Question

Let $A$ and $B$ be two sets defined as given below:
$A=\left\{\right(x,y):|x-3|<1$ and $|y-3|<1\}$
$B=\left\{\right(x,y):4x^2+9y^2-32x-54y+109\le 0\}$
Then,

• A. $A⊏B$
• B. $B⊏A$
• C. $A=B$
• D. None of these

Answer 1

Answer: (A)

$A⊏B$

We have,
$|x-3|<1$ and $|y-3|<1$
$\Rightarrow 2<x<4$ and $2<y<4$
Thus, $A$ is the set of all points $(x,y)$ lying inside the square formed by the lines $x=2$, $x=4$, $y=2$ and $y=4$.
Now,
$4x^2+9y^2-32x-54y+109\le 0$
$\Rightarrow 4(x^2-8x)+9(y^2-6y)+109\le 0$
$\Rightarrow 4(x-4{)}^2+9(y-3{)}^2\le 36$
$\Rightarrow \frac{(x-4{)}^2}{3^2}+\frac{(y-3{)}^2}{2^2}\le 1$.
Thus, $B$ is the set of all points lying inside the ellipse having its centre at $(4,3)$ and of lengths major and minor axes are $3$ and $2$ units.
It can be easily seen by drawing the graphs of two regions that $A\subset B$.