Which of the following function is continuous at for all value of $$x\,?$$
$$\eqalign{ & \left( {\text{i}} \right)\,\,f\left( x \right) = \operatorname{sgn} \left( {{x^3} - x} \right) \cr & \left( {{\text{ii}}} \right)\,\,f\left( x \right) = \operatorname{sgn} \left( {2\,\cos \,x - 1} \right) \cr & \left( {{\text{iii}}} \right)\,\,f\left( x \right) = \operatorname{sgn} \left( {{x^2} - 2x + 3} \right) \cr} $$

Solution :
$$\eqalign{ & \left( {\bf{i}} \right)\,\,f\left( x \right) = \operatorname{sgn} \left( {{x^3} - x} \right) \cr & {\text{Here }}{x^3} - x = 0 \cr & \Rightarrow x = 0,\, - 1,\,1 \cr & {\text{Hence,}}\,f\left( x \right){\text{ is discontinuous at }}x = 0,\, - 1,\,1 \cr & \left( {{\bf{ii}}} \right)\,\,{\text{If }}f\left( x \right) = \operatorname{sgn} \left( {2\,\cos \,x - 1} \right) \cr & {\text{Here, }}2\,\cos \,x - 1 = 0 \cr & \Rightarrow \,\cos \,x = \frac{1}{2} \cr & \Rightarrow x = \,2n\pi + \left( {\frac{\pi }{3}} \right),\,n\, \in \,Z,\,{\text{where }}f\left( x \right){\text{ is discontinuous}}. \cr & \left( {{\bf{iii}}} \right)\,\,f\left( x \right) = \operatorname{sgn} \left( {{x^2} - 2x + 3} \right) \cr & {\text{Here,}}\,{x^2} - 2x + 3 > 0{\text{ for all }}x \cr & {\text{Thus, }}f\left( x \right) = 1{\text{ for all }}x \cr & {\text{Hence, continuous for all }}x. \cr} $$