We know that $$\left[ {\left( {A \cup B} \right) \cap C} \right]' = A' \cap B' \cup C'$$

The given condition is as follows-

$$\eqalign{ & {\text{We know that}} \cr & \left\{ {\left( {a + d + e + g} \right) + \left( {b + d + f + g} \right) + \left( {c + e + f + g} \right)} \right\} - \left( {d + e + f} \right) - 2g = a + b + c + d + e + f + g \cr & {\text{or, }}61x + 46x + 29x - 25x - 2g = 97x \cr & {\text{or, }}2g = 14x \cr & {\text{or, }}g = 7 \cr} $$

\[\begin{array}{l} \left( {fog} \right)\left( x \right) = \left\{ \begin{array}{l} f\left( {x + 3} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \le x + 3 \le 2\\ f\left( { - x + 1} \right),\,\,\,\, - 1 \le - x + 2 \le 2 \end{array} \right.\\ = \left\{ \begin{array}{l} f\left( {x + 3} \right),\,\,\,\,\,\,\,\,\,\,\,\,1 \le x + 3 \le 2\\ f\left( { - x + 1} \right),\,\,\, - 1 \le - x + 1 \le 1\\ f\left( { - x + 1} \right),\,\,\,\,\,\,\,\,1 \le - x + 1 \le 2 \end{array} \right.\\ = \left\{ \begin{array}{l} \,\,\,\,\,x + 1,\,\,\,\,\, - 2 \le x \le - 1\\ - x - 1,\,\,\,\,\, - 1 \le x \le 0\\ \,\,\,\,\,x - 1,\,\,\,\,\,\,\,\,\,\,\,0 \le x \le 2 \end{array} \right. \end{array}\]

$$\eqalign{ & {\text{Given }}n\left( N \right) = 12,\,n\left( P \right) = 16,\,n\left( H \right) = 18,\,n\left( {N \cup P \cup H} \right) = 30 \cr & {\text{From, }}n\left( {N \cup P \cup H} \right) = n\left( N \right) + n\left( P \right) + n\left( H \right) - n\left( {N \cap P} \right) - n\left( {P \cap H} \right) - n\left( {N \cap H} \right) + n\left( {N \cap P \cap H} \right) \cr & \therefore \,n\left( {N \cap P} \right) + n\left( {P \cap H} \right) + n\left( {N \cap H} \right) = 16 \cr} $$
Now, number of pupils taking two subjects
$$\eqalign{ & = n\left( {N \cap P} \right) + n\left( {P \cap H} \right) + n\left( {N \cap H} \right) - 3n\left( {N \cap P \cap H} \right) \cr & = 16 - 0 \cr & = 16 \cr} $$

$$\eqalign{ & \left( {A \cup B} \right)' \cup \left( {A' \cap B} \right) \cr & = \left( {A' \cap B'} \right) \cup \left( {A' \cap B} \right) \cr & = \left( {A' \cup A'} \right) \cap \left( {A' \cup B} \right) \cap \left( {B' \cup A'} \right) \cap \left( {B' \cup B} \right) \cr & = A' \cap \left\{ {A' \cup \left( {B \cap B'} \right)} \right\} \cap U \cr & = A' \cap \left( {A' \cup \phi } \right) \cap U \cr & = A' \cap A' \cap U \cr & = A' \cap U \cr & = A' \cr} $$

$$\eqalign{ & {\text{If }}y = \frac{2}{3}\frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}},\,\,\,{10^{2x}} = \frac{{3y + 2}}{{2 - 3y}} \cr & {\text{or }}x = \frac{1}{2}\,{\log _{10}}\frac{{2 + 3y}}{{2 - 3y}} \cr & \therefore \,{f^{ - 1}}\left( x \right) = \frac{1}{2}\,{\log _{10}}\frac{{2 + 3x}}{{2 - 3x}}\, \cr} $$

Required number of one to one function
$$\eqalign{ & = {}^6{P_4} \cr & = 6 \times 5 \times 4 \times 3 \cr & = 360 \cr} $$

$$\eqalign{ & D\left( f \right) = R;\,D\left( g \right) = R - \left[ { - 1,\,0} \right) \cr & \therefore \,\,D\left( {f + g} \right) \cr & = D\left( f \right) \cap D\left( g \right) \cr & = R \cap \left( {R - \left[ { - 1,\,0} \right)} \right) \cr & = R \cap \left[ { - 1,\,0} \right) \cr & R\left( f \right) = \left[ { - \frac{1}{2},\,\frac{1}{2}} \right];\,\,R\left( g \right) = R - \left\{ 0 \right\} \cr & \therefore \,\,R\left( f \right) \cap R\left( g \right) \cr & = \left[ { - \frac{1}{2},\,\frac{1}{2}} \right] \cap \left( {R - \left\{ 0 \right\}} \right) \cr & = \left[ { - \frac{1}{2},\,\frac{1}{2}} \right] - \left\{ 0 \right\} \cr} $$

