Question

Let $Z$ be the set of integers. If $A=\{x\in Z:\frac{x^2-8x+12}{x^2+8x+17}\le 0\}$ and $B=\{x\in Z:||3x-5|+4|\le 8\},$ then $n\left[\right(A\cup B)\times (A\cap B\left)\right]$ is

• A. $12$
• B. $14$
• C. $10$
• D. $8$

Answer 1

The correct option is A $12$

$A=\{x\in Z:\frac{x^2-8x+12}{x^2+8x+17}\le 0\}$
Now, $\frac{x^2-8x+12}{x^2+8x+17}\le 0$
Since, $x^2+8x+17=(x+4{)}^2+1>0$
$\Rightarrow x^2-8x+12\le 0$
$\Rightarrow (x-2)(x-6)\le 0$
$\Rightarrow x\in [2,6]$
$\Rightarrow A=\{2,3,4,5,6\}$

$B=\{x\in Z:||3x-5|+4|\le 8\}$
$\Rightarrow ||3x-5|+4|\le 8$
$\Rightarrow -8\le |3x-5|+4\le 8$
$\Rightarrow -12\le |3x-5|\le 4$
Since, $|3x-5|$ is always greater than or equal to $0.$
$\therefore |3x-5|\le 4$
$\Rightarrow -4\le 3x-5\le 4$
$\Rightarrow \frac{1}{3}\le x\le 3$
$\Rightarrow B=\{1,2,3\}$

$A\cup B=\{1,2,3,4,5,6\}$
$\Rightarrow n(A\cup B)=6$
$A\cap B=\{2,3\}$
$\Rightarrow n(A\cap B)=2$

$n\left[\right(A\cup B)\times (A\cap B\left)\right]$
$=n(A\cup B)\times n(A\cap B)$
$=6\times 2=12$
Hence the correct answer is Option B.