- #1
klen
- 41
- 1
A function f :[0,1]→ℝ is twice differentiable with f(0)=f(1)=0 such that
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.
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.