined endpoint traces.
Multiplying by $v\in H^1$ and integrating by parts gives
$$
\int_0^1fv=\int_0^1(u'v'+\cos x\,uv)-u'(1)v(1)+u'(0)v(0).
$$
The two Robin conditions give $u'(0)=u(0)$ and $u'(1)=-u(1)$, so define
$$
a(u,v)=\int_0^1(u'v'+\cos x\,uv)+u(0)v(0)+u(1)v(1),\qquad
\ell(v)=\int_0^1fv.
$$
A <weak solution> is $u\in H^1(0,1)$ satisfying $a(u,v)=\ell(v)$ for every $v\in H^1(0,1)$. The <boundary conditions> are natural; no endpoint <derivative> is imposed in the trial space.
Since $\cos x\geq\cos1>0$ on this interval,
$$
\boxed{a(v,v)\geq\cos1\,\|v\|_{H^1}^2.}
$$
This proves <coercivity> on the full space. Trace continuity and <Cauchy-Schwarz> give continuity of the form, so the <Lax-Milgram theorem> also ensures a unique <weak solution>.
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