Question

$A=\left\{\right(x,y):x^2+y^2\le 9;x,y\text{are integers}\}$
$B=\left\{\right(x,y):|x-1-x^2|\le |x^2-3x+4|;x,y\text{are integers}\}$
Number of integers $(x,y)$ in set $A\cap B$ are

• A. $20$
• B. $18$
• C. $22$
• D. $23$

Answer 1

Answer: (D)

$23$

$A=x^2+y^2\le 9\Rightarrow -3\le \times \le 3$
$-3\le y\le 3$
$|x-1-x^2|\le |x^2-3x+4|$
$\Rightarrow -(x^2-3x+4)\le (x-1-x^2)\le (x^2-3x+4)$
$\Rightarrow 3x-4\le x-1$ and $2x^2-4x+5\ge 0$
$\Rightarrow 2x\le 3$
$x\le 3/2$
So possible integral $x$ are $-3,-2,-1,0,1$
at $x=1y=-2,-1,0,1,2$
at $x=0y=-3,-2,-1,0,1,2,3$
at $x=-1y=-2,-1,0,1,2$
at $x=-2y=-2,-1,0,1,2$
at $x=-3y=0$