Let
$
A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\}
$
and
$
B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\}.
$
Then:
Show Hint
- \( B \subset A \)
- \( A \cup B = \{(x, y) : -4 \leq x \leq 4, -1 \leq y \leq 11 \} \)
- neither \( A \subset B \) nor \( B \subset A \)
- \( A \subset B \)
The Correct Option is A
Approach Solution - 1
To solve the given problem, let's analyze the sets \( A \) and \( B \) and determine the relationship between them.
- Start with Set \( A \): \(A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\}\).
- The condition \(|\alpha - 1| \leq 4\) implies that \(-4 \leq \alpha - 1 \leq 4\), or \(-3 \leq \alpha \leq 5\).
- The condition \(|\beta - 5| \leq 6\) implies that \(-6 \leq \beta - 5 \leq 6\), or \(-1 \leq \beta \leq 11\).
- Thus, the set \( A \) represents a rectangle with vertices at \((-3, -1)\), \((5, -1)\), \((5, 11)\), and \((-3, 11)\).
- Next, consider Set \( B \): \(B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\}\).
- This inequality represents an ellipse centered at \((2, 6)\).
- The equation \(\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \leq 1\) shows the semi-major axis along the \( \beta \)-axis with length \(8\) and the semi-minor axis along the \( \alpha \)-axis with length \(3\).
- Let's determine the boundaries of each set:
- The ellipse \( B \) lies within the rectangle \( A \) because:
- For an ellipse centered at \((2, 6)\) with a semi-horizontal axis of length \(3\), the \( \alpha \)-range is \([-1, 5]\), which is within \([-3, 5]\).
- For a vertical semi-axis of \(8\), the \( \beta \)-range is \([-2, 14]\), mostly within \([-1, 11]\).
- Thus, every point of the ellipse \( B \) lies within the rectangle \( A \), confirming \(B \subset A\).
- The ellipse \( B \) lies within the rectangle \( A \) because:
Thus, after analyzing both shapes and their boundaries, it is clear that the correct answer is \( B \subset A \).
Approach Solution -2
We are given two sets \( A \) and \( B \) defined by: \[ A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\} \] This defines a rectangular region where \( \alpha \) lies between \( -3 \) and \( 5 \), and \( \beta \) lies between \( -1 \) and \( 11 \). \[ B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\} \] This defines an ellipse with center \( (2, 6) \), semi-major axis 4 along the \( \beta \)-axis, and semi-minor axis 3 along the \( \alpha \)-axis.
We see that the ellipse \( B \) fits entirely within the rectangle \( A \), meaning that \( B \subset A \).
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Ellipse Questions
- If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:
- If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is.
- A parabola has vertex \((2, 3)\), equation of directrix is \(2x – y = 1\) and equation of ellipse is \(\frac {x^2}{a^2} +\frac {y^2}{b^2} =1\), \(e=\frac {1}{\sqrt 2}\) and ellipse passing through focur of parabola then square of length of latus rectum of ellipse is
- Let \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a>b \), be an ellipse whose eccentricity is \( \frac{1}{\sqrt{2}} \) and the length of the latus rectum is \( \sqrt{14} \). Then the square of the eccentricity of \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is:
- Let \( f(x) = x^2 + 9 \), \( g(x) = \frac{x}{x-9} \), and \[ a = f \circ g(10), \, b = g \circ f(3). \]
If \( e \) and \( l \) denote the eccentricity and the length of the latus rectum of the ellipse \[ \frac{x^2}{a} + \frac{y^2}{b} = 1, \] then \( 8e^2 + l^2 \) is equal to:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,