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#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 17 maths textbook solution

$f(x) \text { is decreasing on }(0,\frac{\pi }{4}) \text { and increasing on } (\frac{\pi }{4},\frac{\pi }{2})$

Given:

$f(x)=cot^{-1}(sin\: x+cos\: x)$

To prove:

$\text { We have to show that } f(x) \text { is decreasing on }(0,\frac{\pi }{4}) \text { and increasing on } (\frac{\pi }{4},\frac{\pi }{2})$

Hint:

1. condition for f(x) to be increasing is f'(x)>0
2. condition for f(x) to be decreasing f'(x)<0

Solution:

Given

$f(x)=cot^{-1}(sin\: x+cos\: x)$

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left[\cot ^{-1}(\sin x+\cos x)\right] \\ &\Rightarrow f^{\prime}(x)=\frac{-1}{1+(\sin x+\cos x)^{2}} \times(\cos x-\sin x),\left[\therefore \frac{d}{d x} \cot ^{-1} x=\frac{-1}{1+x^{2}}\right] \\ &\Rightarrow f^{\prime}(x)=\frac{\sin x-\cos x}{1+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} \end{aligned}

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{\sin x-\cos x}{2+2 \sin x \cos x},\left[\therefore \sin ^{2} x+\cos ^{2} x=1\right] \\ &\Rightarrow f^{\prime}(x)=\frac{\sin x-\cos x}{2(1+\sin x \cos x)} \end{aligned}

Now, as given

\begin{aligned} &x \in\left(0, \frac{\pi}{4}\right) \\ &\Rightarrow \frac{\sin x-\cos x}{2(1+\sin x \cos x)}<0 \: \: \: \text{(for some values of x in first quadrant )}\\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}

$\text { Thus }f(x) \text { is decreasing on interval } x \: \in \: (0,\frac{\pi }{4})$

\begin{aligned} &\text { Now, for } x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\\ &\Rightarrow \sin x-\cos x>0 \: \: \: \:\; (\text { for some values of x in first quadrant })\\ &\Rightarrow \frac{\sin x-\cos x}{2(1+\sin x \cos x)}>0\\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}

$\text { Thus }f(x) \text { is decreasing on interval } x \: \in \: (\frac{\pi }{4},\frac{\pi }{2})$