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Question: Find the values of c that satisfy the MVT for integrals on \[\left[ \dfrac{3\pi }{4},\pi \right]\]. ...

Find the values of c that satisfy the MVT for integrals on [3π4,π]\left[ \dfrac{3\pi }{4},\pi \right].
f(x)=cos⁡(2x−π)f\left( x \right)=\cos \left( 2x-\pi \right)
A. c=5π2−12cos⁡−1(−2π)c=\dfrac{5\pi }{2}-\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{-2}{\pi } \right)
B. c=π−12cos⁡−1(−2π)c=\pi -\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{-2}{\pi } \right)
C. c=π2+12sin⁡−1(−2π)c=\dfrac{\pi }{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{-2}{\pi } \right)
D. c=π2+12sin⁡−1(2π)c=\dfrac{\pi }{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2}{\pi } \right)

Explanation

Solution

Hint:We know that according to mean value theorem i.e. MVT. If we have a function f(x) and a value ‘c’ that satisfy the MVT for integrals on [a,b]\left[ a,b \right] for f(x), then
f′(c)=f(b)−f(a)b−af'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}
when f’(c) is differentiation of f(x) at x=c.
In the question, we have a function as cos⁡(2x−π)\cos \left( 2x-\pi \right) and also the interval, i.e. a=3π4a=\dfrac{3\pi }{4} and b=πb=\pi .
So by using this, we will get the value of ‘c’.

Complete step-by-step answer:
We have been asked to find the value of c that satisfy the values of MVT for integrals on [3π4,π]\left[ \dfrac{3\pi }{4},\pi \right] and f(x)=cos⁡(2x−π)f(x)=\cos \left( 2x-\pi \right).
We know that MVT for integral says that there is a value c which belongs to [a,b]\left[ a,b \right], such that f’(c) is equal to the average value of f over [a,b]\left[ a,b \right].
We have f(x)=cos⁡(2x−π)f(x)=\cos \left( 2x-\pi \right), a=3π4a=\dfrac{3\pi }{4} and b=πb=\pi .
f(a)=cos⁡(2×3π4−π)=cos⁡(3π2−π)=cos⁡(3π−2π2)=cos⁡π2f(a)=\cos \left( 2\times \dfrac{3\pi }{4}-\pi \right)=\cos \left( \dfrac{3\pi }{2}-\pi \right)=\cos \left( \dfrac{3\pi -2\pi }{2} \right)=\cos \dfrac{\pi }{2}
Since we know that cos⁡π2=0\cos \dfrac{\pi }{2}=0
f(b)=cos⁡(2π−π)=cos⁡π=−1f(b)=\cos \left( 2\pi -\pi \right)=\cos \pi =-1
Since we know that cos⁡π=−1\cos \pi =-1
Then according to MVT we have,

& f'(c)=\dfrac{f(b)-f(a)}{b-a} \\\ & f'(x)=\dfrac{d}{dx}\left[ \cos \left( 2x-\pi \right) \right]=-\sin \left( 2x-\pi \right)\times 2 \\\ \end{aligned}$$ Now we have f’(c), and substituting the values of $$f'(c)=-2\sin \left( 2c-\pi \right),f(a)=0,f(b)=-1,a=\dfrac{3\pi }{4},b=\pi $$, we get as follows: $$f'(c)=-2\sin \left( 2c-\pi \right)$$ Applying the MVT we have, $$\begin{aligned} & \Rightarrow -2\sin \left( 2c-\pi \right)=\dfrac{-1-0}{\pi -\dfrac{3\pi }{4}} \\\ & \Rightarrow 2\sin \left( 2c-\pi \right)=\dfrac{1}{\dfrac{4\pi -3\pi }{4}} \\\ & \Rightarrow 2\sin \left( 2c-\pi \right)=\dfrac{4}{\pi } \\\ & \Rightarrow \sin \left( 2c-\pi \right)=\dfrac{2}{\pi } \\\ \end{aligned}$$ On taking sine inverse on both sides, we get as follows: $$\Rightarrow {{\sin }^{-1}}\left( \sin (2c-\pi ) \right)={{\sin }^{-1}}\left( \dfrac{2}{\pi } \right)$$ We know that $${{\sin }^{-1}}\sin x=x$$ $$\begin{aligned} & \Rightarrow 2c-\pi ={{\sin }^{-1}}\left( \dfrac{2}{\pi } \right) \\\ & \Rightarrow 2c=\pi +{{\sin }^{-1}}\left( \dfrac{2}{\pi } \right) \\\ & \Rightarrow c=\dfrac{\pi }{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2}{\pi } \right) \\\ \end{aligned}$$ Therefore, the correct option of the given question is option D. Note:The Mean Value Theorem states that if a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c in the interval (a,b) such that f'(c) is equal to the function's average rate of change over [a,b].Be careful while finding the values of f(a) and f(b) and also take care that sometimes we just forget the minus sign while finding the derivative of cosine function and we write the derivative of cosx is equal to sinx which is wrong and we get incorrect answers.