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

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

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 exists atc \in\left(0, \frac{\pi}{2}\right)

Given:

f(x)=\sin ^{4} x+\cos ^{4} x on \left[0, \frac{\pi}{2}\right] 

Explanation:

f(x)=\sin ^{4} x+\cos ^{4} x on \left[0, \frac{\pi}{2}\right] 

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

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

\begin{array}{l} \Rightarrow \quad f(0)=\sin ^{4}(0)+\cos ^{4}(0) \\\\ \Rightarrow \quad f(0)=0+1 \\\\ \Rightarrow f(0)=1 \\\\ \Rightarrow f\left(\frac{\pi}{2}\right)=\sin ^{4}\left(\frac{\pi}{2}\right)+\cos ^{4}\left(\frac{\pi}{2}\right) \\\\ f\left(\frac{\pi}{2}\right)=1+0 \\\\ \quad f\left(\frac{\pi}{2}\right)=1 \end{array}

We havef (0) = f ( \frac{\pi}{2}) , so there exists atc \in ( 0 , \frac{\pi}{2}) , such thatf ' (c) = 0 

Let’s find the derivative of f (x)

\Rightarrow \quad f^{\prime}(x)=\frac{d\left(\sin ^{4} x+\cos ^{4} x\right)}{d x}\\

\Rightarrow \quad f^{\prime}(x)=4 \sin ^{3} x \frac{d(\sin x)}{d x}+4 \cos ^{3} x \frac{d(\cos x)}{d x} \ \ \ \ \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]\\

                                                                                                                          \left[ \because \frac{d(\sin x)}{d x}=\cos x \right]

                                                                                                                          \left[ \because \frac{d(\cos x)}{d x}=-\sin x \right]

\Rightarrow \quad f^{\prime}(x)=4 \sin ^{3} x \cos x-4 \cos ^{3} x \sin x

\Rightarrow \quad f^{\prime}(x)=4 \sin x \cos x\left(\sin ^{2} x-\cos ^{2} x\right)\\\\ \Rightarrow \quad f^{\prime}(x)=2(2 \sin x \cos x)(-\cos 2 x)\ \ \ \ \quad\left[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right] \\\\ \Rightarrow \quad f^{\prime}(x)=-2(\sin 2 x)(\cos 2 x) \ \ \ \ \ \ \ \ [\because \sin 2 x=2 \sin x \cos x] \\\\ \Rightarrow \quad f^{\prime}(x)=-\sin 4 x
We have

f^{\prime}(c)=0 \\\\ \Rightarrow \quad-\sin 4 c=0 \\\\ \Rightarrow \quad \sin 4 c=0 \\\\ \Rightarrow \quad 4 c=0_{\text {or }} \pi \\\\ \Rightarrow \quad c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)

Hence Rolle’s Theorem is verified.

Posted by

infoexpert22

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