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Question
Show that the function f(x) = cot \[-\] l(sinx + cosx) is decreasing on \[\left( 0, \frac{\pi}{4} \right)\] and increasing on \[\left( 0, \frac{\pi}{4} \right)\] ?
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Solution
\[\text { We have,} \]
\[f\left( x \right) = \cot^{- 1} \left( \sin x + \cos x \right)\]
\[ \Rightarrow f'\left( x \right) = \frac{- 1}{1 + \left( \sin x + \cos x \right)^2} \times \left( \cos x - \sin x \right)\]
\[ = \frac{\sin x - \cos x}{1 + \sin^2 x + \cos^2 x + 2\sin x\cos x}\]
\[ = \frac{\sin x - \cos x}{1 + 1 + 2\sin x\cos x}\]
\[ = \frac{\sin x - \cos x}{2 + 2\sin x\cos x}\]
\[ = \frac{1}{2} \times \frac{\sin x - \cos x}{1 + \sin x\cos x}\]
\[\text { For } f\left( x \right) \text { to be decreasing, we must have }\]
\[f'\left( x \right) < 0\]
\[ \Rightarrow \frac{1}{2} \times \frac{\sin x - \cos x}{1 + \sin x\cos x} < 0\]
\[ \Rightarrow \frac{\sin x - \cos x }{1 + \sin x\cos x} < 0\]
\[ \Rightarrow \sin x - \cos x < 0 \left( \text { In first quadrant } \right)\]
\[ \Rightarrow \sin x < \cos x\]
\[ \Rightarrow \tan x < 1\]
\[ \Rightarrow 0 < x < \frac{\pi}{4}\]
\[So, f\left( x \right) \text { is decreasing on } \left( 0, \frac{\pi}{4} \right) . \]
\[\text { For } f\left( x \right) \text { to be increasing, we must have } \]
\[f'\left( x \right) > 0\]
\[ \Rightarrow \frac{1}{2} \times \frac{\sin x - \cos x}{1 + \sin x\cos x} > 0\]
\[ \Rightarrow \frac{\sin x - \cos x}{1 + \sin x\cos x} > 0\]
\[ \Rightarrow \sin x - \cos x > 0 \left(\text { In first quadrant } \right)\]
\[ \Rightarrow \sin x > \cos x\]
\[ \Rightarrow \tan x > 1\]
\[ \Rightarrow \frac{\pi}{4} < x < \frac{\pi}{2}\]
\[\text { So,} f\left( x \right) \text { is increasing on } \left( \frac{\pi}{4}, \frac{\pi}{2} \right) .\]
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