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
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
- The largest $n \in N$ such that $3^n$ divides $50!$ is :
- The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :
- Let A be the set of all functions $f: Z \to Z$ and R be a relation on A such that $R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R is :
Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
