Question

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

• A. 50
• B. 0
• C. 100
• D. 200

Answer 1

Answer: (C)

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