Let f:[0,1] rarr R be a function.such thatf(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 ?

Let f:[0,1] rarr R be a function.such thatf(0)=f(1)=0 and f''(x)+f(x)

1.	Let f:[0,1] rarr R be a function.such thatf(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 ?
Answer» `f(x)lt 0 f (x) " for " 1/4 lt x lt 3/4` `f(x)ge f(x) for 0 lt x lt 1/4` `f(x)lt f (x) " for " 0 lt x lt 1/4` `f(x)for 3/4 lt x lt` Solution :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 FOR0 ltx lt 1/4 and phi (x) gt 0" for " 1/4 lt x lt 1` `rArre^(-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

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Let f:[0,1] rarr R be a function.such thatf(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 ?

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