We are given that
\[\begin{array}{l} f:R \to R\,{\rm{such}}\,{\rm{that}}\,f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0,}\\ {x,} \end{array}} \right.\begin{array}{*{20}{c}} {x \in {\rm{ rational}}}\\ {x \in {\rm{ irrational}}} \end{array}\\ g:R \to R\,{\rm{such}}\,{\rm{that}}\,g\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0,}\\ {x,} \end{array}} \right.\,\begin{array}{*{20}{c}} {x \in {\rm{ irrational}}}\\ {x \in {\rm{ rational}}} \end{array} \end{array}\]
$$\therefore \left( {f - g} \right):R \to R\,{\text{such}}\,{\text{that}}\,$$
\[\left( {f - g} \right)\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { - x,}\\ x \end{array}\,\,\begin{array}{*{20}{c}} {{\rm{if}}}\\ {{\rm{if}}} \end{array}\,\begin{array}{*{20}{c}} {x \in {\rm{ rational}}}\\ {x \in {\rm{ irrational}}} \end{array}} \right.\]
Since $$f - g:R \to R$$    for any $$x$$ there is only one value of $$\left( {f\left( x \right) - g\left( x \right)} \right)$$     whether $$x$$ is rational or irrational. Moreover as $$x \in R,f\left( x \right) - g\left( x \right)$$     also belongs to $$R.$$ Therefore, $$\left( {f - g} \right)$$   is one-one onto.

For $$f\left( x \right) = \sqrt {{{\sin }^{ - 1}}\left( {2x} \right) + \frac{\pi }{6}} $$      to be defined and real if $$si{n^{ - 1}}2x + \frac{\pi }{6} \geqslant 0$$
$$ \Rightarrow {\sin ^{ - 1}}2x \geqslant - \frac{\pi }{6}\,......\left( 1 \right)$$
But we know that
$$ - \frac{\pi }{2} \leqslant {\sin ^{ - 1}}2x \leqslant \frac{\pi }{2}\,......\left( 2 \right)$$
Combining (1) and (2), we get
$$\eqalign{ & - \frac{\pi }{6} \leqslant {\sin ^{ - 1}}2x \leqslant \frac{\pi }{2} \cr & \Rightarrow \sin \left( { - \frac{\pi }{6}} \right) \leqslant 2x \leqslant \sin \left( {\frac{\pi }{2}} \right) \Rightarrow - \frac{1}{2} \leqslant 2x \leqslant 1 \cr & \Rightarrow - \frac{1}{4} \leqslant x \leqslant \frac{1}{2}\therefore {D_f} = \left[ { - \frac{1}{4},\frac{1}{2}} \right] \cr} $$

$$\eqalign{ & af\left( {x + 1} \right) + bf\left( {\frac{1}{{x + 1}}} \right) = x \cr & {\text{put}}\,t = x + 1 \Rightarrow x = t - 1 \cr & af\left( t \right) + bf\left( {\frac{1}{t}} \right) = t - 1.....\left( 1 \right) \cr & af\left( {\frac{1}{t}} \right) + bf\left( t \right) = \frac{{1 - t}}{t}.....\left( 2 \right) \cr & {\text{Equation }}\left( 1 \right) + \left( 2 \right),{\text{ we get}} \cr & \Rightarrow \left( {a + b} \right)\left( {f\left( t \right) + f\left( {\frac{1}{t}} \right)} \right) = \frac{{{{\left( {t - 1} \right)}^2}}}{t} \cr & {\text{Equation }}\left( 1 \right) - \left( 2 \right),{\text{ we get}} \cr & \Rightarrow \left( {a - b} \right)\left( {f\left( t \right) + f\left( {\frac{1}{t}} \right)} \right) = \frac{{{t^2} - 1}}{t} \cr & f\left( t \right) = \frac{{{{\left( {t - 1} \right)}^2}}}{{t\left( {a + b} \right)}} + \frac{{{t^2} - 1}}{{t\left( {a - b} \right)}} \cr & f\left( 2 \right) = \frac{1}{2}\left( {\frac{1}{{a + b}} + \frac{3}{{a - b}}} \right) \cr & f\left( 2 \right) = \frac{1}{2}\left( {\frac{{a - b + 3a + 3b}}{{{a^2} - {b^2}}}} \right) \cr & f\left( 2 \right) = \frac{{2a + b}}{{{a^2} - {b^2}}} \cr} $$

