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Answer:

f(x) is neither increasing nor decreasing in (0,2$\pi$)

Given:

$f(x)=cos \: x$

To prove:

We have to prove that f(x) is neither increasing nor decreasing in (0,2$\pi$)

Hint:

For f(x) to be increasing we must have f’(x) > 0

and for f(x) to be decreasing we must have f’(x) < 0

Solution:

Given

$f(x)=cos \: x$

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\cos x) \\ &\Rightarrow f^{\prime}(x)=-\sin x \\ &\text { Since for each } x \in(0, \pi), \\ &\sin x>0 \\ &\Rightarrow-\sin x<0 \end{aligned}

By applying negative sign change comparison sign.

$\Rightarrow f'(x)<0$

Hence f(x) is decreasing function in (0, $\pi$)

Again,

\begin{aligned} &f^{\prime}(x)=-\sin x\\ &\text { Since for each } x \in(\pi, 2 \pi), \sin x<0\\ &\Longrightarrow-\sin x>0 \end{aligned}

By applying negative sign change comparison sign.

$\Rightarrow f'(x)>0$

Clearly, from above we get f(x) is neither increasing nor decreasing in (0,2$\pi$)

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