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Question: Answered & Verified by Expert
Show that the function given by $f(x)=\sin x$ is
(a) strictly increasing in $\left(0, \frac{\pi}{2}\right)$
(b) strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$
(c) neither increasing nor decreasing in $(0, \pi)$
MathematicsApplication of Derivatives
Solution:
2401 Upvotes Verified Answer
We have $\mathrm{f}(\mathrm{x})=\sin \mathrm{x} \quad \therefore \mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}$
(a) $f^{\prime}(x)=\cos x$ is $+$ ve in the interval $\left(0, \frac{\pi}{2}\right)$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$
(b) $\mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}$ is a-ve in the interval $\left(\frac{\pi}{2}, \pi\right)$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$
(c) $\mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}$ is +ve in the interval $\left(0, \frac{\pi}{2}\right)$ while $\mathrm{f}^{\prime}(\mathrm{x})$
is -ve in the interval $\left(\frac{\pi}{2}, \pi\right)$
$\therefore \mathrm{f}(\mathrm{x})$ is neither increasing nor decreasing in $(0, \pi)$

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