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

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

Answers (1)

Answer:

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

Hint:

               f(0)=f(\pi) , so there exists atc \in(0, \pi)

Given:

               f(x)=4^{\sin x} on [0, \pi]

Explanation:

Given function is f(x)=4^{\sin x} on [0, \pi]

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

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

\\ \Rightarrow \quad f(0)=4^{\sin (0)} \\\\ f(0)=4^{0}\\\\ \therefore f(0)=1\\\\ \Rightarrow f(\pi)=4^{\sin (x)}\\\\ \Rightarrow \quad f(\pi)=4^{0} \\\\ \therefore f(\pi)=1\\\\

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

Let’s find the derivative of f(x)

\Rightarrow \quad f^{\prime}(x)=\frac{d\left(4^{\operatorname{lin} x}\right)}{d x}\\ \Rightarrow \quad f^{\prime}(x)=4^{\sin x} \log 4 \frac{d(\sin x)}{d x} \ \ \ \ \left[\because \frac{d(\sin x)}{d x}=\cos x\right]\\ \Rightarrow \quad f^{\prime}(x)=4^{\sin x} \log 4 \cos x

We have f ' (c) = 0

\\ \Rightarrow \quad 4^{\text {sinc }} \log 4 \cos c=0\\\\ \Rightarrow \quad \cos c=0\\\\ \Rightarrow \quad c=\frac{\pi}{2} \in(0, \pi)

Hence Rolle’s Theorem is verified

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