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  •  If a function   <img alt="\mathrm{f(x) \text { is defined as } f(x)=\left\{\begin{array}{cc} \frac{x}{\sqrt{x^2}}, & x \neq 0 \\ 0, & x=0 \end{array}\right.}" src="https://e

 If a function   \mathrm{f(x) \text { is defined as } f(x)=\left\{\begin{array}{cc} \frac{x}{\sqrt{x^2}}, & x \neq 0 \\ 0, & x=0 \end{array}\right.}          then

Option: 1

f(x) is continuous at x=0 but not differentiable at x=0


Option: 2

f(x) is continuous as well as differentiable at x=0


Option: 3

f(x) is discontinuous at x=0


Option: 4

none of these.


Answers (1)

best_answer

\text { We have } f(x)=\left\{\begin{array}{cc} \frac{x}{\sqrt{x^2}} & x \neq 0 \\ 0, & x=0 \end{array}\right.

=\left\{\begin{array}{ll} \frac{x}{|x|}, & x \neq 0 \\ 0, & x=0 \end{array}=\left\{\begin{aligned} 1, & x>0 \\ -1, & x<0 \\ 0, & x=0 \end{aligned}\right.\right.

 f(x)  is not continuous at  x=0 

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Divya Prakash Singh

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