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#### Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 3

$f(x) \text { is increasing on } \left(0, \frac{3 \pi}{4}\right) \\ \text { and } f(x) \text { is decreasing on } \left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$

Given:

$\text { given } f(x)=sin\: x-cos\: x$

To find:

We have to interval in which f(x) is increasing or decreasing.

Hint:

f'(x)=0 to find critical point then use increasing or decreasing property.

Solution:

We have

$f(x)=sin\: x-cos\: x$

On differentiating both sides we get

$f'(x)=cos\: x+sin\: x$

For f(x) let us find critical point, we must have

\begin{aligned} &\Rightarrow f^{\prime}(x)=0 \\ &\Rightarrow \cos x+\sin x=0 \end{aligned}

On dividing by cos x we get

\begin{aligned} &\Rightarrow \tan x=-1 \\ &\Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4} \end{aligned}

Here the points divide the angle range from 0 to 2π.

Since we have x as angle.

$\text { Clearly, } f^{\prime}(x)>0 \text { if } 0

$\text {Thus } f(x) \text { increases on }\left(0, \frac{3 \pi}{4}\right) \cup\left(\frac{7 \pi}{4}, 2 \pi\right) \text { and } f(x) \text { is decreasing on interval } x \in\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$