Let $A = \{(\alpha, \beta) \varepsilon R\times R: |\alpha-1|\leq 4$ and $|\beta-5|\leq6\}$ and $B = \{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144 \}$. Then
B ⊂ A
Analyzing Set A Definition
Set A is defined as $A = \{(\alpha, \beta) \varepsilon R\times R: |\alpha-1|\leq 4 \text{ and } |\beta-5|\leq6\}$.
- The condition $|\alpha-1|\leq 4$ implies: $-4 \leq \alpha-1 \leq 4$ Adding 1 to all parts gives: $1-4 \leq \alpha \leq 1+4$ $ -3 \leq \alpha \leq 5$
- The condition $|\beta-5|\leq6$ implies: $-6 \leq \beta-5 \leq 6$ Adding 5 to all parts gives: $5-6 \leq \beta \leq 5+6$ $ -1 \leq \beta \leq 11$
- Therefore, Set A represents a rectangular region defined by $-3 \leq \alpha \leq 5$ and $-1 \leq \beta \leq 11$.
Analyzing Set B Definition
Set B is defined as $B = \{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144 \}$.
To understand the shape of Set B, we can rewrite the inequality:
$ \frac{16(\alpha-2)^2}{144} + \frac{9(\beta-6)^2}{144} \leq 1 $ $ \frac{(\alpha-2)^2}{9} + \frac{(\beta-6)^2}{16} \leq 1 $This is the standard inequality for an ellipse centered at $(2, 6)$.
- The semi-axes are $a = \sqrt{9} = 3$ (in the $\alpha$ direction) and $b = \sqrt{16} = 4$ (in the $\beta$ direction).
- The range of $\alpha$ values within the ellipse is determined by the center plus/minus the semi-axis in the $\alpha$ direction: $ 2 - 3 \leq \alpha \leq 2 + 3 $ $ -1 \leq \alpha \leq 5 $
- The range of $\beta$ values within the ellipse is determined by the center plus/minus the semi-axis in the $\beta$ direction: $ 6 - 4 \leq \beta \leq 6 + 4 $ $ 2 \leq \beta \leq 10 $
- Therefore, Set B represents the region inside and on the boundary of an ellipse with $\alpha$ values ranging from -1 to 5 and $\beta$ values ranging from 2 to 10.
Comparing Sets A and B
We need to determine the relationship between Set A (a rectangle) and Set B (an ellipse).
- Check if B ⊂ A: Does every point in B also lie in A?
- Range of $\alpha$ in B: $[-1, 5]$. Range of $\alpha$ in A: $[-3, 5]$. Since $[-1, 5]$ is contained within $[-3, 5]$, the $\alpha$ condition for B is satisfied within A.
- Range of $\beta$ in B: $[2, 10]$. Range of $\beta$ in A: $[-1, 11]$. Since $[2, 10]$ is contained within $[-1, 11]$, the $\beta$ condition for B is satisfied within A.
- Check if A ⊂ B: Does every point in A also lie in B?
The rectangle A extends from $\alpha=-3$ to $\alpha=5$ and $\beta=-1$ to $\beta=11$. The ellipse B covers $\alpha$ from -1 to 5 and $\beta$ from 2 to 10. Points in A but outside these ranges of B (e.g., $\alpha=-3$) are not in B. Therefore, A is not a subset of B.
Conclusion
Based on the analysis, Set B is entirely contained within Set A.
Final Answer: The final answer is $\boxed{B \subset A}$
Similar Questions
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Important Questions from Algebra
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
