Science talk:Math Exam Resources/Courses/MATH105/April 2011/Question 3
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Thread title | Replies | Last modified |
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Addenda to solution | 1 | 17:17, 20 March 2012 |
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?