Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (ii)

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

Answers (1)

Answer:

               c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)

Hint:

               f(0)=f\left(\frac{\pi}{2}\right) , so there existsc \in\left(0, \frac{\pi}{2}\right)

Given:

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

Explanation:

We have

               f(x)=\sin 2 x \text { on }\left[0, \frac{\pi}{2}\right]             

We know that sine function is continuous and differentiable on

\begin{array}{l} f(0)=\sin 2(0) \\\\ f(0)=\sin (0) \\\\ f(0)=0 \\\\ f\left(\frac{\pi}{2}\right)=\sin 2\left(\frac{\pi}{2}\right) \\\\ f\left(\frac{\pi}{2}\right)=\sin \pi \end{array}.

f\left(\frac{\pi}{2}\right)=0

We havef(0)=f\left(\frac{\pi}{2}\right) 

So there existsc \in\left(0, \frac{\pi}{2}\right)

Such thatf^{\prime}(c)=0

Let find the derivative off(x)

\\ f^{\prime}(x)=\frac{d(\sin 2 x)}{d x} \\\\ f^{\prime}(x)=\cos 2 x \frac{d(2 x)}{d x} \\\\ \qquad f^{\prime}(x)=2 \cos 2 x \\\\

We have

\\f^{\prime}(c)=0 \\\\ 2 \cos 2 c=0 \\\\ 2 c=\frac{\pi}{2} \\\\ c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)

Hence, Rolle’s Theorem is verified.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question