Let $A = {2, 3, 4}$ and $B = {8, 9, 12}$. Then the number of elements in the relation $R = {((a_1, b_1), (a_2, b_2)) \in (A \times B, A \times B) : a_1$ divides $b_2$ and $a_2$ divides $b_1}$ is :

Two sets A and B are as under : A = {$(a, b) ∈ R × R : |a - 5| < 1$ and $|b - 5| < 1$}; B = {$(a, b) ∈ R × R : 4(a - 6)^2 + 9(b - 5)^2 ≤ 36$}; Then

Consider the relations $R_1$ and $R_2$ defined as $aR_1b \Leftrightarrow a^2 + b^2 = 1$ for all $a, b \in R$ and $(a, b)R_2(c, d) \Leftrightarrow a+ d = b + c$ for all $(a, b), (c, d) \in N \times N$. Then:

Let $A = {x \in (0, \pi) - {\frac{\pi}{2}} : \log_{(2/\pi)} |\sin x| + \log_{(2/\pi)} |\cos x| = 2}$ and $B = {x \ge 0 : x(x-4) - 3|x-2| + 6 = 0}$. Then $n(A \cup B)$ is equal to :

Let $\mathrm{A}=\{-4,-3,-2,0,1,3,4\}$ and $\mathrm{R}=\left\{(a, b) \in \mathrm{A} \times \mathrm{A}: b=|a|\right.$ or $\left.b^{2}=a+1\right\}$ be a relation on $\mathrm{A}$. Then the minimum number of elements, that must be added to the relation $\mathrm{R}$ so that it becomes reflexive and symmetric, is __________