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

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               c=\frac{\pi}{6} \in(0, \pi)


               f(0)=f(\pi) , so there exists at C belongs to(0, \pi)


               f(x)=\sin 3 x on [0, \pi] 


We have

        f(x)=\sin 3 x on [0, \pi] 

We know that sine functions are continuous and differentiable on R. 

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

\\ \Rightarrow \quad f(0)=\sin 3(0) \\\\ \Rightarrow \quad f(0)=\sin (0)\\\\ \therefore f(0)=0 \\\\ \Rightarrow \quad f(\pi)=\sin 3(\pi) \\\\ f (\pi) = 0

We havef(0)=f(\pi) , so there exists atc \in(0, \pi) , such that f^{\prime}(c)=0

Let’s find the derivative of f(x) 

\\ \Rightarrow f^{\prime}(x)=\frac{d(\sin 3 x)}{d x} \\\\ \Rightarrow f^{\prime}(x)=\cos 3 x \frac{d(3 x)}{d x}\\\\ \Rightarrow \quad f^{\prime}(x)=3 \cos 3 x

We have \\ f^{\prime}(c)=0

\\\\ \Rightarrow \quad 3 \cos 3 c=0\\\\ \Rightarrow 3c = \frac{\pi}{2}\\\\ \Rightarrow c=\frac{\pi}{6} \in(0, \pi)

Hence, Rolle’s Theorem is verified.

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