Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

The value of c in Rolle’s Theorem for the function $f\left( x \right)=\cos \dfrac{x}{2}$ on $\left[ \pi ,3\pi \right]$ is:
(A)$0$
(B)$2\pi $
(C)$\dfrac{\pi }{2}$
(D)$\dfrac{3\pi }{2}$

Answer
VerifiedVerified
577.8k+ views
Hint: For answering this question we will see if Rolle’s Theorem is applicable or not and then we will solve it get the value of $c$ for the function $f\left( x \right)=\cos \dfrac{x}{2}$ on $\left[ \pi ,3\pi \right]$ . For verifying Rolle’s Theorem we should be aware of its statement. Rolle’s Theorem states that for any $f\left( x \right)$ satisfying the below two conditions there exists a value $c$ satisfying ${{f}^{'}}\left( c \right)=0$ with in the same interval. The conditions are as follows:
(i) $f\left( a \right)=f\left( b \right)$ For the interval $\left[ a,b \right]$.
(ii) $f\left( x \right)$ should be differentiable within the given limit $\left[ a,b \right]$.

Complete step-by-step solution:
Here from the question we have $f\left( x \right)=\cos \dfrac{x}{2}$ on$\left[ \pi ,3\pi \right]$ .
Rolle’s Theorem states that for any $f\left( x \right)$ satisfying the below two conditions there exists a value $c$ satisfying ${{f}^{'}}\left( c \right)=0$ with in the same interval. The conditions are as follows:
(i) $f\left( a \right)=f\left( b \right)$ For the interval $\left[ a,b \right]$ .
(ii) $f\left( x \right)$ should be differentiable within the given limit $\left[ a,b \right]$ .
By applying this theorem for the given function we need to verify the two conditions. For verifying the first condition, $f\left( \pi \right)=\cos \dfrac{\pi }{2}=0$ and $f\left( 3\pi \right)=\cos \dfrac{3\pi }{2}=0$ . Hence the first condition is verified. We need to verify the second condition. As $\cos \theta $ is differentiable $f\left( x \right)$ will also be differentiable within the given limit $\left[ \pi ,3\pi \right]$ . Hence Rolle’s Theorem is applicable.
By applying the Rolle’s Theorem, ${{f}^{'}}\left( c \right)=-\dfrac{1}{2}\sin \dfrac{c}{2}=0$. The solution for this is $\sin \dfrac{c}{2}=0$ .
 As we know that solutions for $\sin \theta =0$ is $\theta =n\pi $ for $n$ be any integer by using it here we will have $\dfrac{c}{2}=n\pi $ .
After simplifying this we will have $c=2n\pi $ .
As $c$ must be in the given interval $\left[ \pi ,3\pi \right]$ it will be $c=2\pi $ .
Hence option B is correct.

Note: While answering questions of this type we should be sure with the statement of Rolle’s Theorem. If by mistake we forget that $c$ should be in the given interval $\left[ \pi ,3\pi \right]$ . If we forget this then we will go with option A because we have \[~c=2n\pi \] for any integer. Then we can $c=0$ which is a wrong answer.