Question

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

• A. $[-3,3]$
• B. $[-\sqrt{3},\sqrt{3}]$
• C. $[-\sqrt{2},\sqrt{2}]$
• D. $[-2,2]$

Answer 1

Answer: (B)

$[-\sqrt{3},\sqrt{3}]$

Domain $=$ value of $x$ in $f=2x^2+3y^2\le 6$ then
$2x^2+3y^2\le 6$ = $\frac{x^2}{3}+\frac{y^2}{2}\le 1$
Since it is an equation of ellipse then value of $x$ $\Rightarrow x\in [-\sqrt{3},\sqrt{3}]$