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#### Provide solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 1 sub question 3

Rolle’s Theorem is not applicable on$f (x)$in$[-1,1]$

Hint:

Let         $\lim _{h \rightarrow 0 } \sin( \frac{1}{h} ) = k$as$k \in [-1,1]$

Given:

$f (x) = \sin \frac{1}{x} , x \in [-1,1]$

Explanation:

\begin{aligned} L . H . S &=\lim _{x \rightarrow(0-h)} \sin \left(\frac{1}{x}\right) \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{1}{0-h}\right) \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{-1}{h}\right) \\ &=-\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right) \\ & = -k \ \ \ \ \ \ \ \ \ \left [ \because \lim _{h \rightarrow 0} \sin \frac{1}{h}= k,k \in [-1,1] \right ]\\\\ R . H . S &=\lim _{x \rightarrow(0+h)} \sin \left(\frac{1}{x}\right) \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right) \ \ \ \ \ \ \\ &=k \\ \Rightarrow \quad & L \cdot H . S \neq R \cdot H . S \end{aligned}

$f(x)$is not continuous at 0.

So, Rolle’s Theorem is not applicable on $f(x)$in$[-1,1]$