Problem 166 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Two sets \(\mathrm{A}\) and \(\mathrm{B}\) are as under : \(\mathrm{A}=\\{(\mathrm{a}, \mathrm{b}) \in \mathrm{R} \times \mathrm{R}:|\mathrm{a}-5|<1\) and \(|\mathrm{b}-5|<1\\}\) \(\mathrm{B}=\left\\{(\mathrm{a}, \mathrm{b}) \in \mathrm{R} \times \mathrm{R}: 4(\mathrm{a}-6)^{2}+9(\mathrm{~b}-5)^{2} \leq 36\right\\}\). Then : (a) \(\mathrm{A} \subset \mathrm{B}\) (b) \(\mathrm{A} \cap \mathrm{B}=\phi\) (an empty set) (c) neither \(\mathrm{A} \subset \mathrm{B}\) nor \(\mathrm{B} \subset \mathrm{A}\) (d) \(\mathrm{B} \subset \mathrm{A}\)

Step-by-Step Solution

Answer

(a) A ⊂ B

Step 1: Analyze Set A

The set \( A = \{ (a, b) \in \mathbb{R} \times \mathbb{R} : |a - 5| < 1 \ \text{and} \ |b - 5| < 1 \} \) represents all points \((a, b)\) such that \( 4 < a < 6 \) and \( 4 < b < 6 \). This defines an open square centered at \((5, 5)\) with side length 2.

Step 2: Analyze Set B

The set \( B = \{(a, b) \in \mathbb{R} \times \mathbb{R} : 4(a - 6)^2 + 9(b - 5)^2 \leq 36 \} \) represents an ellipse centered at \((6, 5)\). To analyze the boundary, divide through by 36 to reshape: \((a - 6)^2/9 + (b - 5)^2/4 \leq 1\). This standard form shows it's an ellipse with radii 3 along the \( a \)-axis and 2 along the \( b \)-axis.

Step 3: Compare Set A and B

Set A is a square centered at \((5, 5)\), and Set B is an ellipse centered at \((6, 5)\) with semi-major and semi-minor axes 3 and 2 respectively. The center of B is 1 unit right of the center of A. Check positions: \(6 - 3 = 3\) and \(5 - 2 = 3\), and notice Set B extends beyond \(A\), particularly towards \(a = 6\), meaning \(A\) is entirely within \(B\).

Key Concepts

Set Theory

Set Theory is a mathematical concept that deals with collections of objects, which we refer to as sets. In this exercise, we have two sets, \(A\) and \(B\). The set \(A\) is defined as a collection of ordered pairs \((a, b)\) such that the absolute differences from 5 are less than 1. This means the x and y values, \(a\) and \(b\), respectively, must satisfy the inequalities: \(|a - 5| < 1\) and \(|b - 5| < 1\).
This definition of a set is essential in understanding how we can group elements based on certain conditions. Set \(B\), on the other hand, is defined by a more complex relationship involving an ellipse equation. We use set theory to determine relationships such as inclusions and intersections in this exercise.
When comparing sets, we look for set relationships:

• Subset: If all elements of set \(A\) are in \(B\), then \(A \subset B\).
• Intersection: Common elements that belong to both sets \(A\) and \(B\), represented as \(A \cap B\).
• Disjoint: Sets with no elements in common, such that \(A \cap B = \emptyset\).

Inequalities

Inequalities are used in this exercise to define the boundaries of our sets. We specifically use them in set \(A\) to establish the range of values \(a\) and \(b\) can take. The expression \(|a - 5| < 1\) denotes the points that fall within the interval \(4 < a < 6\). Similarly, \(|b - 5| < 1\) provides that \(b\) also lies between 4 and 6.
In set \(B\), inequalities appear in the standard form of an ellipse \(4(a - 6)^2 + 9(b - 5)^2 \leq 36\). This inequality shows all the points within the ellipse shape, where the equation modification to \((a - 6)^2/9 + (b - 5)^2/4 \leq 1\) further clarifies the boundary. The use of inequalities here helps us determine whether certain positions fall within, outside, or on the set boundary, vital for comparing \(A\) and \(B\).

Geometry of Shapes

The exercise explores different two-dimensional shapes, namely a square and an ellipse, highlighting the Geometry of Shapes.
Set \(A\) forms an open square centered at \((5, 5)\), with a side length of 2 units. This square is defined by ensuring \(a\) and \(b\) values are strictly between 4 and 6. The square's geometry offers a straightforward boundary and an easy comparison parameter.
Set \(B\), describes an ellipse. Centered at \((6, 5)\), this ellipse has semi-axes of lengths 3 (horizontal) and 2 (vertical). The reshaped equation \((a - 6)^2/9 + (b - 5)^2/4 \leq 1\) depicts this standard form. Understanding how the ellipse's boundary extends shows it encapsulates more area than the square, leading to our conclusion on the set relationships.

Coordinate Geometry

Coordinate Geometry, also known as analytic geometry, allows us to plot the sets in a coordinate system. Here, it plays a crucial role in visually and mathematically analyzing set positions.
For set \(A\), which is an open square, coordinate geometry enables us to identify all possible \((a, b)\) pairs in the designated region defined by \(4 < a < 6\) and \(4 < b < 6\).
The ellipse represented by set \(B\) also benefits from coordinate geometry as it helps draw the set on the plane. Translating \(4(a - 6)^2 + 9(b - 5)^2 \leq 36\) into its standard form using coordinate geometry aids in understanding its boundaries and how it encompasses the square at \((5, 5)\).
Coordinate geometry thus provides a clear and mathematical means to visualize the problem, facilitating the determination that \(A\) is completely inside \(B\).