Let - be the set of integers- If A-x-x2-8x-12x2-8x-17-0 and B-x-3x-5-4-8- then n-A- B-A- B- is

A={x∈ℤ:x^2 8x+12x^2+8x+17≤0} Now, x^2 8x+12x^2+8x+17≤0 Since, x^2+8x+17 gt;0 as D lt;0 ⇒ x^2 8x+12≤0 ⇒ (x 2)(x 6)≤0 ⇒ x∈[2,6] ⇒ A={2,3,4,5,6} B={x∈ℤ:||3x 5 ...

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Standard XII

Mathematics

Cartesian Product

Let ℤ be the ...

Question

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

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Solution

The correct option is B $12$
$A = {x \in Z : \frac{x^{2} - 8 x + 12}{x^{2} + 8 x + 17} \leq 0}$
Now, $\frac{x^{2} - 8 x + 12}{x^{2} + 8 x + 17} \leq 0$
Since, $x^{2} + 8 x + 17 > 0$ as $D < 0$
$\Rightarrow x^{2} - 8 x + 12 \leq 0$
$\Rightarrow (x - 2) (x - 6) \leq 0$
$\Rightarrow x \in [2, 6]$
$\Rightarrow A = {2, 3, 4, 5, 6}$

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

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

$n [(A \cup B) \times (A \cap B)]$
$= n (A \cup B) \times n (A \cap B)$
$= 6 \times 2 = 12$

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Let Z be the set of integers. If A={x∈Z:x2−8x+12x2+8x+17≤0} and B={x∈Z:||3x−5|+4|≤8}, then n[(A∪B)×(A∩B)] is

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