Question

The value of 'c' in Rolle's theorem for the function $f\left(x\right)=\{\begin{array}{c}{x}^2\cos \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}$ in the interval $[-1,1]$ is?

• A. $\frac{-1}{2}$
• B. $\frac{1}{4}$
• C. $0$
• D. Non-existent in the interval

Answer 1

Answer: (C)

$0$

Clearly, Rolle's theorem is applicable for the function $f\left(x\right)$ in $[-1,1]$.
Also, $f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{h^2\cos \left(\frac{1}{h}\right)-0}{h}=0$
$\therefore c=0$.