Let `f(x)` satisfy the requirements of Lagrange's mean value theorem in [0,1], f(0) = 0 and `f'(x)le1-x, AA x in (0,1)` then

Options

A. `f(x)gex`
B. `|f(x)|ge1`
C. `f(x)le x (1-x)`
D. `f(x)le 1//4`

Answer

To solve the problem, we will follow the steps outlined in the video transcript while providing a clear and structured solution.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We are given a function \( f(x) \) that satisfies the conditions of Lagrange's Mean Value Theorem on the interval \([0, 1]\). We know that:
- \( f(0) = 0 \)
- \( f'(x) \leq 1 - x \) for all \( x \in (0, 1) \)

2. **Applying Lagrange's Mean Value Theorem:**
According to Lagrange's Mean Value Theorem, if \( f(x) \) is continuous on \([0, 1]\) and differentiable on \( (0, 1) \), there exists a point \( c \in (0, 1) \) such that:
\[ f'(c) = \frac{f(1) - f(0)}{1 - 0} \]
Since \( f(0) = 0 \), we have:
\[ f'(c) = f(1) \]

3. **Using the Derivative Condition:**
We know from the problem statement that \( f'(x) \leq 1 - x \). Therefore, at the point \( c \):
\[ f'(c) \leq 1 - c \]
Since we established that \( f'(c) = f(1) \), we can write:
\[ f(1) \leq 1 - c \]

4. **Finding a Bound for \( f(x) \):**
To analyze \( f(x) \) further, we can use the fact that:
\[ f'(x) = \frac{f(x) - f(0)}{x - 0} \Rightarrow f'(x) = \frac{f(x)}{x} \]
Substituting this into the inequality \( f'(x) \leq 1 - x \):
\[ \frac{f(x)}{x} \leq 1 - x \]
Multiplying both sides by \( x \) (valid since \( x > 0 \)):
\[ f(x) \leq x(1 - x) \]

5. **Conclusion:**
From the analysis, we conclude that:
\[ f(x) \leq x(1 - x) \]
for all \( x \in (0, 1) \).

### Final Result:
Thus, the function \( f(x) \) satisfies the condition:
\[ f(x) \leq x(1 - x) \quad \text{for all } x \in (0, 1). \]