Two sets A and B are as under: A = ( a,b ) R R: | a - 5 | < 1 , ,… - Vidyadip

MCQ

Two sets $A$ and $B$ are as under:

$A = \{ \left( {a,b} \right) \in R \times R:\left| {a - 5} \right| < 1 \,\,and\,\,\left| {b - 5} \right| < 1\} $; $B = \left\{ {\left( {a,b} \right) \in R \times R:4{{\left( {a - 6} \right)}^2} + 9{{\left( {b - 5} \right)}^2} \le 36} \right\}$ then : . . . . .

✓
$A \subset B$
B
$A \cap B = \emptyset $
C
neither $A \subset B$ nor $B \subset A$
D
$B \subset A$

✓

Answer

Correct option: A.

$A \subset B$

a
$A = \left\{ {\left( {a,b} \right) \in R \times R:\left| {a - 5} \right| < 1,\left| {b - 5} \right| < 1} \right\}$

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, $A \subset B$.

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