If f(x) and g(x) ar edifferentiable function for `0lex le1` such that `f(0)=2,g(0) = 0,f(1)=6,g(1)=2`, then in the interval (0,1)

Options

A. `f'(x)=0` for all x
B. `f'(x)=2g'(x)` for at leaset one x
C. `f'(x)=2g'(x)` for at most one x
D. none of these

Answer

To solve the problem, we will apply Lagrange's Mean Value Theorem (LMVT) to the functions \( f(x) \) and \( g(x) \) on the interval \([0, 1]\).

### Step-by-step Solution:

1. **Identify the Functions and Their Values**:
- We have two differentiable functions \( f(x) \) and \( g(x) \).
- Given values:
- \( f(0) = 2 \)
- \( f(1) = 6 \)
- \( g(0) = 0 \)
- \( g(1) = 2 \)

2. **Apply Lagrange's Mean Value Theorem**:
- According to LMVT, if a function 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:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
- For \( f(x) \):
- Here, \( a = 0 \) and \( b = 1 \).
- So, we calculate:
\[ f'(c) = \frac{f(1) - f(0)}{1 - 0} = \frac{6 - 2}{1 - 0} = \frac{4}{1} = 4 \]

3. **Apply LMVT to \( g(x) \)**:
- Similarly, we apply LMVT to \( g(x) \):
\[ g'(d) = \frac{g(1) - g(0)}{1 - 0} = \frac{2 - 0}{1 - 0} = \frac{2}{1} = 2 \]
- Here, \( d \) is some point in the interval \((0, 1)\).

4. **Relate the Derivatives**:
- From the above calculations, we have:
- \( f'(c) = 4 \)
- \( g'(d) = 2 \)
- We can establish a relationship between the derivatives:
\[ f'(c) = 2 \cdot g'(d) \]
- This implies that at least one point \( x \) in the interval \((0, 1)\) satisfies:
\[ f'(x) = 2g'(x) \]

### Conclusion:
Thus, we conclude that there exists at least one point \( x \) in the interval \((0, 1)\) such that \( f'(x) = 2g'(x) \).