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

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               c=\frac{\pi}{2} \in(0, \pi)


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


               f(x)=\cos 2 x on [0, \pi] 


We have

f(x)=\cos 2 x on [0, \pi] 

We know that exponential and cosine functions are continuous and differentiable on R.

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

\\ \Rightarrow \quad f(0)=\cos 2(0) \\\\ \Rightarrow \quad f(0)=\cos (0)\\\\ \therefore \quad f(0)=1\\\\ \Rightarrow \quad f(\pi)=\cos 2(\pi)\\\\ \therefore f(\pi)=1

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 \quad f^{\prime}(x)=\frac{d(\cos 2 x)}{d x} \\\\ \Rightarrow \quad f^{\prime}(x)=-\sin 2 x \frac{d(2 x)}{d x}\\\\ \Rightarrow \quad f^{\prime}(x)=-2 \sin 2 x

We havef^{\prime}(c)=0

\\ \Rightarrow \quad-2 \sin 2 c=0\\\\ \Rightarrow \quad \sin 2 c=0\\\\ \Rightarrow \quad 2 c=0 \ or \ \pi \\\\

\\ \Rightarrow c = 0 \ or \ \frac{\pi}{2} \\\\ \Rightarrow c = \frac{\pi}{2} \in(0, \pi)

Hence, Rolle’s Theorem is verified.

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