$If f (x) = \frac{\cos x - \sqrt{3} sinx}{\cos 3 x}, x \neq \frac{π}{6} is continuous on [0, \frac{π}{2}], then f (\frac{π}{6}) =$

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$\frac{2}{3}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

$f (\frac{π}{6}) = \lim_{x \to \frac{π}{6}} f (x) = \lim_{x \to \frac{π}{6}} \frac{- \sin x - \sqrt{3} \cos x}{- 3 \sin 3 x} = \frac{\frac{- 1}{2} - \sqrt{3} (\frac{\sqrt{3}}{2})}{- 3 (1)} = \frac{2}{3}$

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