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

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

Answers (1)

Answer:

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})

Posted by

Gurleen Kaur

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