Practicing Success
Let A and B be two sets defined as given below: |
$A ⊂ B$ $B ⊂ A$ A = B none of these |
$A ⊂ B$ |
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 $A ⊂ B$. |