Let (x) satisfy the required of Largrange's Meahn value theorem in [0,3]. If `f(0)=0 and |f'(x)| le (1)/(2) "for all" x in [0,2]` then

Options

A. `f(x)le2`
B. `|f(x)|le1`
C. `f(x)=2x`
D. f(x)=3 for at least one 'x in [0.2]

Answer

B

Explanation

Let `x in(2,0)`. Since f (x)satifies the requirements of Largrange's mean value theorem in [0,2] So, it also satfies in `[0,x]` . Consequenlty , there exist `c in (0,x)` such that
`f'(c)=(f(x)-f(0))/(x-0)`
`rArr f'(c)=(f(x))/(x)`
`rArr |(f(x))/(x)|=|f'(c)|le(1)/(2)" " [ :. |f'(x)|le(1)/(2)]`
`rArr |f(x)|le(|x|)/(2)`
`rArr
|f(x)|le(x)/(2)" " [ :. x le0]`
`rArr |f(x)|le1" " [ :. x in (0,2):.|x| le2]`