Let $$M =$$  set of Mathematics teachers
$$P =$$  set of Physics teachers $$n$$ (only Maths teachers)
$$ = n\left( M \right) - n\left( {M \cap P} \right) = 12 - 4 = 8$$
Also, $$n\left( {M \cup P} \right) = n$$    (only Maths teachers) $$ +\, n$$ (only Physics teachers) $$ +\, n\left( {M \cap P} \right)$$
$$20 = 8 + 4 + n$$    (only Physics teachers)
$$ \Rightarrow \,n = 8.$$

We observe the following properties :
Reflexivity : Let $$\left( {a,\,b} \right)$$  be an arbitrary element of $${\bf{N}} \times {\bf{N}}$$
$$\eqalign{ & {\text{Then, }}\left( {a,\,b} \right) \in \,{\bf{N}} \times {\bf{N}} \Rightarrow a,\,b\, \in \,{\bf{N}} \cr & \Rightarrow ab\left( {b + a} \right) = ba\left( {a + b} \right) \Rightarrow \left( {a,\,b} \right)R\left( {a,\,b} \right) \cr & {\text{Thus, }}\left( {a,\,b} \right)R\left( {a,\,b} \right){\text{ for all }}\left( {a,\,b} \right) \in \,{\bf{N}} \times {\bf{N}} \cr} $$
So, $$R$$ is reflexive on $${\bf{N}} \times {\bf{N}}$$
Symmetry : Let $$\left( {a,\,b} \right),\,\left( {c,\,d} \right)\, \in \,{\bf{N}} \times {\bf{N}}$$      be such that $$\left( {a,\,b} \right)R\left( {c,\,d} \right).$$
$$\eqalign{ & {\text{Then, }}\left( {a,\,b} \right)R\left( {c,\,d} \right)\, \cr & \Rightarrow ad\left( {b + c} \right) = bc\left( {a + d} \right) \cr & \Rightarrow cb\left( {d + a} \right) = da\left( {c + b} \right) \cr & \Rightarrow \left( {c,\,d} \right)R\left( {a,\,b} \right) \cr & {\text{Thus, }}\left( {a,\,b} \right)R\left( {c,\,d} \right) \cr & \Rightarrow \left( {c,\,d} \right)R\left( {a,\,b} \right){\text{ for all }}\left( {a,\,b} \right)\left( {c,\,d} \right)\, \in \,{\bf{N}} \times {\bf{N}} \cr} $$
So, $$R$$ is symmetric on $${\bf{N}} \times {\bf{N}}$$
Transitivity : Let $$\left( {a,\,b} \right),\,\left( {c,\,d} \right),\,\left( {e,\,f} \right)\, \in \,{\bf{N}} \times {\bf{N}}$$       such that $$\left( {a,\,b} \right)R\left( {c,\,d} \right){\text{ and }}\left( {c,\,d} \right)R\left( {e,\,f} \right)$$
$$\eqalign{ & {\text{Then, }}\left( {a,\,b} \right)R\left( {c,\,d} \right)\, \cr & \Rightarrow ad\left( {b + c} \right) = bc\left( {a + d} \right) \cr & \Rightarrow \frac{{b + c}}{{bc}} = \frac{{a + d}}{{ad}} \cr & \Rightarrow \frac{1}{b} + \frac{1}{c} = \frac{1}{a} + \frac{1}{d}.......({\text{i}}) \cr & {\text{and, }}\left( {c,\,d} \right)R\left( {e,\,f} \right) \Rightarrow cf\left( {d + e} \right) = de\left( {c + f} \right) \cr & \Rightarrow \frac{{d + e}}{{de}} = \frac{{c + f}}{{cf}} \cr & \Rightarrow \frac{1}{d} + \frac{1}{e} = \frac{1}{c} + \frac{1}{f}.......({\text{ii}}) \cr & {\text{Adding (i) and (ii), we get}} \cr & \left( {\frac{1}{b} + \frac{1}{c}} \right) + \left( {\frac{1}{d} + \frac{1}{e}} \right) = \left( {\frac{1}{a} + \frac{1}{d}} \right) + \left( {\frac{1}{c} + \frac{1}{f}} \right) \cr & \Rightarrow \frac{1}{b} + \frac{1}{e} = \frac{1}{a} + \frac{1}{f} \cr & \Rightarrow \frac{{b + e}}{{be}} = \frac{{a + f}}{{af}} \cr & \Rightarrow af\left( {b + e} \right) = be\left( {a + f} \right) \cr & \Rightarrow \left( {a,\,b} \right)R\left( {e,\,f} \right) \cr & {\text{Thus, }}\left( {a,\,b} \right)R\left( {c,\,d} \right){\text{ and }}\left( {c,\,d} \right)R\left( {e,\,f} \right) \cr & \Rightarrow \left( {a,\,b} \right)R\left( {e,\,f} \right){\text{ for all }}\left( {a,\,b} \right),\,\left( {c,\,d} \right),\,\left( {e,\,f} \right)\, \in \,{\bf{N}} \times {\bf{N}} \cr} $$
So, $$R$$ is transitive on $${\bf{N}} \times {\bf{N}}.$$
Hence, $$R$$ being reflexive, symmetric and transitive, is an equivalence relation on $${\bf{N}} \times {\bf{N}}.$$