Question

If the function f(x) defined by
$f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+......+\frac{x^2}{2}+x+1$, then f'(0)=

• A. $100$
• B. $-1$
• C. $100f^{\prime }\left(0\right)$
• D. $1$

Answer 1

The correct option is D $1$

We have, $f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+......+\frac{x^2}{2}+x+1$

Since $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Thus $f^{\prime }\left(x\right)=x^{99}+x^{98}+.............+x+1+0$

Now substitute $x=0$

$\Rightarrow f^{\prime }\left(0\right)=0+0+.............+0+1=1$