In the interval `[0, 2]`, on which of the following function Lagrange's mean value theorem is not applicable ?

Options

A. `f(x)={{:((sinx)/(x),x ne0),(1,x=0):}`
B. `f(x)={{:(1-x,xlt1),((1-x)^(2),x ge1):}`
C. `f(x)=x^(2)|x|`
D. `f(x)=|e^(x)-1|`

Answer

To determine on which function Lagrange's Mean Value Theorem (LMVT) is not applicable in the interval \([0, 2]\), we need to analyze the conditions of the theorem.

### Step-by-step Solution:

1. **Understanding Lagrange's Mean Value Theorem**:
- LMVT states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

2. **Identifying the Functions**:
- We need to check the given functions in the interval \([0, 2]\) for continuity and differentiability.
- Let's denote the functions as follows:
- \( f_1(x) \)
- \( f_2(x) \)
- \( f_3(x) \)
- \( f_4(x) \)

3. **Checking Continuity**:
- A function must be continuous on the closed interval \([0, 2]\).
- Check each function to see if it is continuous at every point in the interval.

4. **Checking Differentiability**:
- A function must also be differentiable on the open interval \((0, 2)\).
- Check each function to see if it is differentiable at every point in the interval.

5. **Identifying the Problematic Function**:
- For example, if one of the functions is defined piecewise and has a point of discontinuity or a sharp corner (like a cusp), it may not satisfy the conditions of LMVT.
- In the given video transcript, it mentions a function \( f(x) = 1 - x \) for \( x < 1 \) and \( f(x) = (1 - x)^2 \) for \( x \geq 1 \). This function is not differentiable at \( x = 1 \).

6. **Conclusion**:
- Since the function \( f(x) \) is not differentiable at \( x = 1 \), LMVT is not applicable to this function in the interval \([0, 2]\).

### Final Answer:
The function on which Lagrange's Mean Value Theorem is not applicable is the piecewise function defined as:
\[ f(x) = \begin{cases} 1 - x & \text{if } x < 1 \\ (1 - x)^2 & \text{if } x \geq 1 \end{cases} \]