Question

If $A=\{x:x^2-4\le 0,x\in Z\}$ and $B=\{y:y^2-9\ge 0,y\in Z\}$, then $n(A\cap B)$ is equal to

• A. $1$
• B. $4$
• C. $0$
• D. $10$

Answer 1

The correct option is C $0$

$x^2-4\le 0$
$\Rightarrow (x+2)(x-2)\le 0$
$\Rightarrow (x+2)\le 0$ and $(x-2)\ge 0$
or $(x+2)\ge 0$ and $(x-2)\le 0$
$\Rightarrow x\le -2$ and $x\ge 2$
or $x\ge -2$ and $x\le 2$
$\Rightarrow x\in [-2,2]$
$\therefore A=\{-2,-1,0,1,2\}$

$y^2-9\ge 0$
$\Rightarrow (y+3)(y-3)\ge 0$
$\Rightarrow (y+3)\le 0$ and $(y-3)\le 0$
or $(y+3)\ge 0$ and $(y-3)\ge 0$
$\Rightarrow y\le -3$ and $y\le 3$
or $y\ge -3$ and $y\ge 3$
$\Rightarrow y\in (-\infty ,-3]\cup [3,\infty )$
$\therefore B=\{...-4,-3,3,4,...\}$
$\Rightarrow A\cap B=\varphi$
$\Rightarrow n(A\cap B)=0$