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  • Please solve RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (xiii) maths textbook solution

Please solve RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (xiii) maths textbook solution

Answers (1)

Answer:

               c \in(-1,0), Rolle’s Theorem is verified.

Hint:

               f(-1)=f(0), so there exists atc \in(-1,0)

Given:

f(x)=\frac{x}{2}-\sin \frac{\pi x}{6} on [-1,0]

Explanation:

We have

f(x)=\frac{x}{2}-\sin \frac{\pi x}{6} on [-1,0]

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

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

\\ f(-1)=\frac{-1}{2}-\sin \left(\frac{\pi(-1)}{6}\right)\\\\ f(-1)=\frac{-1}{2}-\sin \left(\frac{-\pi}{6}\right)\\\\ f(-1)=\frac{-1}{2}-\left(-\frac{1}{2}\right)\\\\ \therefore f(-1)=0\\\\ \Rightarrow f(0)=\frac{0}{2}-\sin \left(\frac{\pi(0)}{6}\right)\\\\ \begin{array}{ll}\Rightarrow & f(0)=0-\sin (0) \\\\ \therefore & f(0)=0\end{array}

We havef(-1)=f(0) , so there exists at c \in(-1,0), such thatf^{\prime}(c)=0

Let’s find the derivative of f(x)

\\ \Rightarrow f^{\prime}(x)=\frac{d\left(\frac{x}{2}-\sin \left(\frac{\pi x}{6}\right)\right)}{d x} \\\\ \Rightarrow f^{\prime}(x)=\frac{1}{2}-\cos \left(\frac{\pi x}{6}\right) \frac{d\left(\frac{\pi x}{6}\right)}{d x} \ \ \ \left[\begin{array}{l}\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1} \\ \\\because \frac{d(\sin x)}{d x}=\cos x\end{array}\right] \\\\ f^{\prime}(x)=\frac{1}{2}-\frac{\pi}{6} \cos \left(\frac{\pi x}{6}\right) \\\\

We havef'(c) = 0

\Rightarrow \frac{1}{2}-\frac{\pi}{6} \cos \left(\frac{\pi c}{6}\right)=0\\\\ \Rightarrow \quad \frac{\pi }{6} \cos \left(\frac{-\pi c}{6}\right)=\frac{1 }{2}\\\\ \Rightarrow \quad \cos \left(\frac{\pi c}{6}\right)= \frac{1}{2}\times\frac{6}{\pi}\\\\ \Rightarrow \quad \cos \left(\frac{\pi c}{6}\right)= \frac{3}{\pi} \\\\ \Rightarrow \quad \frac{\pi c}{6}=\cos ^{-1}\left(\frac{3}{\pi}\right)\\\\ \Rightarrow \quad c=\frac{6}{\pi} \cos ^{-1}\left(\frac{3}{\pi}\right)

Cosine is positive between\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2} , for our convenience we take the interval to be\frac{-\pi}{2} \leq \theta \leq 0, since the values of the cosine repeats.

We know that \frac{3}{\pi} value is nearly equal to 1. So, the value of thec nearly equal to 0.

So, we can clearly say that c \in (-1,0)

Hence, Rolle’s Theorem is verified.

Posted by

infoexpert22

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