Question

If $f\left(x\right)={\prod }_{n=1}^{100}{(x-n)}^{n(101-n)}$, then find $\frac{f\left(101\right)}{f^{{}^{\prime }}\left(101\right)}$

Answer 1

$f\left(x\right)={\prod }_{n=1}^{100}{(x-n)}^{n(101-n)}$

$\ln f\left(x\right)={\sum }_{n=1}^{100}n(101-n)\ln (x-n)$

$\ln f\left(x\right)=100\ln (x-1)+2\times 99\ln (x-2)+3\times 98\ln (x-3).........100\ln (x-100)$

Now differentiating,

$\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{100}{x-1}+\frac{2\times 99}{x-2}+\frac{3\times 98}{x-3}.....\frac{100}{x-100}$

$\frac{f^{\prime }\left(101\right)}{f\left(101\right)}=\frac{100}{100}+\frac{2\times 99}{99}+\frac{3\times 98}{98}......\frac{100}{1}$

$\frac{f^{\prime }\left(101\right)}{f\left(101\right)}=1+2+\mathrm{3.....100}$

$\frac{f^{\prime }\left(101\right)}{f\left(101\right)}=\frac{100\left(101\right)}{2}$

$\frac{f^{\prime }\left(101\right)}{f\left(101\right)}=5050$

$\frac{f\left(101\right)}{f^{\prime }\left(101\right)}=\frac{1}{5050}$