Question

If \( f(x) = \log(\sin x) \), \( x \in \left[ \frac{\pi}{6}, \frac{5\pi}{6} \right] \), then the value of \( c \) by applying L.M.V.T. is

• A. \( \frac{\pi}{2} \)
• B. \( \frac{2\pi}{3} \)
• C. \( \frac{3\pi}{4} \)
• D. \( \frac{\pi}{4} \)

Hint:

When applying the Mean Value Theorem, always ensure that the function is continuous and differentiable in the given interval.

Answer 1

The Correct Option is A

Step 1: Applying the Mean Value Theorem.
The Mean Value Theorem (MVT) states that for a function \( f \) continuous on \( [a, b] \) and differentiable on \( (a, b) \), there exists some \( c \in (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] For this problem, \( f(x) = \log(\sin x) \), and we apply the MVT on the interval \( \left[ \frac{\pi}{6}, \frac{5\pi}{6} \right] \).

Step 2: Finding \( f'(x) \).
The derivative of \( f(x) = \log(\sin x) \) is: \[ f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x \] Thus, \( f'(c) = \cot c \). By applying the MVT, we find that \( c = \frac{\pi}{2} \).

Step 3: Conclusion.
Thus, the value of \( c \) is \( \frac{\pi}{2} \), which makes option (A) the correct answer.