Question

If \(\mathrm{A}=\\{(\mathrm{x}, \mathrm{y}) /|\mathrm{x}-3|<1,|\mathrm{y}-3|<1, \mathrm{x}, \mathrm{y} \in \mathrm{R}\\}\) and \(B=\left\\{(x, y) / 4 x^{2}+9 y^{2}-32 x-54 y+109 \leq 0, x, y \in R\right\\}\) then which of the following is true (a) \(\mathrm{A}\) is a proper sub set of \(\mathrm{B}\) (b) B is a proper sub set of \(\mathrm{A}\) (c) \(\mathrm{A}=\mathrm{B}\) (d) \(\mathrm{A}^{\prime}=\mathrm{B}\)

Short answer

The correct answer is (a) A is a proper sub set of B.

Step-by-step solution

Rewrite the conditions for A

Rewrite the conditions for set A given by \(|\mathrm{x}-3|<1,|\mathrm{y}-3|<1\). Using the properties of absolute values, we can say \(\mathrm{-1}<\mathrm{x}-3<1\) and \(\mathrm{-1}<\mathrm{y}-3<1\). Adding 3 to all parts of the inequalities, we get \(2<x<4\) and \(2<y<4\). So, set A consists of all points (x, y) such that \(2<x<4\) and \(2<y<4\).

Rewrite the conditions for B

Rewrite the conditions for set B given by \(4 x^{2}+9 y^{2}-32 x-54 y+109 \leq 0\). Let's complete the square for this equation to better understand its geometric properties. We will have \(4(x^2 - 8x) + 9(y^2 - 6y) \leq -109\). Complete the square for \(x^2 - 8x\): \(x^2 - 8x + 16 = (x-4)^2\), so we need to add 16 to balance. Complete the square for \(y^2 - 6y\): \(y^2 - 6y + 9 = (y-3)^2\), so we need to add 9 to balance. After completing the squares and balancing, we have: \(4(x-4)^2 + 9(y-3)^2 \leq 109 + 16 + 9 = 134\) So, set B consists of all points (x, y) that satisfy the inequality: \(\cfrac{(x-4)^2}{\cfrac{134}{4}}+\cfrac{(y-3)^2}{\cfrac{134}{9}}\leq1\) This inequality represents an ellipse with center (4,3) and semi-major axes \(a=\sqrt{\cfrac{134}{4}}\) and \(b=\sqrt{\cfrac{134}{9}}\)

Compare sets A and B

Now, we'll compare the elements in sets A and B to determine which of the given statements is true. Set A contains points (x, y) such that \(2<x<4\) and \(2<y<4\). It defines a square in the xy-plane. Set B contains points within the specified ellipse. Notice that the center of the ellipse (4,3) does not satisfy the conditions for set A. So B is not a subset of A. However, since the ellipse contains points within the square, A could be a subset of B. Now let's check whether A is a proper subset of B. We can see that the points on the boundary of the square defined by A are not in the ellipse defined by B. So, A is indeed a proper subset of B. Therefore, the correct answer is: (a) \(\mathrm{A}\) is a proper sub set of \(\mathrm{B}\).

Key concepts

Set Theory

Set theory is fundamental in understanding how different collections of objects, or 'sets', relate to each other. Here, we explore the relationship between two sets, A and B, defined in the problem. Set A is defined by pairs \(x, y\) such that the values of x and y are restricted by absolute value inequalities, forming a region on the plane. This kind of region is often bounded, as seen with the square defined by the range \(2 < x < 4\) and \(2 < y < 4\).
Set B is defined by an equation representing an ellipse on the coordinate plane.

Set comparisons in set theory allow us to check if sets share common elements, or if one set is entirely contained within another. Options like "A is a proper subset of B" or "B is a proper subset of A" require careful comparison of these regions. In this context, a proper subset means that all elements of set A are in set B, but not all elements of set B are in set A. This understanding helps solve the problem by revealing the spatial relationship between the given square and elliptical regions.

Ellipses

Ellipses are a central concept in geometry and represent the path traced by a point that moves so that the sum of its distances from two fixed points, called the foci, is constant. In the given problem, set B is defined by an inequality that describes an ellipse. This is rewritten from the original equation by completing the square for both the x and y terms.

An ellipse's standard form is given by the equation \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where \(h, k\) is the center of the ellipse, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.
In this case, the center is at \(4, 3\), and the ellipse is formed around these central coordinates. The task involves determining whether the square region defined by A fits entirely within this elliptical region, guiding us to infer whether A is a subset of B. Understanding ellipses allows a deeper grasp of how these geometrical figures interact spatially.

Absolute Value Inequalities

Absolute value inequalities like \(|x-3| < 1\) are crucial for defining regions on a plane. Absolute value represents the distance from zero and, consequently, these inequalities describe a range of values within a certain distance from a central point. In the problem, the inequalities \(2 < x < 4\) and \(2 < y < 4\) are derived from \(|x-3| < 1\) and \(|y-3| < 1\), respectively.

This effectively creates a box or square region on the plane, centered around \(3, 3\), but stretched within the specified range for x and y. The significance of these inequalities lies in their ability to define bounded regions that can be compared to others, such as an ellipse in this scenario. By solving these inequalities, we establish the specific area in plane geometry where the set A resides, setting the stage for understanding how it is a subset of another spatial region defined by B.