if $$A={x$$:x∈I,−$$2$$≤x≤$$2}$$,$$B={x$$∈I,$$0$$≤x≤$$3}$$,$$C={x$$:x∈N,$$1$$≤x≤$$2}$$ and $$D={x$$,$$y)$$∈N×N;$$x+y=8}$$ Then.

Correct option is 4none of these

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if $A = {x : x \in I, - 2 \leq x \leq 2},$ $B = {x \in I, 0 \leq x \leq 3}, C = {x : x \in N, 1 \leq x \leq 2}$ and $D = {x, y) \in N \times N; x + y = 8}$ Then.

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n(A∪(B∪C)=5

n(D) = 6

n(B∪C)=5

none of these

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detailed solution

Correct option is D

A = {– 2, – 1, 0, 1, 2}, B = {0, 1, 2, 3}, C = {1, 2} so B∪C={0,1,2,3} A∪(B∪C)={−2,−1,0,1,2,3}so n(A∪(B∪C)=6,n(B∪C)=4 and D = {(1, 7), (2, 6), (3, 5), (4, 4), (5,3), (6, 2), (7, 1)} so n(D) = 7.

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if A={x:x∈I,−2≤x≤2},B={x∈I,0≤x≤3},C={x:x∈N,1≤x≤2} and D={x,y)∈N×N;x+y=8} Then.

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if $A = {x : x \in I, - 2 \leq x \leq 2},$ $B = {x \in I, 0 \leq x \leq 3}, C = {x : x \in N, 1 \leq x \leq 2}$ and $D = {x, y) \in N \times N; x + y = 8}$ Then.