Let f:[0,1] `rarr` R be a function . Suppose the fuction f is twice differentiable with f(0) =f(1)=0 and satisfies
`f(x)-2f(x)+f(x) ge e^x " for all " x in [0,1]`.Which of the following is true for x `in (0,1]` ?

Options

A. `0 lt f(x) lt oo`
B. `-1/2 lt f(x)lt 1/2`
C. `-1/4 lt f(x) lt1`
D. `-oo lt f(x) lt 0`

Answer

D

Explanation

We have ,
`f''(x)-2f(x)+f(x) ge e^x " for all " x in [0,1]`
`e^(x)f''(x)-2f''(x)e^(-x)+f(x)e^(-x)f(x)e^(-x) ge 1 " for all " x in [0,1]`
`rArr {e^(-x) f''(x) -e^(-x)f(x)}-{-e^(-x))ge 1 " for all " x in [0,1]`
`rArr d/(dx) f(x)e^)(-x)-f(x)e^(-x) ge 1 " for all " x in [0,1]`
`rArr d/dx{d/dx f(x)e^(-x)} ge 1 " for all " x in [0,1]`
`rArr d^2/dx^2(f(x)d^(-x) ge 1 " for all " x in [0,1]`
`rArr d^2/dx^2( phi (x))ge 1 " for all " x in [0,1] " where " phi (x)=f(x)e^(-x)`
`rArr (x) ` is concave upward on [0,1]
It is given that f(0)=f(1)=0. Therefore `phi(0)= phi(1)=0`
Therefore
`phi(x)lt 0 " for all" x in (0,1) rArr - oo lt f(x) lt 0 " for all " x in (0,1)`