Question

Two sets $A$ and $B$ are defined as
$A=\left\{\right(x,y)\in R\times R:|x-3|+|y-3|\le 3\}$
$B=\left\{\right(x,y)\in R\times R:|x{|}^2+|y{|}^2\le 6^2\}$,
then

• A. $A\subset B$
• B. $B\subset A$
• C. $A\cap B=\varphi$
• D. $A\subset ̸B$ and $B\subset ̸A$

Answer 1

The correct option is D $A\subset ̸B$ and $B\subset ̸A$

$A=\left\{\right(x,y)\in R\times R:|x-3|+|y-3|\le 3\}$
$B=\left\{\right(x,y)\in R\times R:|x{|}^2+|y{|}^2\le 6^2\}$
From graph it is clear that,
$A\subset ̸B$ and $B\subset ̸A$

Alternate solution:
$A=\left\{\right(x,y)\in R\times R:|x-3|+|y-3|\le 3\}$
Vertices $(x,y)=(3,0),(0,3),(6,3)(3,6)$
Since, $(6,3)(3,6)$ do not satisfy the inequality $|x{|}^2+|y{|}^2\le 6^2$
$\Rightarrow A\subset ̸B$
Similarly, $(-6,0)$ satisfies $|x{|}^2+|y{|}^2\le 6^2$ but not $|x-3|+|y-3|\le 3$
$\Rightarrow B\subset ̸A$