Differentiation Question 111
Question: 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
Options:
A) 50
B) 0
C) 100
D) 200
Show Answer
Answer:
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{100x^{99}}{100}+\frac{99x^{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 $
BETA
