Question

In which one of the following intervals Rolle's theorem hold(s) good for $y=x^2\sin \frac{1}{x}+x^3\cos \frac{1}{2x}$.

• A. $\left[\frac{1}{\pi },\frac{2}{\pi }\right]$
• B. $\left[\frac{1}{3\pi },\frac{1}{\pi }\right]$
• C. $\left[\frac{1}{2\pi },\frac{1}{\pi }\right]$
• D. $\left[\frac{1}{\pi },\frac{3}{\pi }\right]$

Answer 1

Answer: (B)

We have $f(x)=x^2\sin \frac{1}{x}+x^3\cos \frac{1}{2x};f^{\prime }(x)$ has opposite signs in $\left[\frac{1}{2\pi },\frac{1}{\pi }\right]$ $\Rightarrow f^{\prime }(x)=0$ atleast once
Clearly $f(x)$ is continuous as well as differentiable in $\left[\frac{1}{3\pi },\frac{1}{\pi }\right]$.
Also $f\left(\frac{1}{3\pi }\right)=0=f\left(\frac{1}{\pi }\right)$ (Only in this interval this will be true)
So, $f(x)$ satisfies Rolle's theorem.
Hence there exist some $c\in \left(\frac{1}{3\pi },\frac{1}{\pi }\right)$ such that $f^{\prime }(c)=0$.