Question

In which one of the following intervals, Rolle's theorem holds good for $f\left(x\right)=x^2\sin \frac{1}{x}+x^3\cos \frac{1}{2x}$?

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

Answer 1

The correct option is D $[\frac{1}{3\pi },\frac{1}{\pi }]$

We have $f\left(x\right)=x^2\sin \frac{1}{x}+x^3\cos \frac{1}{2x}$
Clearly, $f\left(x\right)$ is continuous and differentiable for all intervals given in the options.

Checking values of function at boundary points :
$f\left(\frac{1}{3\pi }\right)=f\left(\frac{1}{\pi }\right)=0$
And for no other options, the values of $f\left(x\right)$ at end points are equal.
Hence, by Rolle's theorem, there exists $c\in (\frac{1}{3\pi },\frac{1}{\pi })$ such that $f^{\prime }\left(c\right)=0$