Application of Derivatives — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
Question
Is the function $cos3x$ decreasing on $(0, \frac{\pi}{2})$?
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Answer
Let f(x) = cos 3x $\therefore$ $\mathrm{f}^{\prime}(\mathrm{x})$ = -3 sin 3x Now, $f^\prime(x)$ = 0 $\Rightarrow$ sin 3x = 0 $\Rightarrow$ 3x = $\pi$, as $x \in\left(0, \frac{\pi}{2}\right)$ $\Rightarrow x=\frac{\pi}{3}$ The point $x=\frac{\pi}{3}$ divides the interval $\left(0, \frac{\pi}{2}\right)$ into two distinct intervals. i.e. $\left(0, \frac{\pi}{3}\right)$ and $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ Now, in the interval, $\left(0, \frac{\pi}{3}\right)$ $f^{\prime}(x)=-3 \sin 3 x<0 \text { as }\left(0<x<\frac{\pi}{3} \Rightarrow 0<3 x<\pi\right)$ Therefore, 'f ' is strictly decreasing in interval $\left(0, \frac{\pi}{3}\right)$ Now, in the interval $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ $f^{\prime}(x)=-3 \sin 3 x>0$ as $\frac{\pi}{3}<\mathrm{x}<\frac{\pi}{2} \Rightarrow \pi<3 \mathrm{x}<\frac{3 \pi}{2}$ Therefore, 'f ' is strictly increasing in the interval $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$.
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