Question

Let $A=\{x:2<|x|\le 5\text{and}x\in Z\}$ and $B$ be the set of values of $a$ for which the equation $||x-1|+a|=4$ can have real solutions. Then $n(A\cap B)$ is

• A. $3$
• B. $2$
• C. $5$
• D. $0$

Answer 1

The correct option is C $5$

$2<|x|\le 5$
$\Rightarrow 2<|x|$ and $|x|\le 5$
$\Rightarrow x\in (-\infty ,-2)\cup (2,\infty )$ and $x\in [-5,5]$
$\Rightarrow x\in [-5,-2)\cup (2,5]$
Given that $x\in Z$
$\therefore A=\{-5,-4,-3,3,4,5\}$

$||x-1|+a|=4$
$\Rightarrow |x-1|+a=\pm 4$
$\Rightarrow |x-1|=(-a\pm 4)$
For real solutions, we must have
$4-a\ge 0$ or $-4-a\ge 0$
$\Rightarrow a\le 4$ or $a\le -4$
$\Rightarrow a\in (-\infty ,4]$

$\therefore A\cap B=\{-5,-4,-3,3,4\}$
$\Rightarrow n(A\cap B)=5$