Question

For the function
$f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\dots \frac{x^2}{2}+x+1$
$f^{\prime }(1)=m{f}^{\prime }(0)$, where $m$ is equal to

• A. 50
• B. 0
• C. 100
• D. 200

Answer 1

Answer: (C)

Given, $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\dots +\frac{x^2}{2}+x+1$
$\Rightarrow f^{\prime }(x)=\frac{100x^{99}}{100}+\frac{99x^{98}}{99}+\dots +\frac{2x}{2}+1+0$
$\left[\because f(x)=x^n\Rightarrow f^{\prime }(x)=n{x}^{n-1}\right]$
$\Rightarrow f^{\prime }(x)=x^{99}+x^{98}+\dots +x+1$...(i)
Putting $x=1$, we get
$f^{\prime }(1)=\underset{100\text{ times }}{\underbrace{(1{)}^{99}+1^{98}+\dots +1+1}}=\underset{100\text{ times }}{\underbrace{1+1+1\dots +1+1}}$
$\Rightarrow f^{\prime }(1)=100$...(ii)
Again, putting $x=0$, we get
$f^{\prime }(0)=0+0+\dots +0+1$
$\Rightarrow f^{\prime }(0)=1$...(iii)
From eqs. (ii) and (iii), we get
$f^{\prime }(1)=100f^{\prime }(0)$
Hence, $m=100$