Question

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

• A. $f$ satisfies the conditions of Rolle’s theorem on $\left[-1,1\right]$
• B. $f$ satisfies the condition of Lagrange’s mean value theorem on $\left[-1,1\right]$
• C. $f$ satisfies the condition of Rolle’s theorem on $\left[0,1\right]$
• D. $f$ satisfies the condition of Lagrange’s mean value theorem on $\left[0,1\right]$

Answer 1

The correct option is D

$f$ satisfies the condition of Lagrange’s mean value theorem on $\left[0,1\right]$

Satisfying the conditions:

Step 1: Check the continuity of the given function.

A function $f$ satisfies Rolle's theorem if it follows the condition that

• The function $f$ is continuous on the closed interval $\left[a,b\right]$
• The function $f$ is differentiable on the open interval $\left(a,b\right)$, such that $f\left(a\right)=f\left(b\right)$,
• There exists for some $x$, with $a\le x\le b$ for which $f^{\prime }\left(x\right)=0$.

A function $f$ satisfies Lagrange’s theorem if it follows conditions that

• The function $f$ is continuous on the closed interval $\left[a,b\right]$ and
• The function $f$ is differentiable on the open interval $\left(a,b\right)$ such that $f\left(a\right)\ne f\left(b\right)$,
• There exists at least one value $c$ in $\left(a,b\right)$ such that $f^{\prime }\left(c\right)={\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}$.

Here, the given function $f\left(x\right)=\left\{\begin{array}{cc}x\sin \frac{1}{x}& x\ne 0\\ 0& x=0\end{array}\right\}$ is continuous at zero, as ${\displaystyle \underset{x\to 0^{-}}{lim}x\sin {\displaystyle \frac{1}{x}}={\displaystyle \underset{x\to 0^{+}}{lim}x\sin {\displaystyle \frac{1}{x}}\Rightarrow 0}}$

Therefore, is continuous for all values of $x$.

Step 2: Check the differentiability condition

At $x=0$

$\begin{array}{c}LHD=\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{h\sin \frac{1}{h}}{h}\\ =\underset{h\to 0}{\lim }\sin \frac{1}{h}\\ =-\sin \left(\frac{1}{0}\right)\\ \end{array}$

Thus, $f\left(x\right)$ is not differentiable at $x=0$,

From the given function $f\left(1\right)=\sin \left(1\right)$ and

$\begin{array}{rcl}f(-1)& =& -\sin (-1)\\ & =& \sin \left(1\right)\end{array}$

Also $f\left(0\right)=0$

Which implies $f\left(1\right)=f\left(-1\right)$ and $f\left(1\right)\ne f\left(0\right)$,

Therefore, Rolle's theorem does not satisfy.

Since the left-hand derivative is not equal to the right-hand derivative, therefore the function is not differentiable at zero.

Step 3: Test the last condition

In the given function, in the interval $\left[0,1\right]$ , the derivative

$\begin{array}{rcl}{f}^{\text{'}}\left(c\right)& =& \frac{f\left(1\right)-f\left(0\right)}{1}\\ & =& \sin 1\end{array}$

Similarly, for the interval $\left[-1,1\right]$

$\begin{array}{rcl}{f}^{\text{'}}\left(c\right)& =& \frac{f\left(1\right)-f(-1)}{1-(-1)}\\ & =& \frac{\sin 1-\sin 1}{2}\\ & =& 0\end{array}$

Therefore, conclude that the function satisfies the condition of Lagrange’s mean value theorem on $\left[0,1\right]$.

Hence, the correct option is (D).