Question

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

• A. 50
• B. 0
• C. 100
• D. 200

Answer 1

Answer: (C)

100

Answer 2

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