Question

Two sets $A$ and $B$ are as under:
$A=\left\{\right(a,b)ϵR\times R:|a-5|<1and|b-5|<1\}$;
$B=\{a,b)ϵR\times R:4(a-6{)}^2+9(b-5{)}^2\le 36\}$. Then.

• A. $A\cap B=\Phi$ (an empty set)
• B. Neither $A\subset B$ not $B\subset A$
• C. $B\subset A$
• D. $A\subset B$

Answer 1

The correct option is D $A\subset B$

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

Now, $|a-5|<1⟹a\in (4,6)$ and $|b-5|<1⟹b\in (4,6)$

$\therefore A=\left\{\right(a,b):a\in (4,6),b\in (4,6\left)\right\}$

Now, we have $B=\left\{\right(a,b)\in R\times R:4(a-6{)}^2+9(b-5{)}^2\le 36\}$

Considering conditions from set $A$, we can clearly see that they satisfy conditions in set $B$ as maximum value of

$4(a-6{)}^2+9(b-5{)}^2=4(4-6{)}^2+9(6-5{)}^2=16+9=25\le 36$

Apart from these values of $a$ and $b$, we have $a=6,b=7$ satisfying the equation and many more such values.

Hence, $A\subset B$.