Question

If a relation ${}^{\prime }{R}^{\prime }$ is defined by $R=\left\{\left(x,y\right)/2x^2+3y^2\le 6\right\},$ then the domain of $R(x,y)$ is

• A. $x\in \left[-3,3\right]$
• B. $x\in \left[-\sqrt{3},\sqrt{3}\right]$
• C. $y\in \left[-\sqrt{2},\sqrt{2}\right]$
• D. $y\in \left[-2,2\right]$

Answer 1

Answer: (B), (C)

$x\in \left[-\sqrt{3},\sqrt{3}\right]$
$y\in \left[-\sqrt{2},\sqrt{2}\right]$

$2x^2+3y^2\le 6$

$⟹\frac{x^2}{3}+\frac{y^2}{2}\le 1$

This is an ellipse with ends being ($-\sqrt{3},0$) , ($\sqrt{3},0$) , ($0,-\sqrt{2}$) , ($0,\sqrt{2}$)

Hence the domain for $x=(-\sqrt{3},\sqrt{3})$

Hence the domain for $y=(-\sqrt{2},\sqrt{2})$