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  • <img alt="\mathrm{\text { Consider } f(x)=\left\{\begin{array}{cc} \frac{x^2}{|x|}, & x \neq 0 \\ 0 & x=0 \end{array}\right.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

\mathrm{\text { Consider } f(x)=\left\{\begin{array}{cc} \frac{x^2}{|x|}, & x \neq 0 \\ 0 & x=0 \end{array}\right.}

Option: 1

f(x) is discontinuous everywhere
 


Option: 2

f(x) is continuous everywhere
 


Option: 3

\mathrm{f^{\prime}(x)}  exists in (-1,1)


Option: 4

\mathrm{f^{\prime \prime}(x)}  exists in (-2,2).


Answers (1)

We have 

\mathrm{f(x)=\left\{\begin{array}{cc} \frac{x^2}{|x|}, x \neq 0 \\ 0, x=0 \end{array}=\left\{\begin{array}{cc} \frac{x^2}{x}=x, & x>0 \\ 0, & x=0 \\ \frac{x^2}{-x}=-x, & x<0 \end{array}\right.\right.}

\mathrm{\Rightarrow \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}-x=0, \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} x=0 \text { and } f(0)=0.}

So f(x) is continuous at x=0. Also f(x) is continuous for all other values of x. Hence, f(x) is everywhere continuous. Clearly. \mathrm{L f^{\prime}(0)=-1~ and ~R f^{\prime}(0)=1.}  Therefore, f(x) is not differentiable at x=0.

Posted by

Sumit Saini

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