Prove that the function f(x)= is: Strictly increasing in ( ,2 ) - Vidyadip

Question

Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Strictly increasing in $(\pi,2\pi)$

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Answer

$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
Here,
$\pi<\text{x}<2\pi$
$\Rightarrow\sin\text{x}<0$ $[\because$ sine function is negative in third and fourth quadrent$]$
$\Rightarrow-\sin\text{x}>0$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in(\pi,2\pi)$
So, f(x) is strictly decreasing on $(\pi,2\pi).$

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