If `f(x)a n dg(x)` are differentiable functions for `0lt=xlt=1` such that `f(0)=10 ,g(0)=2,f(1)=2,g(1)=4,` then in the interval `(0,1)dot` (a) `f^(prime)(x)=0 for a l lx` (b) `f^(prime)(x)+4g^(prime)(x)=0` for at least one `x` (c) `f(x)=2g'(x)` for at most one `x` (d) none of these

Options

A. `f(x)=0` for all x
B. `f(x)+4g'(x)=0` for at least one x
C. `f(x)=2g'(x)` for at most one x
D. none of these

Answer

To solve the problem, we will use the Mean Value Theorem (MVT) for the differentiable functions \( f(x) \) and \( g(x) \) over the interval \([0, 1]\).

### Step-by-Step Solution:

1. **Identify the given values:**
- \( f(0) = 10 \)
- \( g(0) = 2 \)
- \( f(1) = 2 \)
- \( g(1) = 4 \)

2. **Apply the Mean Value Theorem for \( f(x) \):**
According to MVT, there exists at least one \( c \in (0, 1) \) such that:
\[ f'(c) = \frac{f(1) - f(0)}{1 - 0} \]
Substituting the values:
\[ f'(c) = \frac{2 - 10}{1 - 0} = \frac{-8}{1} = -8 \]

3. **Apply the Mean Value Theorem for \( g(x) \):**
Similarly, for \( g(x) \), there exists at least one \( d \in (0, 1) \) such that:
\[ g'(d) = \frac{g(1) - g(0)}{1 - 0} \]
Substituting the values:
\[ g'(d) = \frac{4 - 2}{1 - 0} = \frac{2}{1} = 2 \]

4. **Relate \( f'(c) \) and \( g'(d) \):**
From the results of MVT, we have:
\[ f'(c) = -8 \quad \text{and} \quad g'(d) = 2 \]
We can express \( f'(c) \) in terms of \( g'(d) \):
\[ f'(c) = -4 \cdot g'(d) \]
This implies:
\[ f'(c) + 4g'(d) = 0 \]

5. **Conclusion:**
Thus, we have shown that:
\[ f'(c) + 4g'(c) = 0 \]
for at least one \( c \in (0, 1) \).

### Answer:
The correct option is (b) \( f'(x) + 4g'(x) = 0 \) for at least one \( x \).