Q100 / 200 · WBJEE 2014 · Mathematics
Let $R$ be the set of all real numbers and $f:[-1,1] \rightarrow R$ be defined $f(x)=\left\{\begin{array}{ll}x \sin \frac{1}{x}, & x \neq 0 \text { . Then } \\ 0, & x=0\end{array}\right.$
- A$f$ satisfies the conditions of Rolle's theorem on [-1,1]
- B$f$ satisfies the conditions of Lagrange's mean value theorem on [-1,1]
- C$f$ satisfies the conditions of Rolle's theorem on [0,1]
- Correct answerD$f$ satisfies the conditions of Lagrange's mean value theorem on [0,1]
Solution
Given, $\quad f(x)=\left\{\begin{array}{cl}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.$
Continuity at $x=0$. $\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} x \sin \frac{1}{x}=0$
$\mathrm{RHL}=\lim _{x \rightarrow 0^{*}} x \sin \frac{1}{x}=0$
and
$\mathrm{LHL}=\mathrm{RHL}=f(0)$
Hence, $f(x)$ is continuous for all values of $x$. Differentiability at $x=0$
$$
\begin{aligned}
\mathrm{LHD} &=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}}{h} \\
&=-\lim _{h \rightarrow 0} \sin \frac{1}{h} \\
&=-\sin \left(\frac{1}{0}\right)=\text { some definite value }
\end{aligned}
$$
Thus, $f(x)$ is not differentiable at $x=0 .$ Now, $f(0)=0$ and $f(1)=1 \sin 1$
$\therefore f(0) \neq f(1)$. Rolle's theorem is not satisfy.
Now.
$$
\begin{aligned}
f^{\prime}(c) &=\frac{f(1)-f(0)}{b-0} \\
&=\frac{1 \sin (1)-0}{1-0}=\sin 1=\sin 1
\end{aligned}
$$
which lies between 0 to 1 .
Hence, there exist a real number $c \in(0,1)$ Hence,
Lagrange's mean value theorem is satisfy on $(0,1)$.
Hence, the correct answer is option: 'D'.