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For $n geq 2$, let $phi =|t(t-1)...(t-n)|; forall; t in [0,n]$.

1/ Show that $phi$ peaks at a point belonging to $[0,1]$

2/ Evaluate $dfrac{phi'}{phi}$ in accordance with $displaystyle g(t)=sum_{k=0}^n frac{1}{(t-k)}$

Any help would be appreciated.

Well, assuming that I understand you correctly, the first part is the product:
$$phi_n(t)=left|prod_{k=0}^nt-kright|implies{left(phi(t)=0iff t-k=0;\thereforephi(t)=0iff t=kright)}$$
You can evaluate this product using the falling factorial, provided that you appropriately restrict the domain of $n$ to $ngeq2$:
$$phi_n(t)=left|(t)_{n+1}right| : ngeq2$$
By definition, this is the same as
$$phi_n(t)=left|frac{Gamma(t+1)}{Gamma(t-n)}right|; text{if }tinmathbb{Z}, text{then }phi_n(t)=left|frac{t!}{Gamma(t-n)}right|$$
Now, I'm not sure what you mean by "peaks at a point belonging to $[0,1]$," but if you mean that $phi_n(t)$ has a local maximum at $0leq tleq 1$ then you only need to show that $phi_n(0)=0$, $phi_n(1)=0$, and $exists tin(0,1):phi_n(t)neq0$.

From the original product, you know that $phi_n(k)=0$, where $kinleft{xinmathbb{Z}mid0leq xleq nright}$, so the first two statements are true (you can also show this using the limit of the gamma function in the denominator). For the final statement, consider that $Gamma(x)$ has no zeros (so the numerator is never zero), and is finite if $x$ is not a negative integer (so the denominator is finite if $x$ is not an integer). Thus if $tnotinmathbb{Z}$, then $phi_n(t)neq0$.

Hope this helps.

As for the second part to your question, I'm not sure how to interpret it. Is $frac{phiprime}{phi}$ meant to denote $frac{partialphi_n(t)}{partial t}frac{1}{phi_n(t)}$? And if so, what does $g(t)$ have to do with it?