Question

Mark the correct alternative in each of the following:

If $f\left(x\right)=1+x+\frac{x^2}{2}+...+\frac{x^{100}}{100}$, then $f\text{'}\left(1\right)$ is equal to

(a) $\frac{1}{100}$ (b) 100 (c) 50 (d) 0

Answer 1

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

Differentiating both sides with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\left(1+x+\frac{x^2}{2}+...+\frac{x^{100}}{100}\right) \\ =\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2}\right)+...+\frac{d}{dx}\left(\frac{x^{100}}{100}\right) \\ =\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{1}{2}\frac{d}{dx}\left(x^2\right)+...+\frac{1}{100}\frac{d}{dx}\left(x^{100}\right) \\ =0+1+\frac{1}{2}\times 2x+...+\frac{1}{100}\times 100x^{99}\left(y=x^n\Rightarrow \frac{dy}{dx}=n{x}^{n-1}\right) \\ =1+x+x^2+...+x^{99} \end{gathered}$$

Putting x = 1, we get

$$\begin{gathered} f\text{'}\left(1\right)=1+1+1+...+1\left(100\mathrm{terms}\right) \\ =100 \end{gathered}$$

Hence, the correct answer is option (b).