Let f(x) satisfy the requirements of Lagrang's mean value theorem in [0, 2]. If f(0)=0 and `|f'(x)|le1//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]"`

Answer

B

Explanation

`(f(2)-f_(0))/(2-0)=f' (x) rArr (f(2)-0)/(2)=f'(x)`
`rArr (df(x))/(dx)=(f(2))/(2) rArr f(x)=(f(2))/(2)x+c`
`therefore f(0)=0 rArr c=0, therefore f(x)=(f(2))/(2)x ......(i)`
Given `|f'(x) le 1/2 rArr |(f(2))/(2)| le 1/2 ........(ii)`
(i) `|f(x)=|(f(2))/(2)x|=|(f(2))/(2)| |x| le 1/2 |x| "(from (ii))"`
In [0,2] for maximum `x(x=2) rArr |f(x)| le1`