Question

For the function $f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+....+\frac{x^2}{2}+1$. Prove that $f^{\prime }\left(1\right)=100f^{\prime }\left(0\right)$

Answer 1

Given, $f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+....+\frac{x^2}{2}+1$
$\Rightarrow f^{\prime }\left(x\right)=\frac{100x^{99}}{100}+\frac{x^{99}}{99}+.....+\frac{2x}{2}+1+0$ $(\because )f\left(x\right)=x^n\Rightarrow f^{\prime }\left(x\right)=n{x}^{n-1}$
$\Rightarrow f^{\prime }\left(x\right)=x^{99}+x^{98}+....+x+1$ ...(i)
Putting x=1, we get
$f^{\prime }\left(1\right)=\frac{(1{)}^{99}+1^{98}+1...+1+1}{100\text{times}}$
$=\frac{1+1+\mathrm{1...}+1+1}{\text{times}}$
$\Rightarrow f^{\prime }\left(1\right)=100$ ...(ii)
Again putting x=0, we get
$f^{\prime }\left(0\right)=0+0+....+0+1$
$\Rightarrow f^{\prime }\left(0\right)=1$
From eqs. (ii) and (iii), $f^{\prime }\left(1\right)=100f^{\prime }\left(0\right)$
Hence proved.