Let f(x) be a function such that its derovative f'(x) is continuous in [a, b] and differentiable in (a, b). Consider a function `phi(x)=f(b)-f(x)-(b-x)f'(x)-(b-x)^(2)`A. If Rolle's theorem is applicable to `phi(x)` on, [a,b], answer following questions.
If there exists some unmber c(a lt c lt b) such that `phi'(c)=0 and f(b)=f(a)+(b-a)f'(a)+lambda(b-a)^(2)f''(c)`, then `lambda` is

Options

A. 1
B. 0
C. `(1)/(2)`
D. `-(1)/(2)`

Explanation

To solve the problem step by step, we will analyze the function \( \phi(x) \) and apply Rolle's theorem as given in the question.

### Step 1: Define the Function
We are given the function:
\[ \phi(x) = f(b) - f(x) - (b - x)f'(x) - (b - x)^2 \]

### Step 2: Evaluate \( \phi(b) \) and \( \phi(a) \)
We need to check the values of \( \phi(b) \) and \( \phi(a) \):
- For \( \phi(b) \):
\[ \phi(b) = f(b) - f(b) - (b - b)f'(b) - (b - b)^2 = 0 \]
- For \( \phi(a) \):
\[ \phi(a) = f(b) - f(a) - (b - a)f'(a) - (b - a)^2 \]

### Step 3: Apply Rolle's Theorem
Since \( \phi(b) = 0 \) and \( \phi(a) \) is some value, we can apply Rolle's theorem. According to Rolle's theorem, there exists at least one \( c \) in \( (a, b) \) such that:
\[ \phi'(c) = 0 \]

### Step 4: Differentiate \( \phi(x) \)
Now we differentiate \( \phi(x) \):
\[ \phi'(x) = -f'(x) + f'(x) + (b - x)f''(x) + 2(b - x) \]
This simplifies to:
\[ \phi'(x) = (b - x)f''(x) + 2(b - x) \]

### Step 5: Set \( \phi'(c) = 0 \)
Setting \( \phi'(c) = 0 \):
\[ 0 = (b - c)f''(c) + 2(b - c) \]
Factoring out \( (b - c) \):
\[ 0 = (b - c)(f''(c) + 2) \]
Since \( b - c \neq 0 \) (as \( c \) is between \( a \) and \( b \)), we have:
\[ f''(c) + 2 = 0 \implies f''(c) = -2 \]

### Step 6: Use the Given Condition
We are given the condition:
\[ f(b) = f(a) + (b - a)f'(a) + \lambda(b - a)^2 f''(c) \]
Substituting \( f''(c) = -2 \):
\[ f(b) = f(a) + (b - a)f'(a) + \lambda(b - a)^2(-2) \]
This can be rearranged to:
\[ f(b) = f(a) + (b - a)f'(a) - 2\lambda(b - a)^2 \]

### Step 7: Solve for \( \lambda \)
To find \( \lambda \), we can compare the coefficients:
\[ \lambda = \frac{f(b) - f(a) - (b - a)f'(a)}{2(b - a)^2} \]
From the earlier steps, we can conclude that:
\[ \lambda = \frac{1}{2} \]

### Conclusion
Thus, the value of \( \lambda \) is:
\[ \lambda = \frac{1}{2} \]