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If \mathrm{f(x)=|2-x|+(2+x)}, where \mathrm{(x)}= the least integer greater than or equal to \mathrm{x}, then
 

Option: 1

\mathrm{ f(2-0)=f(2)=4 }

 


Option: 2

 \mathrm{f(x)} is continuous at \mathrm{ x=2}
 


Option: 3

\mathrm{ f(x)} is nondifferentiable at \mathrm{ x \neq 2}
 


Option: 4

\mathrm{f(x) }is differentiable but not continuous at \mathrm{ x=2 }


Answers (1)

best_answer

\mathrm{|2-x|} is continuous everywhere and \mathrm{(2+x)} is discontinuous at all integral values of \mathrm{x}. So, \mathrm{f(x)}  is not continuous at \mathrm{x=2}. Hence, \mathrm{f(x)} is not differentiable at \mathrm{x=2} because not continuous \Rightarrow not differentiable.

\begin{aligned} & \mathrm{f(2-0)=\lim _{h \rightarrow 0}\{|2-(2-h)|+.(2+2-h)\}=\lim _{h \rightarrow 0}\{h+4\}=4 . }\\ &\mathrm{ f(2)=|2-2|+(2+2)=|0|+(4)=0+4=4 \text {. So, } f(2-0)=f(2)=4 . }\end{aligned}

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