Let f:[0,1]→ℝ (the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0)=f(1)=0 and satisfies f''(x) - f'(x) + f(x) ≥ eˣ, x∈[0,1]. Which of the following is true for 0<x<1?

Define a function g(x) = e⁻ˣf(x) ⇒ g'(x) = e⁻ˣ(f'(x) - f(x)) ⇒ g''(x) = e⁻ˣ[f''(x) - f'(x) + f(x)]
Given that f''(x) - f'(x) + f(x) ≥ eˣ, x∈[0,1] ⇒ e⁻ˣ(f''(x) - f'(x) + f(x)) ≥ 1
Hence g''(x) > 1 > 0
So, g(x) is concave upward and g(0) = g(1) = 0
Hence, g(x) < 0 ∀x∈(0,1) ⇒ e⁻ˣf(x) < 0
f(x) < 0 ∀x∈(0,1)