JEE Main & Advanced Mathematics Differentiation Question Bank Self Evaluation Test - Limits and Derivatives

For the function \[f(x)=\frac{{{x}^{100}}}{100}+\frac{{{x}^{99}}}{99}+...\frac{{{x}^{2}}}{2}+x+1.\] \[f'(1)=mf'(0),\] Where m is equal to

A) 50

B) 0

C) 100

D) 200

Correct Answer: C

Solution :

[c] Given,

\[f(x)=\frac{{{x}^{100}}}{100}+\frac{{{x}^{99}}}{99}+...+\frac{{{x}^{2}}}{2}+x+1\]

\[\Rightarrow f'(x)=\frac{100{{x}^{99}}}{100}+\frac{99{{x}^{98}}}{99}+..+\frac{2x}{2}+1+0\]

\[[\because \,\,\,\,f(x)={{x}^{n}}\Rightarrow f'(x)=n{{x}^{n-1}}]\]

\[\Rightarrow f'(x)={{x}^{99}}+{{x}^{98}}+...+x+1\] ? (i)

Putting \[x=1,\] we get

\[f'(1)=\]

\[\underbrace{{{(1)}^{99}}+{{1}^{98}}+...+1+1}_{100times}=\underbrace{1+1+1...+1+1}_{100times}\]

\[\Rightarrow f'(1)=100\] ? (ii)

Again, Putting \[x=0,\]we get

\[f'(0)=0+0+...+0+1\Rightarrow f'(0)=1\] ? (iii)

From eqs. (ii) and (iii), we get; \[f\,'\,(1)=100f\,'\,(0)\]

Hence, \[m=100\]