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Let f:[0,1]rarrR be a function. Suppose ...

Let `f:[0,1]rarrR` be a function. Suppose the function `f` is twice differentiable, `f(0)=f(1)=0` and satisfies `f\'\'(x)-2f\'(x)+f(x) ge e^x, x in [0,1]` Which of the following is true for `0 lt x lt 1 ?`

A

`0ltf(x)ltoo`

B

`-1/2ltf(x)lt1/2`

C

`-1/4lt f(x)lt1`

D

`-ooltf(x)lt0`

Text Solution

Verified by Experts

The correct Answer is:
4
Give `f(x)-2f(x)+f(x)ge^(x)`
The terms to the L.H.S suggest htat we have to consides function `g(x)=e^(-x)f(x)`
Now `g(X)=f(x)e^(-x)-e^(-x)f(x)`
and `g(X)=f(x)e^(-x)-f(x)e^(-x)-e^(-x)f(x)+e^(-x)f(x)`
`=e^(-x)f(x)-2f(x)+f(x)`
Give that f(x)-2f(x)-2f(x)+f(x)
Given that `f(x)-2f(x)+f(x)gee^(x)`
or `g(x)=e^(-x)(fx)-2f(x)ge1ge0`
or g(X) is concave upwared and g(0)=g(1)=0
Hence `g(X)lt0forallx in (0,1) or e^(-x)f(x)lt 0` or
`f(X)lt0forall x in (0,1)`
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