Let `f:[0,1]rarrR` (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)-2f\'(x)+f(x) ge e^x, x in [0,1]` Which of the following is true for `0 lt x lt 1` ? (A) `0 lt f(x) lt oo` (B) `-1/2 lt f(x) lt 1/2` (C) `-1/4 lt f(x) lt 1` (D) `-oo lt f(x) lt 0`

Options

A. g is increasing on `(1,oo)`
B. g is decreasing on `(0,1)`
C. g is increasing on (1,2) and decreasing on `(2,oo)`
D. g is decreasing on (1,2) and increasing on `(2,oo)`

Answer

To solve the problem, we need to analyze the given differential inequality and the conditions on the function \( f \).

### Step 1: Understand the given conditions
We have a function \( f: [0, 1] \rightarrow \mathbb{R} \) that is twice differentiable, with boundary conditions:
- \( f(0) = 0 \)
- \( f(1) = 0 \)

Additionally, the function satisfies the inequality:
\[ f''(x) - 2f'(x) + f(x) \geq e^x \quad \text{for } x \in [0, 1] \]

### Step 2: Rewrite the inequality
We can rewrite the inequality as:
\[ f''(x) - 2f'(x) + f(x) - e^x \geq 0 \]

### Step 3: Multiply by \( e^{-x} \)
To simplify the analysis, we multiply the entire inequality by \( e^{-x} \) (which is positive for \( x \in [0, 1] \)):
\[ e^{-x} f''(x) - 2 e^{-x} f'(x) + e^{-x} f(x) - 1 \geq 0 \]

### Step 4: Define a new function
Let:
\[ g(x) = e^{-x} f(x) \]
Now we need to find the first and second derivatives of \( g(x) \).

### Step 5: Compute the first derivative \( g'(x) \)
Using the product rule:
\[ g'(x) = e^{-x} f'(x) - e^{-x} f(x) = e^{-x} (f'(x) - f(x)) \]

### Step 6: Compute the second derivative \( g''(x) \)
Again using the product rule:
\[ g''(x) = e^{-x} (f''(x) - f'(x)) - e^{-x} (f'(x) - f(x)) = e^{-x} (f''(x) - 2f'(x) + f(x)) \]

### Step 7: Analyze \( g''(x) \)
From our earlier inequality, we know:
\[ g''(x) \geq e^{-x} \]
This implies that:
\[ g''(x) \geq 0 \]
indicating that \( g(x) \) is concave upward.

### Step 8: Evaluate \( g(x) \) at the boundaries
Since \( f(0) = 0 \) and \( f(1) = 0 \):
\[ g(0) = e^{0} f(0) = 0 \]
\[ g(1) = e^{-1} f(1) = 0 \]

### Step 9: Analyze the behavior of \( g(x) \)
Since \( g(x) \) is concave upward and \( g(0) = g(1) = 0 \), it must be that \( g(x) \leq 0 \) for \( x \in (0, 1) \). This implies:
\[ e^{-x} f(x) \leq 0 \implies f(x) \leq 0 \quad \text{for } x \in (0, 1) \]

### Step 10: Conclusion
Since \( f(x) \) is non-positive and \( f(0) = f(1) = 0 \), we conclude that:
\[ -\infty < f(x) < 0 \quad \text{for } 0 < x < 1 \]

Thus, the correct option is:
**(D) \(-\infty < f(x) < 0\)**.