Question

Prove that the function f(x) = cos x is:
(i) strictly decreasing in (0, π)
(ii) strictly increasing in (π, 2π)
(iii) neither increasing nor decreasing in (0, 2π)

Answer 1

$$\begin{gathered} f\left(x\right)=\cos x \\ f\text{'}\left(x\right)=-\sin x \\ \left(\mathrm{i}\right) \\ \text{Here,} \\ 0<x<\mathrm{\pi } \\ \Rightarrow \sin x>0\left[\because \mathrm{Sine}\mathrm{function}\mathrm{is}\mathrm{positive}\mathrm{in}\mathrm{first}\mathrm{and}\mathrm{second}\mathrm{quadrant}\right] \\ \Rightarrow -\sin x<0 \\ \Rightarrow f\text{'}\left(x\right)<0,\forall x\in \left(0,\mathrm{\pi }\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is strictly decreasing on}\left(0,\mathrm{\pi }\right). \\ \left(\mathrm{ii}\right) \\ \text{Here,} \\ \mathrm{\pi }<x<2\mathrm{\pi } \\ \Rightarrow \sin \mathrm{x}<0\left[\because \mathrm{Sine}\mathrm{function}\mathrm{is}\mathrm{negative}\mathrm{in}\mathrm{third}\mathrm{and}\mathrm{fourth}\mathrm{quadrant}\right] \\ \Rightarrow -\sin \mathrm{x}>0 \\ \Rightarrow f\mathit{\text{'}}\left(x\right)>0,\forall x\in \left(\mathrm{\pi },2\mathrm{\pi }\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is strictly increasing on}\left(\mathrm{\pi },2\mathrm{\pi }\right). \\ \left(\mathrm{iii}\right) \\ \text{From eqs. (1) and (2), we get} \\ \mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is strictly decreasing on}\left(0,\mathrm{\pi }\right)\text{and is strictly increasing on}\left(\mathrm{\pi },2\mathrm{\pi }\right). \\ \text{So,}f\left(x\right)\text{is neither increasing nor decreasing on}\left(0,2\mathrm{\pi }\right)\text{.} \end{gathered}$$