Question

Let $R$ be the set of all real numbers and $f:[-1,1]\to R$ be defined by $f\left(x\right)=\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}$. Then

• A. $f$ satisfies the conditions of Rolle's theorem on $[-1,1]$
• B. $f$ satisfies conditions of Lagrange's Mean Value Theorem on $[-1,1]$
• C. $f$ satisfies the conditions of Rolle's theorem on $[0,1]$
• D. $f$ satisfies the conditions of Lagrange's Mean Value Theorem on $[0,1]$

Answer 1

The correct option is D $f$ satisfies the conditions of Lagrange's Mean Value Theorem on $[0,1]$

$\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow \frac{f\left(1\right)-f\left(1\right)}{1-(-1)}$

$\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow \frac{\sin \left(1\right)-(-1)\sin (-1)}{2}$

$f\left(x\right)=\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}$

$\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow \sin 1+\sin (-1)=0$

$f^{\prime }\left(x\right)=x\cos \left(\frac{1}{x}\right)\times \frac{-1}{x^2}+\sin \left(\frac{1}{x}\right)$

$=\frac{-1}{x}\cos \frac{1}{x}+\sin \frac{1}{x}$

$f^{\prime }\left(x\right)=0$ satisfies LMVT