Question

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

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

Answer 1

The correct option is C

$1$

Explanation for the correct option

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

Upon differentiating, we get,

$\Rightarrow f\text{'}\left(x\right)=\frac{100x^{99}}{100}+\frac{99x^{98}}{99}+...+\frac{2x}{2}+1+0$ $\left[\because \frac{d}{dx}\left(x^n\right)=n{\left(x\right)}^{n-1}\right]$

$\Rightarrow f\text{'}\left(x\right)=x^{99}+x^{98}...+x+1$

Now we need to evaluate for $x=0$, so upon substituting we get,

$\begin{array}{rcl}& \Rightarrow & f\text{'}\left(0\right)=0+0+...+0+1\\ & \Rightarrow & f\text{'}\left(0\right)=1\end{array}$

Hence, option (C), i.e. $1$ is the correct answer.