Question
Let $$f\left( x \right) = \left[ {{x^2}} \right] - {\left[ x \right]^2},$$ where $$\left[ \cdot \right]$$ denotes the greatest integer function. Then :
A.
$$f\left( x \right)$$ is discontinuous for all integral values of $$x$$
B.
$$f\left( x \right)$$ is discontinuous only at $$x=0,\,1$$
C.
$$f\left( x \right)$$ is continuous only at $$x=1$$
D.
none of these
Answer :
$$f\left( x \right)$$ is continuous only at $$x=1$$
Solution :
$$\eqalign{
& f\left( {1 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\left[ {{{\left( {1 + h} \right)}^2}} \right] - {{\left[ {1 + h} \right]}^2}} \right\} = \mathop {\lim }\limits_{h \to 0} \left\{ {1 - 1} \right\} = 0 \cr
& f\left( {1 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\left[ {{{\left( {1 - h} \right)}^2}} \right] - {{\left[ {1 - h} \right]}^2}} \right\} = \mathop {\lim }\limits_{h \to 0} \left\{ {0 - 0} \right\} = 0 \cr
& {\text{Also }}f\left( 1 \right) = 0.{\text{ So, }}\,f\left( x \right)\,\,{\text{is continuous at }}x = 1. \cr} $$