Let X-x- x- and -3-x-4 and Y-x- x- and x2-x-6 - 0 - Then the number of elements in X - Y - Y- X is

X= {1,2,3} x^2 +x 6 ≤ 0 ⇒ (x+3)(x 2) ≤ 0 ⇒ x ∈ [ 3,2] So, Y={1,2} n(X∩ Y)=2 Hence, n(X × Y) ∩ (Y× X)=(n(X∩ Y))^2 =4 nbsp; ...

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Byju's Answer

Standard XIII

Mathematics

Cartesian Product

Let X=x: x∈ℕ ...

Question

Let $X = {x : x \in N and - 3 < x < 4}$ and $Y = {x : x \in N and x^{2} + x - 6 \leq 0} .$ Then the number of elements in $(X \times Y) \cap (Y \times X)$ is

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Solution

The correct option is D $4$
$X = {1, 2, 3}$
$x^{2} + x - 6 \leq 0$
$\Rightarrow (x + 3) (x - 2) \leq 0$
$\Rightarrow x \in [- 3, 2]$
So, $Y = {1, 2}$

$n (X \cap Y) = 2$
Hence, $n (X \times Y) \cap (Y \times X) = (n (X \cap Y))^{2} = 4$

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Let X={x:x∈N and −3<x<4} and Y={x:x∈N and x2+x−6≤0}. Then the number of elements in (X×Y)∩(Y×X) is

The correct option is D 4X={1,2,3} x2+x−6≤0 ⇒(x+3)(x−2)≤0 ⇒x∈[−3,2] So, Y={1,2} n(X∩Y)=2 Hence, n(X×Y)∩(Y×X)=(n(X∩Y))2=4

Let $X = {x : x \in N and - 3 < x < 4}$ and $Y = {x : x \in N and x^{2} + x - 6 \leq 0} .$ Then the number of elements in $(X \times Y) \cap (Y \times X)$ is

The correct option is D $4$
$X = {1, 2, 3}$
$x^{2} + x - 6 \leq 0$
$\Rightarrow (x + 3) (x - 2) \leq 0$
$\Rightarrow x \in [- 3, 2]$
So, $Y = {1, 2}$

$n (X \cap Y) = 2$
Hence, $n (X \times Y) \cap (Y \times X) = (n (X \cap Y))^{2} = 4$