Mathematics - Functions Question with Solution | TestHub

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