Question

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

• A. $x^{100}$
• B. $100$
• C. $101$
• D. None of these

Answer 1

Answer: (B)

$100$

$Given:\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\therefore forf\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+...............+\frac{x^2}{2}+1f^{\prime }\left(x\right)=\frac{100x^{99}}{100}+99\frac{x^{98}}{99}+.............+\frac{2x}{2}+1Here{f}^{\prime }\left(1\right)=1+1+.........to100term=100Hencef\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+...........+\frac{x^2}{2}+x+1=100$