The value of c in Largrange's theorem for the function `f(x)= log_(e) sin x` in the interval `[ pi//6,5pi//6]` is

Options

A. `pi//4`
B. `pi//2`
C. `2pi//3`
D. none of these

Answer

To find the value of \( c \) in Lagrange's theorem for the function \( f(x) = \log(\sin x) \) in the interval \([ \frac{\pi}{6}, \frac{5\pi}{6} ]\), we will follow these steps:

### Step 1: Identify the endpoints of the interval
Let \( a = \frac{\pi}{6} \) and \( b = \frac{5\pi}{6} \).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
We need to find \( f(a) \) and \( f(b) \):

- For \( f(a) \):
\[ f\left(\frac{\pi}{6}\right) = \log\left(\sin\left(\frac{\pi}{6}\right)\right) = \log\left(\frac{1}{2}\right) = -\log(2) \]

- For \( f(b) \):
\[ f\left(\frac{5\pi}{6}\right) = \log\left(\sin\left(\frac{5\pi}{6}\right)\right) = \log\left(\sin\left(\frac{\pi}{6}\right)\right) = \log\left(\frac{1}{2}\right) = -\log(2) \]

### Step 3: Apply Lagrange's Mean Value Theorem
According to Lagrange's Mean Value Theorem, there exists at least one \( c \) in the interval \( (a, b) \) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Substituting the values we found:
\[ f'(c) = \frac{-\log(2) - (-\log(2))}{\frac{5\pi}{6} - \frac{\pi}{6}} = \frac{0}{\frac{4\pi}{6}} = 0 \]

### Step 4: Find the derivative \( f'(x) \)
Now we need to find \( f'(x) \):
\[ f(x) = \log(\sin x) \]
Using the chain rule, we have:
\[ f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x \]

### Step 5: Set \( f'(c) = 0 \)
From our earlier calculation:
\[ f'(c) = \cot(c) = 0 \]

### Step 6: Solve for \( c \)
The cotangent function is zero when:
\[ c = \frac{\pi}{2} \]

### Conclusion
Thus, the value of \( c \) in Lagrange's theorem for the function \( f(x) = \log(\sin x) \) in the interval \([ \frac{\pi}{6}, \frac{5\pi}{6} ]\) is:
\[ c = \frac{\pi}{2} \]