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  • Provide solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (xiv)

Provide solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (xiv)

Answers (1)

Answer:

               c<\frac{\pi}{12} \in\left(0, \frac{\pi}{6}\right)

Hint:

               f(0)=f\left(\frac{\pi}{6}\right), so there exists atc \in\left(0, \frac{\pi}{6}\right)

Given:

               f(x)=\frac{6 x}{\pi}-4 \sin ^{2} x on\left[0, \frac{\pi}{6}\right]

Explanation:

We have

f(x)=\frac{6 x}{\pi}-4 \sin ^{2} x on\left[0, \frac{\pi}{6}\right]

We know that sine function is continuous and differentiable on R.

Let’s find the values of the function at an extreme

\begin{array}{l} \Rightarrow \quad f(0)=\frac{6(0)}{\pi}-4 \sin ^{2}(0) \\\\ \Rightarrow \quad f(0)=0-4(0) \\\\ \therefore \quad f(0)=0 \\\\ \Rightarrow \quad f\left(\frac{\pi}{6}\right)=\frac{6\left(\frac{\pi}{6}\right)}{\pi}-4 \sin ^{2}\left(\frac{\pi}{6}\right) \\\\ \Rightarrow \quad f\left(\frac{\pi}{6}\right)=\frac{\pi}{\pi}-4\left(\frac{1}{2}\right)^{2} \\\\ \Rightarrow \quad f\left(\frac{\pi}{6}\right)=1-4\left(\frac{1}{4}\right) \\\\ \therefore f\left(\frac{\pi}{6}\right)=0 \end{array}

We havef(0)=f\left(\frac{\pi}{6}\right) , so there exists at c \in\left(0, \frac{\pi}{6}\right) , such thatf'(c) = 0

Let’s find the derivative of f(x) 

\\ \Rightarrow f^{\prime}(x)=\frac{d\left(\frac{6 x}{\pi}-4 \sin ^{2} x\right)}{d x} \\\\ \Rightarrow f^{\prime}(x)=\frac{6}{\pi}-4 \times 2 \sin x \frac{d(\sin x)}{d x} \ \ \ \ \ \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right] \\\\\\ \Rightarrow \quad f^{\prime}(x)=\frac{6}{\pi}-8 \sin x \cos x \left[\because \frac{d(\sin x)}{d x}=\cos x\right] \\\\ \Rightarrow \quad f^{\prime}(x)=\frac{6}{\pi}-4(2 \sin x \cos x) \\\\ \Rightarrow f^{\prime}(x)=\frac{6}{\pi}-4 \sin 2 x \ \ \ \ \ \ \ \ \quad[\because \sin 2 x=2 \sin x \cos x]

We havef'(c) = 0

\begin{array}{l} \Rightarrow \quad \frac{6}{\pi}-4 \sin 2 c=0\\\\ \Rightarrow \quad 4 \sin 2 c=\frac{6}{\pi}\\\\ \Rightarrow \quad \sin 2 c=\frac{6}{4 \pi}\\\\ \end{array}

We know 

\begin{array}{l} \frac{6}{4 \pi}<\frac{1}{2}\\\\ \Rightarrow \quad \sin 2 c<\frac{1}{2}\\ \Rightarrow \quad 2 c<\sin ^{-1}\left(\frac{1}{2}\right)\\ \Rightarrow \quad 2 c<\frac{\pi}{6}\\ \Rightarrow c<\frac{\pi}{12} \in\left(0, \frac{\pi}{6}\right)\\ \end{array}

Hence Rolle’s Theorem is verified.

Posted by

infoexpert22

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