Question

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

• A. $\frac{1}{4950}$
• B. $\frac{102}{5050}$
• C. $\frac{101}{4950}$
• D. $\frac{1}{5050}$

Answer 1

Answer: (D)

$\frac{1}{5050}$

Given, $f\left(x\right)={\Pi }_{n=1}^{100}(x-n{)}^{n(101-n)}$
$\Rightarrow logf\left(x\right)=\left\{n\right(101-n\left)\left\{{\Pi }_{n=1}^{100}log\right(x-n\left)\right\}\right\}$
$\{Here,\Pi \text{changes to}\sum \text{when taken log}\}$
$\Rightarrow logf\left(x\right)=\sum _{n=1}^{100}n(101-n)log(x-n)$
Differentiating both the sides, we get
$\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\sum _{n=1}^{100}n(101-n)\cdot \frac{1}{x-n}$
$\therefore \frac{f^{\prime }\left(101\right)}{f\left(101\right)}=\sum _{n=1}^{100}\frac{n(101-n)}{(101-n)}=\sum _{n=1}^{100}n=5050$

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