Question

Show that the function f(x) = cot^$-$l(sinx + cosx) is decreasing on $\left(0,\frac{\mathrm{\pi }}{4}\right)$ and increasing on $\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right)$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)={\cot }^{-1}\left(\sin x+\cos x\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{-1}{1+{\left(\sin x+\cos x\right)}^2}\times \left(\cos x-\sin x\right) \\ =\frac{\sin x-\cos x}{1+{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x} \\ =\frac{\sin x-\cos x}{1+1+2\sin x\cos x} \\ =\frac{\sin x-\cos x}{2+2\sin x\cos x} \\ =\frac{1}{2}\times \frac{\sin x-\cos x}{1+\sin x\cos x} \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{decreasing},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow \frac{1}{2}\times \frac{\sin x-\cos x}{1+\sin x\cos x}<0 \\ \Rightarrow \frac{\sin x-\cos x}{1+\sin x\cos x}<0 \\ \Rightarrow \sin x-\cos x\mathit{<}0\left(\mathrm{In}\mathrm{first}\mathrm{quadrant}\right) \\ \Rightarrow \sin x<\cos x \\ \Rightarrow \tan x<1 \\ \Rightarrow 0<x<\frac{\mathrm{\pi }}{4} \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{decreasing}\mathrm{on}\left(0,\frac{\mathrm{\pi }}{4}\right). \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{increasing},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \frac{1}{2}\times \frac{\sin x-\cos x}{1+\sin x\cos x}>0 \\ \Rightarrow \frac{\sin x-\cos x}{1+\sin x\cos x}>0 \\ \Rightarrow \sin x-\cos x>0\left(\mathrm{In}\mathrm{first}\mathrm{quadrant}\right) \\ \Rightarrow \sin x>\cos x \\ \Rightarrow \tan x>1 \\ \Rightarrow \frac{\mathrm{\pi }}{4}<\mathrm{x}<\frac{\mathrm{\pi }}{2} \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{on}\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right). \end{gathered}$$