The graph of the function f(x) = frac{cos x}{ | vedclass

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(C) Given $f(x) = \frac{\cos x}{\cos 2x}$.Using the quotient rule,$f'(x) = \frac{-\sin x \cos 2x - \cos x (-2 \sin 2x)}{(\cos 2x)^2} = \frac{-\sin x \

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decreasing on $\left(-\frac{\pi}{4}, 0\right)$ and increasing on $\left(0, \frac{\pi}{4}\right)$

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(C) Given $f(x) = \frac{\cos x}{\cos 2x}$.Using the quotient rule,$f'(x) = \frac{-\sin x \cos 2x - \cos x (-2 \sin 2x)}{(\cos 2x)^2} = \frac{-\sin x \cos 2x + 2 \sin 2x \cos x}{(\cos 2x)^2}$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have $f'(x) = \frac{\sin(2x-x) + \sin 2x \cos x}{(\cos 2x)^2} = \frac{\sin x + 2 \sin x \cos^2 x}{(\cos 2x)^2}$.$f'(x) = \frac{\sin x (1 + 2 \cos^2 x)}{(\cos 2x)^2}$.Since $(1 + 2 \cos^2 x) > 0$ and $(\cos 2x)^2 > 0$ for all $x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$,the sign of $f'(x)$ depends only on $\sin x$.For $x \in \left(-\frac{\pi}{4}, 0\right)$,$\sin x For $x \in \left(0, \frac{\pi}{4}\right)$,$\sin x > 0$,so $f'(x) > 0$,meaning $f(x)$ is increasing.Thus,the correct option is $C$.

The graph of the function $f(x) = \frac{\cos x}{\cos 2x}$ in the domain $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ is

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