Let $f(x)=e^{v+1 / n}-e^{x}$ for real number $p>0$.
The value of $x=s$, for which $f(x)$ is minimum, is
(A) $-\frac{\ln (\mathrm{p}+1)}{\mathrm{p}}$
(B) $-\ln (\mathrm{p}+1)$
(C) $-\ln \mathrm{p}$
(D) $\ln \left(\frac{\mathrm{p}+\mathrm{l}}{\mathrm{p}}\right)$