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B
limx→2−f(x)≠0
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C
limx→2+f(x)≠limx→2−f(x)
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D
f(x)iscontinuousatx=2
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Solution
The correct option is Df(x)iscontinuousatx=2 Here f(2) = 0 limx→2−f(x)=limk→0f(2−h)=limk→0|2−h−2|=0 limx→2+f(x)=limk→0f(2+h)=limk→0|2+h−2|=0 Hence it is continuous at x = 2