$If f (x) = |\begin{matrix} \cos (2 x) & \cos (2 x) & \sin (2 x) \\ - \cos x & \cos x & - \sin x \\ \sin x & \sin x & \cos x \end{matrix}|, ther$

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$f (x) attains its minimum at x = 0$

$f (x) attains its maximum at x = 0$

$f^{'} (x) = 0 at more than three points in (- π, π)$

$f^{'} (x) = 0 at exactly three points in (- π, π)$

answer is A, C.

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Detailed Solution

$\begin{array}{l} f (x) = |\begin{matrix} 0 & \cos 2 x & \sin 2 x \\ - 2 \cos x & \cos x & - \sin x \\ 0 & \sin x & \cos x \end{matrix}| (C_{1} \to C_{1} - C_{2}) \\ f (x) = 2 \cos x \cos 3 x = \cos 4 x + \cos 2 x \\ f^{'} (x) = - 4 \sin 4 x - 2 \sin 2 x \\ = - 2 \sin 2 x [1 + 4 \cos 2 x] \\ = - 4 \sin x \cos x (8 \cos^{2} x - 3) \end{array}$

$\begin{array}{l} f^{'} (0 +) = - v e; f^{'} (0 -) i s + v e \Rightarrow f (x) has a maximum at x = 0 \\ f^{'} (x) = 0 \Rightarrow \sin 2 x = 0; \cos^{2} x = 3 / 8 \Rightarrow 4 s o l u t i o n s i n (- π, π) \\ f^{'} (x) = 0 \Rightarrow \sin 2 x = 0 \\ \Rightarrow 2 x = n π \\ x = \frac{- π}{2}, \frac{π}{2} \end{array}$