A) discontinuous at origin because \[\left| x \right|\] is discontinuous there
B) continuous at origin
C) discontinuous at origin because both \[\left| x \right|\] and \[\frac{\left| x \right|}{x}\] are discontinuous there
D) discontinuous at the origin because is discontinuous there.
Correct Answer: D
Solution :
Let \[f(x)=|x|+\frac{|x|}{x}\] Let us consider \[f(x)={{f}_{1}}(x)+{{f}_{2}}(x)\] Now at \[x=0\] \[{{f}_{1}}(0)=0\] LHL \[=\lim \,\,|x|=0\] \[x\to {{0}^{-}}\] and RHL \[=\lim \,\,|x|=0\] \[x\to {{0}^{+}}\] \[\therefore \,\,{{f}_{1}}(x)\] is continuous at \[x=0\] Now consider \[{{f}_{2}}(x)\] at \[x=0\] \[{{f}_{2}}(0)=0\] Now LHL \[\lim \,\frac{|x|}{x}\] \[x\to {{0}^{-}}\] Put \[x=0-h\] \[\lim \,\frac{|0-h|}{0-h}\] \[x\to {{0}^{-}}\] \[=\lim -\frac{h}{h}=-1\] \[x\to {{0}^{-}}\] now RHL \[\lim \,\frac{|x|}{x}\] \[x\to {{0}^{+}}\] \[=\lim \,\left| \frac{0+h}{0+h} \right|\] [Put \[x=0+h\]] \[x\to {{0}^{+}}\] \[=\lim \,\frac{h}{h}=I\] \[x\to 0\] \[\therefore \] \[LHL\ne RHL\] \[\therefore \,\,{{f}_{2}}(x)\] is discontinuous at origin. \[\therefore \,\,f(x)=|x|+\frac{|x|}{x}\] is discontinuous at \[x=0\] because \[\frac{|x|}{x}\] is discontinuous there.
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