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  • Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 15

Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 15

Answers (1)

Answer:

f(x) \text { is decreasing function on } (\frac{\pi }{4},\frac{\pi }{2})

Given:

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

To prove:

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

Hint:

Condition for decreasing function f’(x)<0.

Solution:

Given

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

Now, as given

\begin{aligned} &x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ &\Rightarrow \cos x-\sin x<0 \\ &\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}

\begin{aligned} &x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ &\Rightarrow \cos x-\sin x<0 \\ \end{aligned}

as here cosine values are smaller than sine values for same angle.

\begin{aligned} &\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}

Hence condition for f(x) to be decreasing.

\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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