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  • Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14 point 1 question 3 sub question (iii)

Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14 point 1 question 3 sub question (iii)

Answers (1)

Answer:

               c \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)

Hint:

f\left(-\frac{\pi}{4}\right)=f\left(\frac{\pi}{4}\right), so there existsc \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)

Given:

               f(x)=\cos 2 x on \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]

Explanation:

We have

                      f(x)=\cos 2 x on \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]

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

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

 \begin{array}{l} f\left(\frac{-\pi}{4}\right)=\cos 2\left(\frac{-\pi}{4}\right) \\\\ f(0)=\cos \left(\frac{-\pi}{2}\right) \end{array} 

We know that

\cos (-x)=\cos x

\begin{array}{l} f(0)=0 \\\\ f\left(\frac{\pi}{4}\right)=\cos 2\left(\frac{\pi}{4}\right) \\\\ f\left(\frac{\pi}{2}\right)=\cos \left(\frac{\pi}{2}\right) \\\\ f\left(\frac{\pi}{2}\right)=0 \end{array}

We havef\left(\frac{-\pi}{4}\right)=f\left(\frac{\pi}{4}\right) , so there existsc \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)

Such thatf '(c) = 0

Let’s find the derivative of f(x)

\begin{array}{l} f^{\prime}(x)=\frac{d(\cos 2 x)}{d x} \\\\ f^{\prime}(x)=-\sin 2 x \frac{d(2 x)}{d x} \\\\ f^{\prime}(x)=-2 \sin 2 x \\\\ \end{array}

We havef '(c) = 0       

\begin{array}{l} -2 \sin 2 c=0 \\\\ \sin 2 c=0 \\\\ 2 c=0 \\\\ c=0_{\mathrm{as}} \quad c \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \end{array}

Hence, Rolle’s Theorem is verified

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