To solve the problem, we need to analyze the properties of the function \( f \) given the conditions stated in the question.
### Step 1: Understand the given information
We have a function \( f: [a, b] \to \mathbb{R} \) that is continuous and differentiable on the interval \([a, b]\). We know:
- \( f(a) = 5 \)
- \( f'(x) \leq 0 \) for all \( x \in [a, b] \)
### Step 2: Apply the Mean Value Theorem (MVT)
Since \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), we can apply the Mean Value Theorem. According to MVT, there exists some \( c \in (a, b) \) such that:
\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]
Given that \( f'(c) \leq 0 \), we can deduce:
\[
\frac{f(b) - f(a)}{b - a} \leq 0
\]
### Step 3: Analyze the inequality
Since \( b - a > 0 \) (as \( b > a \)), we can multiply both sides of the inequality by \( b - a \) without changing the inequality:
\[
f(b) - f(a) \leq 0
\]
This implies:
\[
f(b) \leq f(a)
\]
Given that \( f(a) = 5 \), we have:
\[
f(b) \leq 5
\]
### Step 4: Apply MVT again for \( f(\lambda) \)
Now, we apply the Mean Value Theorem again for the interval \([a, \lambda]\) where \( \lambda \in (a, b)\). There exists some \( d \in (a, \lambda) \) such that:
\[
f'(d) = \frac{f(\lambda) - f(a)}{\lambda - a}
\]
Again, since \( f'(d) \leq 0 \):
\[
\frac{f(\lambda) - f(a)}{\lambda - a} \leq 0
\]
This leads to:
\[
f(\lambda) - f(a) \leq 0
\]
Thus:
\[
f(\lambda) \leq f(a) = 5
\]
### Step 5: Combine results
Now we have:
- \( f(b) \leq 5 \)
- \( f(\lambda) \leq 5 \)
Adding these two inequalities gives:
\[
f(b) + f(\lambda) \leq 5 + 5 = 10
\]
### Step 6: Determine the interval
Since \( f(b) \) and \( f(\lambda) \) can take any value less than or equal to 5, the sum \( f(b) + f(\lambda) \) can take values from \(-\infty\) up to 10. Therefore, the final result is:
\[
f(b) + f(\lambda) \in (-\infty, 10]
\]
### Conclusion
The correct answer is:
**B. \((- \infty, 10]\)**
