Which of the following functions are strictly | vedclass

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(A, B) Let $f_{1}(x) = \cos x$. Then $f_{1}^{\prime}(x) = -\sin x$. In the interval $\left(0, \frac{\pi}{2}\right)$,$\sin x > 0$,so $f_{1}^{\prime}(x)

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$\cos 2x$

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(A, B) Let $f_{1}(x) = \cos x$. Then $f_{1}^{\prime}(x) = -\sin x$. In the interval $\left(0, \frac{\pi}{2}ight)$,$\sin x > 0$,so $f_{1}^{\prime}(x) = -\sin x Thus,$f_{1}(x) = \cos x$ is strictly decreasing on $\left(0, \frac{\pi}{2}ight)$. Let $f_{2}(x) = \cos 2x$. Then $f_{2}^{\prime}(x) = -2 \sin 2x$. For $0  0$. Thus,$f_{2}^{\prime}(x) = -2 \sin 2x Therefore,$f_{2}(x) = \cos 2x$ is strictly decreasing on $\left(0, \frac{\pi}{2}ight)$. Let $f_{3}(x) = \cos 3x$. Then $f_{3}^{\prime}(x) = -3 \sin 3x$. For $x \in \left(0, \frac{\pi}{2}ight)$,$3x \in (0, \frac{3\pi}{2})$. In $\left(0, \frac{\pi}{3}ight)$,$\sin 3x > 0$,so $f_{3}^{\prime}(x) In $\left(\frac{\pi}{3}, \frac{\pi}{2}ight)$,$\sin 3x  0$. Thus,$f_{3}(x)$ is not strictly decreasing on $\left(0, \frac{\pi}{2}ight)$. Let $f_{4}(x) = 	an x$. Then $f_{4}^{\prime}(x) = \sec^{2} x > 0$ for all $x \in \left(0, \frac{\pi}{2}ight)$. Thus,$f_{4}(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}ight)$. Conclusion: Both $\cos x$ and $\cos 2x$ are strictly decreasing on $\left(0, \frac{\pi}{2}ight)$.

Which of the following functions are strictly decreasing on $\left(0, \frac{\pi}{2}\right)?$

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