If A - x - y - x 2- y 2- 1 - x - y - R and B - x - y - x 2- y 2- 4 - x - y - R - thenA- A - B -B- B - A -C- A - B -D- A - B -

The correct option is D A∩ B=ϕThe graph for A={(x,y):x^2+y^2≤1;x,y∈ R} and B={(x,y):x^2+y^2≥4;x,y∈ R} is There is no common region of A and B Hence, nbsp;A∩ ...

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Byju's Answer

Standard XII

Mathematics

Intersection

If A = x , y ...

Question

If $A = {(x, y) : x^{2} + y^{2} \leq 1; x, y \in R}$ and $B = {(x, y) : x^{2} + y^{2} \geq 4; x, y \in R}$ , then

A - B = ϕ

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B - A = ϕ

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A \cap B \neq ϕ

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A \cap B = ϕ

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Solution

The correct option is D $A \cap B = ϕ$
The graph for $A = {(x, y) : x^{2} + y^{2} \leq 1; x, y \in R}$ and $B = {(x, y) : x^{2} + y^{2} \geq 4; x, y \in R}$ is

There is no common region of $A$ and $B$
Hence, $A \cap B = ϕ$

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If A={(x,y):x2+y2≤1;x,y∈R} and B={(x,y):x2+y2≥4;x,y∈R}, then

The correct option is D A∩B=ϕThe graph for A={(x,y):x2+y2≤1;x,y∈R} and B={(x,y):x2+y2≥4;x,y∈R} is There is no common region of A and B Hence, A∩B=ϕ

If $A = {(x, y) : x^{2} + y^{2} \leq 1; x, y \in R}$ and $B = {(x, y) : x^{2} + y^{2} \geq 4; x, y \in R}$ , then

The correct option is D $A \cap B = ϕ$
The graph for $A = {(x, y) : x^{2} + y^{2} \leq 1; x, y \in R}$ and $B = {(x, y) : x^{2} + y^{2} \geq 4; x, y \in R}$ is

There is no common region of $A$ and $B$
Hence, $A \cap B = ϕ$