The function $$f\left( x \right) = {\left[ x \right]^2} - \left[ {{x^2}} \right]$$    (where $$\left[ y \right]$$ is the greatest integer less than or equal to $$y$$ ), is discontinuous at-

Solution :
$$\eqalign{ & {\text{We have }}f\left( x \right) = {\left[ x \right]^2} - \left[ {{x^2}} \right] \cr & {\text{At }}x = 0, \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} f\left( { - h} \right) = \mathop {\lim }\limits_{h \to 0} {\left[ { - h} \right]^2} - \left[ {{{\left( { - h} \right)}^2}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} f{\left( { - 1} \right)^2} - \left[ {{h^2}} \right] = \mathop {\lim }\limits_{h \to 0} 1 - 0 = 1 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} f\left( h \right) = \mathop {\lim }\limits_{h \to 0} {\left[ h \right]^2} - \left[ {{h^2}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,0 - 0 = 0 \cr & \therefore \,{\text{L}}{\text{.H}}{\text{.L}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} \cr & \therefore f\left( x \right){\text{is not continuous at }}x = 0 \cr & {\text{At }}x = 1 \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right) = \mathop {\lim }\limits_{h \to 0} {\left[ {1 - h} \right]^2} - \left[ {{{\left( {1 - h} \right)}^2}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,0 - 0 = 0 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. }} = \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right) = \mathop {\lim }\limits_{h \to 0} {\left[ {1 + h} \right]^2} - \left[ {{{\left( {1 + h} \right)}^2}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \,1 - 1 = 0 \cr & f\left( 1 \right) = {\left[ 1 \right]^2} - \left[ {{1^2}} \right] = 1 - 1 = 0 \cr & \therefore \,{\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( 1 \right) \cr & \therefore f\left( x \right)\,\,{\text{is continuous at }}x = {\text{ }}1{\text{ }} \cr} $$
Clearly at other integral pts $$f\left( x \right)$$  is not continuous.