A function $$y = f\left( x \right)$$  is symmetrical about the origin if $$\left( {x,\,y} \right)$$  is on the graph implies that $$\left( { - x,\, - y} \right)$$  is also on the graph, i.e., $$ - y = f\left( { - x} \right).$$    In other words, $$f\left( { - x} \right) = - f\left( x \right).$$

In $$R - \left[ {0,\,2} \right],\,f\left( {x + 2} \right) = f\left( x \right)$$
The graph of $$f\left( x \right) = x\left( {2 - x} \right),\,0 \leqslant x \leqslant 2$$      is shown.
Clearly, $$f$$ is a periodic function of period 2.

$$\eqalign{ & f\left( x \right) = \frac{{{{\sin }^{ - 1}}\left( {x - 3} \right)}}{{\sqrt {9 - {x^2}} }}\,{\text{is}}\,{\text{defined}} \cr & {\text{if }}\left( {\text{i}} \right) - 1 \leqslant x - 3 \leqslant 1 \Rightarrow 2 \leqslant x \leqslant 4 \cr & {\text{and}}\,\left( {{\text{ii}}} \right)\,9 - {x^2} > 0 \Rightarrow - 3 < x < 3 \cr & {\text{we}}\,{\text{get}}\,2 \leqslant x < 3\,\,\therefore {\text{Domain}} = \left[ {2,3} \right) \cr} $$

$$\eqalign{ & x > 0{\text{ and }}{\log _e}x \ne 0 \cr & \therefore x > 0{\text{ and }}x \ne 1 \cr & \therefore {\text{domain}} = \left( {0,\,1} \right) \cup \left( {1,\, + \infty } \right) \cr} $$

$$f\left( x \right)$$  is continuous and defined for all $$x > 0$$  and $$f\left( {\frac{x}{y}} \right) = f\left( x \right) - f\left( y \right)$$
Also $$f\left( e \right) = 1$$
$$ \Rightarrow $$ Clearly $$f\left( x \right) = \ell n\,x$$    which satisfies all these properties
$$\therefore f\left( x \right) = \ell n\,x$$

The given relation is
$$\eqalign{ & \left( {{f_2}oJo{f_1}} \right)\left( x \right) = {f_3}\left( x \right) = \frac{1}{{1 - x}} \cr & \Rightarrow \left( {{f_2}oJ} \right)\left( {{f_1}\left( x \right)} \right) = \frac{1}{{1 - x}} \cr & \Rightarrow \left( {{f_2}oJ} \right) = \left( {\frac{1}{x}} \right) = \frac{1}{{1 - x}}\,\left[ {\because {f_1}\left( x \right) = \frac{1}{x}} \right] \cr & \Rightarrow {f_2}\left( {J\left( {\frac{1}{x}} \right)} \right) = \frac{1}{{1 - x}} \cr & \Rightarrow \left( {{f_2}J\left( x \right)} \right) = \frac{1}{{1 - \frac{1}{x}}} = \frac{x}{{x - 1}}\left[ {\frac{1}{x}{\text{ is replaced by }}x} \right] \cr & \Rightarrow 1 - J\left( x \right) = \frac{x}{{x - 1}}\left[ {\lambda {f_2}\left( x \right) = 1 - x} \right] \cr & \therefore J\left( x \right) = 1 - \frac{x}{{x - 1}} = \frac{1}{{1 - x}} = {f_3}\left( x \right) \cr} $$

Given that $$X$$ and $$Y$$ are two sets and $$f:X \to Y.$$
$$\eqalign{ & \left\{ {f\left( c \right) = y;c \subset X,y \subset Y} \right\}\,{\text{and}} \cr & \left\{ {{f^{ - 1}}\left( d \right) = x:d \subset Y,x \subset X} \right\} \cr} $$
The pictorial representation of given information is as shown:

$$\eqalign{ & {\text{Since }}{f^{ - 1}}\left( d \right) = x \Rightarrow f\left( x \right) = d{\text{ Now if }}a \subset x \cr & \Rightarrow f\left( a \right) \subset f\left( x \right) = d \Rightarrow \,{f^{ - 1}}\left[ {f\left( a \right)} \right] = a \cr & \therefore {f^{ - 1}}\left( {f\left( a \right)} \right) = a,a \subset x{\text{ is the correct option}}{\text{.}} \cr} $$