Show that the function f(x)= ^ -1 ( + ) is decreasing on (0, 4 ) … - Vidyadip

Question

Show that the function $\text{f}(\text{x})=\cot^{-1}(\sin\text{x}+\cos\text{x})$ is decreasing on $\Big(0,\frac{\pi}{4}\Big)$ and increasing on $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$

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Answer

$\text{f}(\text{x})=\cot^{-1}(\sin\text{x}+\cos\text{x})$
$\text{f}'(\text{x})=\frac{1}{1+(\sin\text{x}+\cos\text{x})^2}\times(\cos\text{x}-\sin\text{x})$
$\text{f}'(\text{x})=-\frac{(\cos\text{x}-\sin\text{x})}{2+2\sin\text{x}\cos\text{x}}$
For f(x) to be increasing, we must have f'(x) > 0
$\Rightarrow-\frac{(\cos\text{x}-\sin\text{x})}{2+2\sin\text{x}\cos\text{x}}>0$
$\Rightarrow-(\cos\text{x}-\sin\text{x})>0$
$\Rightarrow\sin\text{x}>\cos\text{x}$
$\Rightarrow\text{x}\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$
For f(x) to be decreasing, we must have f'(x) > 0
$\Rightarrow-\frac{(\cos\text{x}-\sin\text{x})}{2+2\sin\text{x}\cos\text{x}}<0$
$\Rightarrow-(\cos\text{x}-\sin\text{x})<0$
$\Rightarrow\sin\text{x}<\cos\text{x}$
$\Rightarrow\text{x}\in\Big(0,\frac{\pi}{4}\Big)$

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