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]` If the function `e^(-x)f(x)` assumes its minimum in the interval `[0,1]` at `x=1/4`, which of the following is true? (A) `f\'(x) lt f(x), 1/4 lt x lt 3/4` (B) `f\'(x) gt f(x), `0 lt x lt 1/4` (C) `f\'(x) lt f(x), 0 lt x lt 1/4` (D) `f\'(x) lt f(x), 3/4 lt x lt 1`

Options

A. `f(x) lt f(x) , (1)/(4) lt x lt (3)/(4)`
B. `f(x)' gt f (x) 0 lt x lt 1/4`
C. `f(x) lt F( x) , 0 lt x lt 1/4`
D. ` f'(x) lt f (x) , (3) /(4)lt x lt 1`

Answer

To solve the problem, we need to analyze the given conditions and derive the necessary conclusions step by step.

### Step 1: Understand the Given Information
We have a function \( f: [0, 1] \to \mathbb{R} \) that is twice differentiable, with boundary conditions \( f(0) = f(1) = 0 \). The function satisfies the inequality:
\[ f''(x) - 2f'(x) + f(x) \geq e^x \quad \text{for } x \in [0, 1]. \]
Additionally, the function \( g(x) = e^{-x} f(x) \) attains its minimum at \( x = \frac{1}{4} \).

### Step 2: Analyze the Minimum Condition
Since \( g(x) \) has a minimum at \( x = \frac{1}{4} \), we know that:
\[ g'(x) = 0 \quad \text{at } x = \frac{1}{4}. \]
Calculating \( g'(x) \):
\[ g'(x) = e^{-x} f'(x) - e^{-x} f(x). \]
Setting this equal to zero gives:
\[ e^{-x} (f'(x) - f(x)) = 0. \]
Since \( e^{-x} \) is never zero, we have:
\[ f'(x) - f(x) = 0 \quad \Rightarrow \quad f'(x) = f(x). \]

### Step 3: Determine the Sign of \( f'(x) - f(x) \)
To determine the behavior of \( f'(x) \) relative to \( f(x) \) around \( x = \frac{1}{4} \), we analyze the second derivative:
\[ g''(x) = -e^{-x} f'(x) + e^{-x} f(x) - e^{-x} f'(x) + e^{-x} f(x) = -2e^{-x} f'(x) + 2e^{-x} f(x). \]
Since \( g(x) \) has a minimum at \( x = \frac{1}{4} \), we have:
\[ g''\left(\frac{1}{4}\right) \geq 0. \]
This implies:
\[ -2f'\left(\frac{1}{4}\right) + 2f\left(\frac{1}{4}\right) \geq 0 \quad \Rightarrow \quad f'\left(\frac{1}{4}\right) \leq f\left(\frac{1}{4}\right). \]

### Step 4: Analyze the Intervals
1. For \( 0 < x < \frac{1}{4} \):
- Since \( g(x) \) is decreasing before \( x = \frac{1}{4} \), we have \( g'(x) < 0 \), which implies \( f'(x) < f(x) \).

2. For \( \frac{1}{4} < x < 1 \):
- Since \( g(x) \) is increasing after \( x = \frac{1}{4} \), we have \( g'(x) > 0 \), which implies \( f'(x) > f(x) \).

### Conclusion
From our analysis, we conclude:
- For \( 0 < x < \frac{1}{4} \), \( f'(x) < f(x) \).
- For \( \frac{1}{4} < x < 1 \), \( f'(x) > f(x) \).

Thus, the correct option is:
**(C) \( f'(x) < f(x) \) for \( 0 < x < \frac{1}{4} \)**.