Addenda to solution

The solution is slightly incomplete, because the logic is only valid for positive k (for instance, when k=0, the partial fraction decomposition becomes invalid). This issue can be skirted around by pointing out that the PDF will be identically 0 if k is non-positive, and therefore only k>0 has any chance of working.

This is a rather technical point that may only serve to confuse the reader. Maybe it belongs in an Advanced subsection?

Why not adding this remark at the end of the solution with an Advanced remark warning or something like this. We'll get feedback from students eventually to help us settle this in an appropriate way I think.