Let `f: [0, 8]to R` be differentiable function such that `f (0) =0, f (4)=1, f (8) =1,` then which of the following hold(s) good ?

Options

A. (a) There exist some `c _(1) in (0,8)` where `f '(c _(1)) =1/4 `
B. (b) There exist some `c in (0,8)` where `f '(c ) =1/12`
C. (c) There exist `c _(1), c _(2) in [0,8]` where `8f ' (c _(1)) f (c _(2)) =1`
D. (d) There exist some `alpha, beta=(0,2)` such that `int _(0)^(8)f(t) dt =3 ( alpha ^(2) f (alpha ^(3)) + beta ^(2) (beta ^(3)))`

Answer

To solve the problem, we need to analyze the given differentiable function \( f: [0, 8] \to \mathbb{R} \) with the specified values \( f(0) = 0 \), \( f(4) = 1 \), and \( f(8) = 1 \). We will use the Mean Value Theorem (MVT) and properties of continuous functions to determine which options hold true.

### Step-by-step Solution:

1. **Understanding the Function**:
- We know that \( f \) is differentiable on \( [0, 8] \), which implies that it is also continuous on this interval.

2. **Applying the Mean Value Theorem (MVT)**:
- According to the MVT, if \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \in (a, b) \) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

3. **Checking Option (a)**:
- For \( a = 0 \) and \( b = 8 \):
\[ f'(c) = \frac{f(8) - f(0)}{8 - 0} = \frac{1 - 0}{8} = \frac{1}{8} \]
- Therefore, there exists some \( c_1 \in (0, 8) \) such that \( f'(c_1) = \frac{1}{8} \).
- Now, we check the interval \( [0, 4] \):
\[ f'(c_1) = \frac{f(4) - f(0)}{4 - 0} = \frac{1 - 0}{4} = \frac{1}{4} \]
- Hence, there exists some \( c_1 \in (0, 4) \) such that \( f'(c_1) = \frac{1}{4} \).
- **Conclusion**: Option (a) is correct.

4. **Checking Option (b)**:
- We found that \( f'(c) = \frac{1}{8} \) for some \( c \in (0, 8) \).
- There is no information supporting \( f'(c) = \frac{1}{12} \) since \( \frac{1}{12} \) is not achieved based on the values we calculated.
- **Conclusion**: Option (b) is incorrect.

5. **Checking Option (c)**:
- We need to find \( c_1 \) and \( c_2 \) such that \( 8 \cdot f'(c_1) \cdot f(c_2) = 1 \).
- We already established that \( f'(c_1) = \frac{1}{8} \) for some \( c_1 \in (0, 8) \).
- Since \( f(4) = 1 \) and \( f(8) = 1 \), we can choose \( c_2 = 4 \) or \( c_2 = 8 \).
- Therefore:
\[ 8 \cdot \frac{1}{8} \cdot 1 = 1 \]
- **Conclusion**: Option (c) is correct.

6. **Checking Option (d)**:
- We need to find \( \alpha \) and \( \beta \) in \( (0, 2) \) such that:
\[ \int_0^8 f(t) dt = \alpha^3 + \beta^3 \]
- By applying the Fundamental Theorem of Calculus and properties of integrals, we can find such \( \alpha \) and \( \beta \) that satisfy this equation.
- Since \( f \) is continuous and differentiable, we can find \( \alpha \) and \( \beta \) such that the integral holds true.
- **Conclusion**: Option (d) is correct.

### Final Conclusion:
The correct options are (a), (c), and (d).