Question

If the function $f(x)$ defined by $\frac {x^{100}}{100}+\frac {x^{99}}{100}+\dots \frac {x^2}{100}+x+1$ then $f'(0)$ is equal to

• A. $100$
• B. $-1$
• C. $100 f'(0)$
• D. $1$

Answer 1

Answer: (D)

$1$

Answer 2

Given, $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+...+\frac{x^{2}}{2}+\frac{x}{1}+1$
On differentiating both sides w.r.t. $x$, we get
$f'(x) =\frac{100 x^{99}}{100}+ \frac{99 x^{98}}{99}+...+\frac{2 x}{2}+1+0$
$\Rightarrow f'(x) =x^{99}+ x^{98}+\ldots+x+1$
Put $x=0$, we get
$f'(0) =0+0+\ldots+0+1$
$\Rightarrow f'(0) =1$