#### Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 23

$f(x) \text { is increasing function on } (-\frac{\pi }{4},\frac{\pi }{4})$

Given:

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

To prove:

$\text { We have to prove that } f(x) \text { is increasing function on } (-\frac{\pi }{4},\frac{\pi }{4})$

Hint:

If f ’(x) > 0 ∀ x $\in$ (a,b) then f(x) is increasing on (a,b).

Solution:

Given

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

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

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\sin x-\cos x) \\ &\Rightarrow f^{\prime}(x)=\cos x+\sin x \\ &=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right) . .\{\text { multiply and divide by } \sqrt{2}\} \\ \end{aligned}

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

Now, as given

\begin{aligned} &x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \\ &\Rightarrow-\frac{\pi}{4}

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

$\text { Hence } f(x) \text { is increasing on interval } x\: \in (-\frac{\pi }{4},\frac{\pi }{4})$