#### Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 16 maths

$f(x) \text { is decreasing function on }(\frac{3\pi }{8},\frac{5\pi }{8})$

Given:

$f(x)=sin(2x+\frac{\pi }{4})$

To prove:

$\text { We have to show that } f(x) \text { is decreasing function on }(\frac{3\pi }{8},\frac{5\pi }{8})$

Hint:

Condition for f(x) to be decreasing is f’(x)<0.

Solution:

Given

$f(x)=sin(2x+\frac{\pi }{4})$

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

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x} \sin \left(2 x+\frac{\pi}{4}\right) \\ &\Rightarrow f^{\prime}(x)=\cos \left(2 x+\frac{\pi}{4}\right) \cdot 2 \\ &\Rightarrow f^{\prime}(x)=2 \cos \left(2 x+\frac{\pi}{4}\right) \end{aligned}

Now, as given

\begin{aligned} &x \in\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right) \\ &\Rightarrow \frac{3 \pi}{8}

\begin{aligned} &\text { As here } 2 x+\frac{\pi}{4} \text { lies in } 3 \mathrm{rd} \text { quadrant. }\\ &\Rightarrow \cos \left(2 x+\frac{\pi}{4}\right)<0\\ &\Rightarrow 2 \cos \left(2 x+\frac{\pi}{4}\right)<0\\ &\Longrightarrow f^{\prime}(x)<0 \end{aligned}

Hence condition for f(x) to be decreasing.

$\text { Thus } f(x) \text { is decreasing on interval }x\: \in \: (\frac{3\pi }{8},\frac{5\pi }{8})$