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\mathrm{\text { If } f(x)=\left\{\begin{array}{r} x^2 \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0 \end{array}\right. \text {, then }}

Option: 1

f and f are continuous at x=0


Option: 2

f is derivable at x=0


Option: 3

f is derivable at x=0 and f is not continuous at x= -1


Option: 4

f is derivable at x=0.


Answers (1)

best_answer

We have,   \mathrm{\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}}
\mathrm{ =\lim _{x \rightarrow 0} \frac{x^2 \sin \left(\frac{1}{x}\right)}{x}=\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0 } 
So, f(x) is differentiable at x=0 such that  \mathrm{ f^{\prime}(0)=0.~ For ~x \neq 0 } , we have

\mathrm{ f^{\prime}(x)=2 x \sin \left(\frac{1}{x}\right)+x^2 \cos \left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right) \\ } 
\mathrm{ \Rightarrow f^{\prime}(x)=2 x \sin \frac{1}{x}-\cos \frac{1}{x} } 
\mathrm{ \Rightarrow \lim _{x \rightarrow 0} f^{\prime}(x)=\lim _{x \rightarrow 0} 2 x \sin \frac{1}{x}-\lim _{x \rightarrow 0} \cos \left(\frac{1}{x}\right)=\underset{x \rightarrow 0}{0-\lim _{x \rightarrow 0} \cos \left(\frac{1}{x}\right)} }

Since \mathrm{ \lim _{x \rightarrow 0} \cos \left(\frac{1}{x}\right)~ does~ not ~exist, therefore \lim _{x \rightarrow 0} f^{\prime}(x)~ does~ not~ exist.~ Hence, ~f^{\prime}(x) } is not continuous at x=0.

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