Let `f(x)=8x^(3)-6x^(2)-2x+1,` then

Options

A. `f(x)=0` has no root in (0, 1)
B. `f(x)=0` has atleast one root in (0, 1)
C. f'(c) vanishes for some `c in (0, 1)`
D. None of tha above

Answer

To solve the problem, we need to analyze the function \( f(x) = 8x^3 - 6x^2 - 2x + 1 \) and determine the behavior of this function in the interval \( [0, 1] \).

### Step-by-Step Solution:

1. **Evaluate \( f(0) \)**:
\[ f(0) = 8(0)^3 - 6(0)^2 - 2(0) + 1 = 1 \]

2. **Evaluate \( f(1) \)**:
\[ f(1) = 8(1)^3 - 6(1)^2 - 2(1) + 1 = 8 - 6 - 2 + 1 = 1 \]

3. **Compare \( f(0) \) and \( f(1) \)**:
\[ f(0) = 1 \quad \text{and} \quad f(1) = 1 \]
Since \( f(0) = f(1) \), we can apply Rolle's Theorem.

4. **Apply Rolle's Theorem**:
- According to Rolle's Theorem, if \( f(a) = f(b) \) for some interval \( [a, b] \), then there exists at least one \( c \) in \( (a, b) \) such that \( f'(c) = 0 \).

5. **Conclusion**:
- Since \( f(0) = f(1) \), there exists at least one \( c \) in the interval \( (0, 1) \) such that \( f'(c) = 0 \).

### Final Answer:
The correct option is that \( f'(c) \) vanishes for some \( c \) belonging to \( (0, 1) \).