Hello! I believe there's a mistake in the 5th exam paper's solution. Probably I am wrong, but at least I want to understand why.
In the 3rd exercise, part (a), case [latex:2bgrzuxi]l=\pi[/latex:2bgrzuxi], there is the inequality [latex:2bgrzuxi]-\cos&#40;a&#41;+\cos&#40;b&#41; \geq \frac{a^2-b^2}{8}[/latex:2bgrzuxi] for any [latex:2bgrzuxi]&#40;a,b&#41;[/latex:2bgrzuxi] in a suitable neighborhood of [latex:2bgrzuxi]&#40;0,0&#41;[/latex:2bgrzuxi].
I believe this is wrong. 
Indeed, considering WLOG neighborhood = [latex:2bgrzuxi]B_{\delta} &#40;&#40;0,0&#41;&#41;[/latex:2bgrzuxi], we can swap [latex:2bgrzuxi]a[/latex:2bgrzuxi] and [latex:2bgrzuxi]b[/latex:2bgrzuxi]. 
Therefore we have [latex:2bgrzuxi]-\cos&#40;a&#41;+\cos&#40;b&#41; \geq \frac{a^2-b^2}{8}[/latex:2bgrzuxi] and [latex:2bgrzuxi]-\cos&#40;b&#41;+\cos&#40;a&#41; \geq \frac{b^2-a^2}{8}[/latex:2bgrzuxi], thus it must be an equality. 
So there exists a constant [latex:2bgrzuxi]k[/latex:2bgrzuxi] such that for any [latex:2bgrzuxi]a[/latex:2bgrzuxi] with [latex:2bgrzuxi]|a|[/latex:2bgrzuxi] small enough we have [latex:2bgrzuxi]\cos&#40;a&#41; +\frac{a^2}{8} = k[/latex:2bgrzuxi], which is clearly false.

Moreover I believe that, for [latex:2bgrzuxi]l=\pi, u_0[/latex:2bgrzuxi] is not a (WLM). 

That's because (just outlining the main points) if [latex:2bgrzuxi]u_0[/latex:2bgrzuxi] were a (WLM), then [latex:2bgrzuxi]u_{\epsilon}&#40;x&#41; = \epsilon \cdot sin&#40;x&#41;[/latex:2bgrzuxi] would be another (WLM) for [latex:2bgrzuxi]\epsilon[/latex:2bgrzuxi] small enough ([latex:2bgrzuxi]F&#40;u_{\epsilon}&#41; = F&#40;u_0&#41;[/latex:2bgrzuxi] because of the parity of [latex:2bgrzuxi]cos&#40;\cdot&#41;[/latex:2bgrzuxi]), but [latex:2bgrzuxi]u_{\epsilon}[/latex:2bgrzuxi] doesn't satisfy (ELE) for any [latex:2bgrzuxi]\epsilon \neq 0[/latex:2bgrzuxi] (but it should be easier to prove there is a small enough  value of [latex:2bgrzuxi]\epsilon[/latex:2bgrzuxi]).