\[{\rm{Let }}f\left( x \right) = \left\{ \begin{array}{l} \sqrt {1 + {x^2}} ,\,x < \sqrt 3 \\ \sqrt 3 x - 1,\,\sqrt 3 \le x < 4\\ \left[ x \right],\,4 \le x < 5\\ \left| {1 - x} \right|,\,x \ge 5 \end{array} \right.,\]     $$\eqalign{ & {\text{where}}\,\,\left[ x \right]{\text{ is the greatest integer }} \leqslant x \cr} $$
The number of points of discontinuity of $$f\left( x \right)$$  in $$R$$ is :

Solution :
We know that $$\left| {1 - x} \right|$$  is continuous everywhere, $$\left[ x \right]$$ is continuous everywhere except at integers, $$\sqrt {1 + {x^2}} $$   and $$\sqrt 3 x - 1$$   are continuous in their respective intervals of definition. So, the only doubtful points of continuity are $$x = \sqrt 3 ,\,4,\,5$$
$$\eqalign{ & f\left( {\sqrt 3 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \sqrt {1 + {{\left( {\sqrt 3 - h} \right)}^2}} = 2\,; \cr & f\left( {\sqrt 3 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\sqrt 3 \left( {\sqrt 3 + h} \right) - 1} \right\} = 2\,;\,\,\,\,\,f\left( {\sqrt 3 } \right) = 2 \cr & f\left( {4 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {\sqrt 3 \left( {4 - h} \right) - 1} \right\} = 4\sqrt 3 - 1\,;\,\,\,f\left( {4 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] = 4 \cr & f\left( {5 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \left[ {5 - h} \right] = \mathop {\lim }\limits_{h \to 0} \,4 = 4\,; \cr & f\left( {5 + 0} \right) = \mathop {\lim }\limits_{h \to 0} \left| {1 - \left( {5 + h} \right)} \right| = 4\,;\,\,\,\,\,f\left( 5 \right) = 4 \cr & \therefore f\left( x \right){\text{ is discontinuous at }}x = 4{\text{ only}}{\text{.}} \cr} $$