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If $f(x)=\cos x, g(x)=\cos 2 x, h(x)=\cos 3 x$ and $I(x)$ $=\tan x$, then which of the following option is correct?
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All are correct
All are correct
(a) Let $f(x)=\cos x$, then $f^{\prime}(x)=-\sin x$.
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x) < 0$
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$
(b) Let $f(x)=\cos 2 x \Rightarrow f^{\prime}(x)=-2 \sin x 2 x$
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x) < 0$
Because $\sin 2 x$ will either lie in the first or second quadrant which will give a positive value.
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$
(c) Let $f^{\prime}(x)=\cos 3 x$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-3 \sin 3 \mathrm{x}$. In Interval $\left(0, \frac{\pi}{3}\right), \mathrm{f}^{\prime}(\mathrm{x}) < 0$
Because $\sin 3 \mathrm{x}$ will either lie in the first or second quadrant which will give a positive value.
Therefore, $\mathrm{f}(\mathrm{x})$ irstrictly decreasing on $\left(0, \frac{\pi}{3}\right)$.
When $\mathrm{x} \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$, then $\mathrm{f}^{\prime}(\mathrm{x})>0$
Because $\sin 3 x$ will lie in the third quadrant. Therefore, $f(x)$ is not strictly decreasing on
$\left(0, \frac{\pi}{2}\right)$
(d) Let $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^2 x$.
In Interval $\mathrm{x} \in\left(0, \frac{\pi}{2}\right), \mathrm{f}^{\prime}(\mathrm{x})>0$
Therefore, $f(x)$ is not strictly decreasing on
$\left(0, \frac{\pi}{2}\right)$
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x) < 0$
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$
(b) Let $f(x)=\cos 2 x \Rightarrow f^{\prime}(x)=-2 \sin x 2 x$
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x) < 0$
Because $\sin 2 x$ will either lie in the first or second quadrant which will give a positive value.
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$
(c) Let $f^{\prime}(x)=\cos 3 x$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-3 \sin 3 \mathrm{x}$. In Interval $\left(0, \frac{\pi}{3}\right), \mathrm{f}^{\prime}(\mathrm{x}) < 0$
Because $\sin 3 \mathrm{x}$ will either lie in the first or second quadrant which will give a positive value.
Therefore, $\mathrm{f}(\mathrm{x})$ irstrictly decreasing on $\left(0, \frac{\pi}{3}\right)$.
When $\mathrm{x} \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$, then $\mathrm{f}^{\prime}(\mathrm{x})>0$
Because $\sin 3 x$ will lie in the third quadrant. Therefore, $f(x)$ is not strictly decreasing on
$\left(0, \frac{\pi}{2}\right)$
(d) Let $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^2 x$.
In Interval $\mathrm{x} \in\left(0, \frac{\pi}{2}\right), \mathrm{f}^{\prime}(\mathrm{x})>0$
Therefore, $f(x)$ is not strictly decreasing on
$\left(0, \frac{\pi}{2}\right)$
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