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दो समुच्चय $A$ तथा $B$ निम्न प्रकार के हैं
$A=\{(a, b) \in R \times R:|a-5|< 1$ तथा $|b-5|< 1\}$
$B=\left\{(a, b) \in R \times R: 4(a-6)^{2}+9(b-5)^{2} \leq 36\right\}$ तो
$A \subset B$
$A \cap B=\phi$ (एक रिक्त समुच्चय)
न तो $A \subset B$ और न ही $B \subset A$
$B \subset A$
Solution
$A = \left\{ {\left( {a,b} \right) \in R \times R:\left| {a – 5} \right| < 1,\left| {b – 5} \right| < 1} \right\}$
Let $a – 5 = x,b – 5 = y$
set $A$ contains all points inside $\left| x \right| < 1,\left| y \right| < 1$
$B = \left\{ {\left( {a,b} \right) \in R \times R:4{{\left( {a – 6} \right)}^2} + 9{{\left( {B – 5} \right)}^2} \le 36} \right\}$
St $B$ contains all points inside or on
$\frac{{{{\left( {x – 1} \right)}^2}}}{9} + \frac{{{y^2}}}{4} = 1$
$\left( { \pm 1, \pm 1} \right)$ lies inside the ellipse.
Hence, $A \subset B$.
