- #1

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f''(x) - 2f'(x) + f(x) ≥ e^x, x ε [0,1].

Then for 0<x<1 which of the following is true

0<f(x)<∞, -1/2<f(x)<1/2, -1/4<f(x)<1, -∞<f(x)<0.

I tried to solve it in the following way:

I tried to take the e^x to the left side and found that the equation becomes,

[(e^-x)f'(x) - (e^-x)f(x)]' ≥ 1, thus on integration,

(e^-x)f'(x) - (e^-x)f(x) -f'(0) ≥ x which is same as

[(e^-x)f(x)]' ≥ x on integration

(e^-x)f(x) ≥ (x^2)/2 + f'(0)x.

I don't know how to proceed from here.