Let f:[0,1] `rarr` R be a function.such that `f(0)=f(1)=0 and f''(x)+f(x) ge e^x` for all `x in [0,1]`.If the fucntion `f(x)e^(-x)` assumes its minimum in the interval [0,1] at `x=1/4` which of the following is true ?

Options

A. `f(x)lt 0 f (x) " for " 1/4 lt x lt 3/4`
B. `f(x)ge f(x) for 0 lt x lt 1/4`
C. `f(x)lt f (x) " for " 0 lt x lt 1/4`
D. `f(x)for 3/4 lt x lt`

Explanation

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

### Step 1: Define the function
Let \( f: [0, 1] \to \mathbb{R} \) be a function such that:
- \( f(0) = 0 \)
- \( f(1) = 0 \)
- \( f''(x) + f(x) \geq e^x \) for all \( x \in [0, 1] \)

### Step 2: Define a new function
We define a new function:
\[ \phi(x) = e^{-x} f(x) \]
We know that \( \phi(x) \) attains its minimum at \( x = \frac{1}{4} \).

### Step 3: Analyze the behavior of \( \phi(x) \)
Since \( \phi(x) \) has a local minimum at \( x = \frac{1}{4} \), we can conclude that:
- For \( x < \frac{1}{4} \), \( \phi'(x) \leq 0 \) (decreasing)
- For \( x > \frac{1}{4} \), \( \phi'(x) \geq 0 \) (increasing)

### Step 4: Differentiate \( \phi(x) \)
To find \( \phi'(x) \), we apply the product rule:
\[ \phi'(x) = e^{-x} f'(x) - e^{-x} f(x) \]
This can be simplified to:
\[ \phi'(x) = e^{-x} (f'(x) - f(x)) \]

### Step 5: Analyze the sign of \( \phi'(x) \)
From the conditions of the local minimum at \( x = \frac{1}{4} \):
- For \( x < \frac{1}{4} \):
\[ \phi'(x) < 0 \implies f'(x) - f(x) < 0 \implies f'(x) < f(x) \]
- For \( x > \frac{1}{4} \):
\[ \phi'(x) > 0 \implies f'(x) - f(x) > 0 \implies f'(x) > f(x) \]

### Step 6: Conclusion
From the analysis above, we can conclude that:
- \( f'(x) < f(x) \) for \( x < \frac{1}{4} \)
- \( f'(x) > f(x) \) for \( x > \frac{1}{4} \)

This indicates that the function \( f(x) \) is decreasing until \( x = \frac{1}{4} \) and then increasing thereafter.

### Final Answer
Based on the analysis, the correct option that reflects the behavior of \( f(x) \) in the interval [0, 1] is option number 3.

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