Question

If $R$ be the set of all real numbers and $f:\left[ -1,1 \right]\to R$ be defined by $f\left( x \right)=\left\{ \begin{matrix}
   x\sin \dfrac{1}{x} & x\ne 0 \\
   0 & x=0 \\
\end{matrix} \right\}$ then:
1) $f$ satisfies the conditions of Rolle’s theorem on $\left[ -1,1 \right]$
2) $f$ satisfies the condition of Lagrange’s mean value theorem on $\left[ -1,1 \right]$
3) $f$ satisfies the condition of Rolle’s theorem on $\left[ 0,1 \right]$
4) $f$ satisfies the condition of Lagrange’s mean value theorem on $\left[ 0,1 \right]$

Answer 1

Hint: Here in this question we have been asked to find the conditions which are satisfied by the given function $f\left( x \right)=\left\{ \begin{matrix}
   x\sin \dfrac{1}{x} & x\ne 0 \\
   0 & x=0 \\
\end{matrix} \right\}$ . We know that the Rolle’s theorem is stated as that if a function $f$ is continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ such that $f\left( a \right)=f\left( b \right)$ , then $f'\left( x \right)=0$ for some $x$ with $a\le x\le b$ .

Complete step by step answer:
Now considering from the question we have been asked to find the conditions which are satisfied by the given function $f\left( x \right)=\left\{ \begin{matrix}
   x\sin \dfrac{1}{x} & x\ne 0 \\
   0 & x=0 \\
\end{matrix} \right\}$ given that $f:\left[ -1,1 \right]\to R$ .
From the basic concepts we know that the Rolle’s theorem is stated as that if a function $f$ is continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ such that $f\left( a \right)=f\left( b \right)$ , then $f'\left( x \right)=0$ for some $x$ with $a\le x\le b$ .
The Lagrange’s theorem is stated as that if a function $f$ is continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ such that $f\left( a \right)=f\left( b \right)$ , then there exists at least one value $c$ in $\left( a,b \right)$ such that $f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$.
Here the given function $f\left( x \right)=\left\{ \begin{matrix}
   x\sin \dfrac{1}{x} & x\ne 0 \\
   0 & x=0 \\
\end{matrix} \right\}$ is continuous at zero because $\displaystyle \lim_{x \to {{0}^{-}}}x\sin \dfrac{1}{x}=\displaystyle \lim_{x \to {{0}^{+}}}x\sin \dfrac{1}{x}\Rightarrow 0$ and it is not differentiable at zero because the left hand derivative is not equal to the right hand derivative.
Here $f\left( 1 \right)=\sin \left( 1 \right)$ and $f\left( -1 \right)=-\sin \left( -1 \right)=\sin \left( 1 \right)$ . Hence we can say that $f\left( 1 \right)=f\left( -1 \right)$ and $f\left( 1 \right)\ne f\left( 0 \right)$ .
Here in the interval $\left[ 0,1 \right]$ , $f'\left( c \right)=\dfrac{f\left( 1 \right)-f\left( 0 \right)}{1}=\sin 1$ and in the interval $\left[ -1,1 \right]$ , $f'\left( c \right)=\dfrac{f\left( 1 \right)-f\left( -1 \right)}{1-\left( -1 \right)}=\sin 1$ . Now we can say that for both the cases there is a valid value for $c$ in the particular interval.
Therefore we can conclude that the given function $f$ satisfies the condition of Lagrange’s mean value theorem on $\left[ 0,1 \right]$ .

So, the correct answer is “Option 4”.

Note: During the process of answering questions of this type we should be very sure with the concepts that we are going to apply. Here if someone had a misconception and considered that the given function is differentiable at zero because both the left hand derivative and right hand derivative exists then they will end up concluding that 1,2 and 4 are correct options which is a wrong conclusion.