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

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

Answers (1)

Answer:

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}<x<\frac{\pi}{4} \\ &\Rightarrow 0<\frac{\pi}{4}+x<\frac{\pi}{2} \\ &\Rightarrow \sin 0^{\circ}<\sin \left(\frac{\pi}{4}+x\right)<\sin \frac{\pi}{2} \\ \end{aligned}

\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})

Posted by

Gurleen Kaur

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