Step-by-step Solution:

1. Analyze the Differential Inequality

The given differential inequality is $f''(x) - 2f'(x) + f(x) ≥ e^x$ for x ∈ [0, 1]. The left-hand side (LHS) resembles the result of applying a linear differential operator. Specifically, it can be expressed as the second derivative of a product involving f(x) and an exponential function.

2. Introduce an Auxiliary Function

Let's define an auxiliary function g(x) to simplify the inequality. A good choice for g(x) is inspired by the method of solving second-order linear ODEs. Let's define: $g(x) = e^(-x) f(x)$.

3. Rewrite the Inequality in Terms of g(x)

We need to find the second derivative of g(x). First derivative: $g'(x) = d/dx (e^(-x) f(x)) = -e^(-x) f(x) + e^(-x) f'(x) = e^(-x) (f'(x) - f(x))$

Second derivative: $g''(x) = d/dx [e^(-x) (f'(x) - f(x))]$ Using the product rule: $g''(x) = -e^(-x) (f'(x) - f(x)) + e^(-x) (f''(x) - f'(x))$ $g''(x) = e^(-x) [-f'(x) + f(x) + f''(x) - f'(x)]$ $g''(x) = e^(-x) [f''(x) - 2f'(x) + f(x)]$

Now, we can use the original inequality $f''(x) - 2f'(x) + f(x) ≥ e^x$. Multiplying both sides by $e^(-x)$ (which is always positive, so the inequality sign is preserved): $e^(-x) [f''(x) - 2f'(x) + f(x)] ≥ e^(-x) * e^x$ Substituting g''(x) for the LHS, we get: g''(x) ≥ 1

4. Analyze the Auxiliary Function g(x)

We have established that g''(x) ≥ 1 for x ∈ [0, 1]. Let's find the values of g(x) at the boundaries of the interval. We are given f(0) = 0 and f(1) = 0.

$g(0) = e^(-0) f(0) = 1 * 0 = 0$ $g(1) = e^(-1) f(1) = e^(-1) * 0 = 0$

So, our auxiliary function g(x) has the following properties:

• g(0) = 0
• g(1) = 0
• g''(x) ≥ 1 for x ∈ [0, 1]

The condition g''(x) ≥ 1 > 0 implies that g(x) is a strictly convex function on the interval [0, 1] (its graph is concave up).

5. Apply Properties of Convex Functions

A property of a convex function on an interval [a, b] is that its graph lies on or below the line segment (chord) connecting the points (a, g(a)) and (b, g(b)).

For g(x) on [0, 1], the endpoints are (0, g(0)) = (0, 0) and (1, g(1)) = (1, 0). The chord connecting these points is the line segment y = 0 for x ∈ [0, 1].

Therefore, for any x ∈ (0, 1), the value of g(x) must be less than or equal to the value on the chord, which is 0. So, g(x) ≤ 0 for x ∈ (0, 1).

6. Prove Strict Inequality

Let's check if g(x) can be equal to 0 for some x₀ ∈ (0, 1). Assume, for the sake of contradiction, that g(x₀) = 0 for some x₀ ∈ (0, 1).

We would have g(0) = g(x₀) = g(1) = 0. By applying Rolle's Theorem to g(x) on the intervals [0, x₀] and [x₀, 1], we can conclude that there exist c₁ ∈ (0, x₀) and c₂ ∈ (x₀, 1) such that g'(c₁) = 0 and g'(c₂) = 0.

Now, applying Rolle's Theorem again, this time to the function g'(x) on the interval [c₁, c₂], there must exist some c₃ ∈ (c₁, c₂) ⊂ (0, 1) such that g''(c₃) = 0.

This contradicts our established result that g''(x) ≥ 1 for all x ∈ [0, 1]. Therefore, our assumption that g(x₀) = 0 for some x₀ ∈ (0, 1) must be false.

Combining g(x) ≤ 0 and g(x) ≠ 0, we conclude that g(x) < 0 for x ∈ (0, 1).

7. Relate Back to f(x)

We defined $g(x) = e^(-x) f(x)$. Since we found that g(x) < 0 for x ∈ (0, 1): $e^(-x) f(x) < 0$

The exponential function $e^(-x)$ is always positive for any real x. Therefore, for the product to be negative, f(x) must be negative.

f(x) < 0 for all x ∈ (0, 1).

8. Evaluate the Options

The result f(x) < 0 means that for any x in the open interval (0, 1), the value of f(x) is a negative real number. This can be expressed as -∞ < f(x) < 0.

Let's check the given options:

• A: 0 < f(x) < ∞ - Incorrect, as f(x) is negative.
• B: -1/2 < f(x) < 1/2 - Incorrect, as it allows for positive values of f(x).
• C: -1/4 < f(x) < 1 - Incorrect, as it allows for positive values of f(x).
• D: -∞ < f(x) < 0 - This is a true statement that correctly describes our finding that f(x) is always negative for x ∈ (0, 1).

Therefore, the correct option is D.