Let f:[0,1] `rarr` R be a function.such that `f(0)=f(1)=0 and f''(x)+f(x) ge e^x` for all `x in [0,1]`.If the fucntion `f(x)e^(-x)` assumes its minimum in the interval [0,1] at `x=1/4` which of the following is true ?

Options

A. `f(x)lt 0 f (x) " for " 1/4 lt x lt 3/4`
B. `f(x)ge f(x) for 0 lt x lt 1/4`
C. `f(x)lt f (x) " for " 0 lt x lt 1/4`
D. `f(x)for 3/4 lt x lt`

Answer

C

Explanation

Let `phit(x)=e^(-x)f(x)`.In example 65, we have shown that `phi` (x) is convave upward o [0,1] .It is given that `phi(x)` attains a local minimum at `x=1/4` Therefore ,
`phi(x) le for 0 lt x lt 1/4 and phi (x) gt 0 " for " 1/4 lt x lt 1`
`rArr e^(-x)(f'(x) -f(x)) gt 0 " for " lt x lt 1/4`
and, `e^(-x)(f'(x) -f(x)) gt 0 for 1//4 lt x lt 1 `
`rArr f(x)lt f(x)for lt x lt 1/4 and f(x) gt f(x)for 1/4 lt x lt 1`
Hence ,option (c ) is correct