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# Solution for Show that the Function F(X) = Cot − L(Sinx + Cosx) is Decreasing on ( 0 , π 4 ) and Increasing on ( 0 , π 4 ) ? - CBSE (Commerce) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

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

#### 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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Solution Show that the Function F(X) = Cot − L(Sinx + Cosx) is Decreasing on ( 0 , π 4 ) and Increasing on ( 0 , π 4 ) ? Concept: Increasing and Decreasing Functions.
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