Let A and B be two sets defined as given below:
$A = \{(x, y):|x - 3|<1$ and $|y-3|<1\}$
$B=\{(x, y): 4x^2+9y^2-32x-54y+109<0\}$
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $A ⊂ B$
We have,
$|x-3] <1$ and $|y-3|<1$
$⇒2<x<4$ 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 $y = 4$.
We have,
$4x^2+9y^2-32x-54y+ 109 ≤0$
$⇒4(x^2-8x)+9 (y^2 - 6y) + 109 ≤0$
$⇒4(x-4)^2+9 (y-3)^2 ≤ 36$
$⇒\frac{(x-4)^2}{3^2}+\frac{(y-3)^2}{2^2}≤1$
Thus, B is the set of all points lying inside the ellipse having its centre at (4, 3) and major and minor axes of lengths 3 and 2 units.
It can be easily seen by drawing the graphs of two regions that $A ⊂ B$.