Let A_1, A_2, A_3 be regions in the XY-plane | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The regions are defined as:$A_1: x^2 + 2y^2 \leq 1$ (an ellipse)$A_2: |x|^3 + 2\sqrt{2}|y|^3 \leq 1$$A_3: \max(|x|, \sqrt{2}|y|) \leq 1$ (a rectan

Vedclass · Accepted Answer

$A_3 \supset A_2 \supset A_1$

Vedclass · Answer

(D) The regions are defined as:$A_1: x^2 + 2y^2 \leq 1$ (an ellipse)$A_2: |x|^3 + 2\sqrt{2}|y|^3 \leq 1$$A_3: \max(|x|, \sqrt{2}|y|) \leq 1$ (a rectangle with vertices at $(\pm 1, \pm 1/\sqrt{2})$)Let us test a point $(x, y)$ such that $|x| \leq 1$ and $\sqrt{2}|y| \leq 1$. If $(x, y) \in A_3$,then $|x| \leq 1$ and $|y| \leq 1/\sqrt{2}$.For $A_1$,we have $x^2 + 2y^2 \leq 1^2 + 2(1/\sqrt{2})^2 = 1 + 1 = 2$. This does not guarantee $A_3 \subset A_1$.However,consider the boundary curves. The region $A_3$ is the rectangle $|x| \leq 1, |y| \leq 1/\sqrt{2}$.The region $A_1$ is the ellipse $x^2 + 2y^2 \leq 1$. The region $A_2$ is $|x|^3 + 2\sqrt{2}|y|^3 \leq 1$.By comparing the growth of the powers,for $0 Thus,$x^2 + 2y^2 \leq 1 \implies |x|^3 + 2\sqrt{2}|y|^3 \leq 1$ is not necessarily true,but checking the containment: $A_1 \subset A_2 \subset A_3$ is the correct inclusion order,which means $A_3 \supset A_2 \supset A_1$.Therefore,the correct option is $(d)$.

Let $A_1, A_2, A_3$ be regions in the $XY$-plane defined by:
$A_1 = \{(x, y) : x^2 + 2y^2 \leq 1\}$
$A_2 = \{(x, y) : |x|^3 + 2\sqrt{2}|y|^3 \leq 1\}$
$A_3 = \{(x, y) : \max(|x|, \sqrt{2}|y|) \leq 1\}$
Then,

Similar Questions

The locus of a point such that the sum of its distances from the points $(0, 2)$ and $(0, -2)$ is $6$ is:

What is the equation of the chord of the ellipse $\frac{x^2}{36} + \frac{y^2}{9} = 1$ that is bisected at the point $(2, 1)$?

The product of the perpendiculars from the two foci of the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$ on the tangent at any point on the ellipse is:

The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9x^2 + 16y^2 = 144$ is

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $A_1, A_2, A_3$ be regions in the $XY$-plane defined by:$A_1 = \{(x, y) : x^2 + 2y^2 \leq 1\}$$A_2 = \{(x, y) : |x|^3 + 2\sqrt{2}|y|^3 \leq 1\}$$A_3 = \{(x, y) : \max(|x|, \sqrt{2}|y|) \leq 1\}$Then,