1.

Let f:[0,1] rarr R be a function . Suppose the fuction f is twice differentiable with f(0) =f(1)=0and 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] ?

Answer»

`0 lt F(x) lt oo`
`-1/2 lt f(x)lt 1/2`
`-1/4 lt f(x) lt1`
`-oo lt f(x) lt 0`

Solution :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]`
`rArrd/(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]`
`rArrd^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)`


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