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

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

Answers (1)

Answer:

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}<x<\frac{5 \pi}{8} \\ &\Rightarrow \frac{3 \pi}{4}<2 x<\frac{5 \pi}{4} \\ &\Rightarrow \pi<2 x+\frac{\pi}{4}<\frac{3 \pi}{2} \end{aligned}

\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})

Posted by

Gurleen Kaur

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