Rolle’s theorem can not applicable for :

Options

A. `f(x)=x^(3)-6x^(2)+11x-6" in "[1,3]`
B. `f(x)=sin x" in "[0,pi]`
C. `f(x)=1-(x-1)^(2//3)" in "[0,2]`
D. `f(x)=x^(2)-3x+2" in "[1,2]`

Answer

To determine which option does not satisfy Rolle's Theorem, we need to check each option against the three conditions of the theorem:

1. The function must be continuous on the closed interval [a, b].
2. The function must be differentiable on the open interval (a, b).
3. The values of the function at the endpoints must be equal, i.e., f(a) = f(b).

Let's analyze each option step by step:

### Option 1:
**Function:** \( f(x) = x^3 - 6x^2 + 11x - 6 \) on the interval [1, 3].

1. **Continuity:** Polynomial functions are continuous everywhere. Thus, \( f(x) \) is continuous on [1, 3].
2. **Differentiability:** Polynomial functions are also differentiable everywhere. Thus, \( f(x) \) is differentiable on (1, 3).
3. **Endpoint Values:**
- \( f(1) = 1^3 - 6(1^2) + 11(1) - 6 = 0 \)
- \( f(3) = 3^3 - 6(3^2) + 11(3) - 6 = 0 \)
- Since \( f(1) = f(3) \), the conditions are satisfied.

**Conclusion:** Option 1 satisfies Rolle's Theorem.

### Option 2:
**Function:** \( f(x) = \sin(x) \) on the interval [0, π].

1. **Continuity:** The sine function is continuous everywhere, so it is continuous on [0, π].
2. **Differentiability:** The sine function is differentiable everywhere, so it is differentiable on (0, π).
3. **Endpoint Values:**
- \( f(0) = \sin(0) = 0 \)
- \( f(π) = \sin(π) = 0 \)
- Since \( f(0) = f(π) \), the conditions are satisfied.

**Conclusion:** Option 2 satisfies Rolle's Theorem.

### Option 3:
**Function:** \( f(x) = 1 - (x - 1)^{2/3} \) on the interval [0, 2].

1. **Continuity:** The function \( f(x) \) is continuous on [0, 2].
2. **Differentiability:** We need to check the differentiability:
- The derivative is given by \( f'(x) = -\frac{2}{3}(x - 1)^{-1/3} \).
- This derivative is not defined at \( x = 1 \) (since it leads to division by zero), hence \( f(x) \) is not differentiable at \( x = 1 \).
3. **Endpoint Values:**
- \( f(0) = 1 - (0 - 1)^{2/3} = 1 - 1 = 0 \)
- \( f(2) = 1 - (2 - 1)^{2/3} = 1 - 1 = 0 \)
- Since \( f(0) = f(2) \), the condition for equal endpoint values is satisfied.

**Conclusion:** Option 3 does not satisfy the differentiability condition of Rolle's Theorem.

### Option 4:
**Function:** \( f(x) = x^2 - 5x + 6 \) on the interval [1, 2].

1. **Continuity:** Polynomial functions are continuous everywhere, so \( f(x) \) is continuous on [1, 2].
2. **Differentiability:** Polynomial functions are also differentiable everywhere, so \( f(x) \) is differentiable on (1, 2).
3. **Endpoint Values:**
- \( f(1) = 1^2 - 5(1) + 6 = 2 \)
- \( f(2) = 2^2 - 5(2) + 6 = 0 \)
- Since \( f(1) \neq f(2) \), the condition for equal endpoint values is not satisfied.

**Conclusion:** Option 4 does not satisfy Rolle's Theorem.

### Final Answer:
Rolle's Theorem is not applicable for **Option 3** and **Option 4**. However, since the question asks for the option that cannot be applicable, we conclude that **Option 3** is the most relevant answer as it fails the differentiability condition.

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