Question

The value of $c$ in Rolle's theorem for the function $f\left(x\right)=\cos \frac{x}{2}$ on $[\pi ,3\pi ]$ is

• A. $0$
• B. $2\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{2}$

Answer 1

Answer: (B)

$2\pi$

According to Rolle's theorem, for a function $f\left(x\right)$ continuous in the interval $x\in [a,b]$ and differentiable in the interval $(a,b)$ such that $f\left(a\right)=f\left(b\right)$ then, there exists a unique number $a<c<b$ such that $f^{\prime }\left(c\right)=0$.

Now $f\left(x\right)=\cos \left(\frac{x}{2}\right)$ and $x\in [\pi ,3\pi ]$.

Now $f\left(\pi \right)=f\left(3\pi \right)=0$.

Thus $f^{\prime }\left(c\right)=\frac{-1}{2}\sin \left(\frac{c}{2}\right)=0$

Or $\sin \left(\frac{c}{2}\right)=0$ or $\frac{c}{2}=\pi$, $c=2\pi$.

$\pi <2\pi <3\pi$. Hence Rolle's theorem is satisfied.