If f(x) = left| begin{array}{ccc} cos(2x) & c | vedclass

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(D) Expanding the determinant $f(x)$:$f(x) = \cos(2x)(\cos x \cos x - (-\sin x \sin x)) - \cos(2x)(-\cos x \cos x - (-\sin x \sin x)) + \sin(2x)(-\cos

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$B, C$

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(D) Expanding the determinant $f(x)$:$f(x) = \cos(2x)(\cos x \cos x - (-\sin x \sin x)) - \cos(2x)(-\cos x \cos x - (-\sin x \sin x)) + \sin(2x)(-\cos x \sin x - \cos x \sin x)$$f(x) = \cos(2x)(\cos^2 x + \sin^2 x) - \cos(2x)(-\cos^2 x + \sin^2 x) + \sin(2x)(-2 \sin x \cos x)$$f(x) = \cos(2x)(1) - \cos(2x)(-\cos 2x) + \sin(2x)(-\sin 2x)$$f(x) = \cos 2x + \cos^2 2x - \sin^2 2x = \cos 2x + \cos 4x$.Now,$f'(x) = -2 \sin 2x - 4 \sin 4x = -2 \sin 2x - 8 \sin 2x \cos 2x = -2 \sin 2x (1 + 4 \cos 2x)$.Setting $f'(x) = 0$,we get $\sin 2x = 0$ or $\cos 2x = -1/4$.In $(-\pi, \pi)$,$\sin 2x = 0$ at $x = 0, \pm \pi/2$. ($3$ points)$\cos 2x = -1/4$ has $4$ solutions in $(-\pi, \pi)$.Thus,$f'(x) = 0$ at $3 + 4 = 7$ points,which is more than $3$. So $B$ is correct.For $f(x) = \cos 2x + \cos 4x$,at $x = 0$,$f(0) = \cos 0 + \cos 0 = 2$.Since $\cos 	heta \le 1$,the maximum value of $f(x)$ is $2$,which occurs at $x = 0$. So $C$ is correct.Therefore,the correct options are $B$ and $C$.

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