If f(x) = log(sin x),x in left[frac{pi}{6}, f | vedclass

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(A) According to Lagrange's Mean Value Theorem,there exists at least one $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = \l

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$\frac{\pi}{2}$

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(A) According to Lagrange's Mean Value Theorem,there exists at least one $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = \log(\sin x)$ on $\left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$,we have $a = \frac{\pi}{6}$ and $b = \frac{5\pi}{6}$.First,calculate the derivative: $f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.Thus,$f'(c) = \cot c$.Next,calculate $f(a)$ and $f(b)$:$f\left(\frac{\pi}{6}\right) = \log(\sin(\frac{\pi}{6})) = \log(\frac{1}{2})$$f\left(\frac{5\pi}{6}\right) = \log(\sin(\frac{5\pi}{6})) = \log(\frac{1}{2})$Now,apply the formula:$f'(c) = \frac{\log(\frac{1}{2}) - \log(\frac{1}{2})}{\frac{5\pi}{6} - \frac{\pi}{6}} = \frac{0}{\frac{4\pi}{6}} = 0$.Therefore,$\cot c = 0$,which implies $c = \frac{\pi}{2}$.

If $f(x) = \log(\sin x)$,$x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$,then the value of $c$ by applying Lagrange's Mean Value Theorem $(LMVT)$ is:

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