$$\eqalign{ & {\text{Given }}f\left( x \right) = x - {x^2} + {x^3} - {x^4} + .....\,{\text{to }}\infty \cr & \Rightarrow y = \frac{x}{{1 + x}}\,\,\,\,\,\left( {{\text{Infinite}}\,{\text{G}}{\text{.P}}{\text{.}}} \right) \cr & \Rightarrow y + xy = x \cr & \Rightarrow y = x\left( {1 - y} \right) \cr & \Rightarrow x = \frac{y}{{1 - y}} \cr & \Rightarrow {f^{ - 1}}\left( y \right) = \frac{y}{{1 - y}} \cr & \Rightarrow {f^{ - 1}}\left( x \right) = \frac{x}{{1 - x}} \cr} $$

$$\eqalign{ & {\text{Suppose }}a\, \in \,X{\text{ and }}a\, \in \,A \cr & \Rightarrow a\, \in \,X \cup A \cr & \Rightarrow a\, \in \,Y{\text{ and }}a\, \in \,A\,\,\left( {\because \,X \cup A = Y \cup A} \right) \cr & \Rightarrow a\, \in \,Y \cap A \cr & \Rightarrow Y \cap A{\text{ is non - empty}} \cr & {\text{This contradicts that }}Y \cap A = \phi \cr & {\text{So, }}X = Y \cr} $$

$${x^2} + {y^2} \leqslant 4,$$   represents all points interior to the circle $${x^2} + {y^2} = 4,$$   hence $$ - 2 \leqslant x \leqslant 2$$   and $$ - 2 \leqslant y \leqslant 2$$
$$\therefore $$  integral values of $$x$$ are $$ - 2,\, - 1,\,0,\,1,\,2$$

$$\eqalign{ & g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right) = \left[ {\left| x \right|} \right]; \cr & f\left( {g\left( x \right)} \right) = f\left( {\left[ x \right]} \right) = \left| {\left[ x \right]} \right| \cr & {\text{When }}x \geqslant 0,\,\left[ {\left| x \right|} \right] = \left[ x \right] = \left| {\left[ x \right]} \right| \cr & \therefore \,f\left( {g\left( x \right)} \right) = g\left( {f\left( x \right)} \right) \cr & {\text{When }}x < 0,\,\left[ x \right] \leqslant x < 0 \Rightarrow \left| {\left[ x \right]} \right| \geqslant \left| x \right| \cr & \therefore \,\left| {\left[ x \right]} \right| \geqslant \left| x \right| \geqslant \left[ {\left| x \right|} \right] \cr & \Rightarrow f\left( {g\left( x \right)} \right) \geqslant g\left( {f\left( x \right)} \right) \cr & {\text{Thus, }}g\left( {f\left( x \right)} \right) \leqslant f\left( {g\left( x \right)} \right){\text{ for all }}x\, \in \,R \cr} $$

If $$A \subset B$$  and $$B \subset C,$$   then these sets is represented in Venn diagram as

$$\eqalign{ & {\text{Clearly }}A \cup B = B{\text{ and }}B \cap C = B \cr & {\text{Hence}}{\text{,}}\,\,A \cup B = B \cap C \cr} $$

$$\eqalign{ & {\text{Let }}A = \left\{ {{n^2}:n\, \in \,N} \right\}{\text{ and }}B = \left\{ {{n^3}:n\, \in \,N} \right\} \cr & A = \left\{ {1,\,4,\,9,\,16,.....} \right\}{\text{ and }}B = \left\{ {1,\,8,\,27,\,64,.....} \right\} \cr & {\text{Now, }}A \cap B = \left\{ 1 \right\}{\text{ which is a finite set}} \cr & {\text{Also, }}A \cup B = \left\{ {1,\,4,\,8,\,9,\,27,.....} \right\} \cr & {\text{So, complement of }}A \cup B\,\,{\text{is infinite set}}{\text{.}} \cr & {\text{Hence, }}A \cup B \ne N \cr} $$

As given :
$$S =$$  the set of all triangles
$$P =$$  the set of all isosceles triangles
$$Q =$$  the set of all equilateral triangles
$$R =$$  the set of all right angled triangles
$$\therefore \,\,P \cap Q$$   represents the set of isosceles triangles and $$R - P$$  represents the set of non-isosceles right angled triangles.

$$P$$ : there is a rational number $$x \in S$$  such that $$x$$ > 0
$$ \sim P$$  : Every rational number $$x \in S$$  satisfies $$x \leqslant 0$$

$$\eqalign{ & f\left( x \right) = 5\,{\log _5}x \Rightarrow {f^{ - 1}}\left( x \right) = {5^{\frac{x}{5}}} \cr & {f^{ - 1}} \left( {\alpha - \beta } \right) = {5^{\frac{{\alpha - \beta }}{5}}} = \frac{{{5^{\frac{\alpha }{5}}}}}{{{5^{\frac{\beta }{5}}}}} = \frac{{{f^{ - 1}}\left( \alpha \right)}}{{{f^{ - 1}}\left( \beta \right)}} \cr} $$

Since, the given function has minimum value 75 which is attained at $$x = 2$$  and maximum value 89 which is attained at $$x = 3.$$   Hence, the range of $$f$$ is $$\left[ {75,\,89} \right].$$