Question

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

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

Answer 1

Answer: (D)

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
$f^{\prime }(x)=\frac{100x^{99}}{100}+\frac{99x^{98}}{99}+...+\frac{2x}{2}+1+0$
$\Rightarrow f^{\prime }(x)=x^{99}+x^{98}+\dots +x+1$
Put $x=0$, we get
$f^{\prime }(0)=0+0+\dots +0+1$
$\Rightarrow f^{\prime }(0)=1$