$$\eqalign{ & f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_3}\left( {\frac{x}{3}} \right)} \right){\text{ exists}} \cr & {\text{if }} - 1 \leqslant {\log _3}\left( {\frac{x}{3}} \right) \leqslant 1 \Leftrightarrow {3^{ - 1}} \leqslant \frac{x}{3} \leqslant {3^1} \cr & \Leftrightarrow 1 \leqslant x \leqslant 9{\text{ or }}x \in \left[ {1,9} \right] \cr} $$

Let us consider a graph symmetric with respect to line $$x = 2$$  as shown in the figure.

$$\eqalign{ & {\text{From the figure :}} \cr & f\left( {{x_1}} \right) = f\left( {{x_2}} \right),{\text{ where }}{x_1} = 2 - x{\text{ and }}{x_2} = 2 + x \cr & \therefore \,f\left( {2 - x} \right) = f\left( {2 + x} \right) \cr} $$

$$\eqalign{ & f\left( {\frac{\pi }{2}} \right) = \cos \,\left[ \pi \right].\frac{\pi }{2} + \cos \,\left[ {\pi .\frac{\pi }{2}} \right] \cr & = \cos \,\frac{{3\pi }}{2} + \cos \,4 \cr & = \cos \,4 \cr} $$

$$\eqalign{ & {\text{We have }}\left[ {x + y} \right] = \left[ x \right] + \left[ {y + x - \left[ x \right]} \right]{\text{ for all }}x,\,y\, \in \,R. \cr & {\text{It is proved as below}}{\text{.}} \cr & {\text{If }}x{\text{ is integer, }}\left[ {x + y} \right] = x + \left[ y \right] = \left[ x \right] + \left[ {y + x - \left[ x \right]} \right],{\text{ because }}x - \left[ x \right] = 0 \cr & {\text{If }}n < x < n + 1,\,x = n + k,\,0 < k < 1.{\text{ Then}} \cr & \left[ {x + y} \right] = \left[ {n + k + y} \right] = n + \left[ {y + k} \right] = \left[ x \right] + \left[ {y + x - \left[ x \right]} \right],{\text{ because}}\,x - \left[ x \right] = k. \cr} $$

$$\eqalign{ & {\text{We have}}\,f\left( x \right) = {x^2} + 2bx + 2{c^2};g\left( x \right) = - {x^2} - 2cx + {b^2} \cr & \Rightarrow f\left( x \right) = {\left( {x + b} \right)^2} + 2{c^2} - {b^2} \cr & {\text{and }}g\left( x \right) = - {\left( {x + c} \right)^2} + {b^2} + {c^2} \cr & \Rightarrow {f_{\min }} = 2{c^2} - {b^2}{\text{ and }}{g_{\max }} = {b^2} + {c^2} \cr & {\text{For }}{f_{\min }} > {g_{\max }} \Rightarrow 2{c^2} - {b^2} > {b^2} + {c^2} \cr & \Rightarrow {c^2} > 2{b^2} \Rightarrow \left| c \right| > \left| b \right|\sqrt 2 \cr} $$

$$\eqalign{ & f\left( {2a - x} \right) = f\left( {a - \left( {x - a} \right)} \right) = f\left( a \right)f\left( {x - a} \right) - f\left( 0 \right) \cr & f\left( x \right) = f\left( a \right)f\left( {x - a} \right) - f\left( x \right) = - f\left( x \right) \cr & \left[ {\because \,x = 0,\,y = 0,\,f\left( 0 \right) = {f^2}\left( 0 \right) - {f^2}\left( a \right) \Rightarrow {f^2}\left( a \right) = 0 \Rightarrow f\left( a \right) = 0} \right] \cr & \Rightarrow f\left( {2a - x} \right) = - f\left( x \right) \cr} $$

$${x^2} - {\left[ x \right]^2} \geqslant 0\,\, \Rightarrow {x^2} \geqslant {\left[ x \right]^2}$$
This is true for all positive $$x$$ and all negative integer $$x.$$

Here, $$f\left( x \right)$$  is defined only when $$x + 3 \ne 0,$$   i.e., when $$x \ne - 3$$
$$\therefore \,\,D\left( f \right) = R - \left\{ { - 3} \right\}$$

$$f\left( x \right) = \sqrt {\left( {x - 2} \right)\left( {x - 3} \right)} + \sqrt { - \left( {x - 4} \right)\left( {x + 2} \right)} $$
The first part is real outside $$\left( {2,\,3} \right)$$  and the second is real in $$\left[ { - 2,\,4} \right]$$  so that the domain is $$\left[ { - 2,\,2} \right] \cup \left[ {3,\,4} \right].$$