Applications Of Derivatives Question 396
Question: Let $ f(x) $ satisfy all the conditions of mean value theorem in [0, 2]. If f (0) = 0 and $ |f’(x)|,\le \frac{1}{2} $ for all x, in [0, 2] then
Options:
A) $ f(x)\le 2 $
B) $ |f(x)|\le 1 $
C) $ f(x)=2x $
D) $ f(x)=3 $ for at least one x in [0, 2]
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{f(2)-f(0)}{2-0}=f’(x)\Rightarrow \frac{f(2)-0}{2}=f’(x) $
$ \Rightarrow \frac{df(x)}{dx}=\frac{f(2)}{2}\Rightarrow f(x)=\frac{f(2)}{2}x+c $
$ \therefore f(0)=0\Rightarrow c=0 $ ;
$ \therefore f(x)=\frac{f(2)}{2}x $ …..(i)
Given $ |f’(x)|\le \frac{1}{2}\Rightarrow | \frac{f(2)}{2} |\le \frac{1}{2} $ …………..(ii) (i)
therefore $ |f(x)|=| \frac{f(2)}{2}x |=| \frac{f(2)}{2} ||x|\le \frac{1}{2}|x| $ [from (ii)]
In [0, 2], for maximum $ x(x=2) $
$ |f(x)|\le \frac{1}{2}.,2\Rightarrow |f(x)|\le 1 $ .
BETA
