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New answer posted•a year ago
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a year ago

37. (i) L.H.S = (A∪ B)' = U – (A ∪B)

= {1,2,3,4,5,6,7,8,9} – [ {2,4,6,8)∪ {2,3,5,7}]

= {1,2,3,4,5,6,7,8,9} – {2,3,4,5,6,7,8}

= {1,9}

R.H.S. = A' ∩ B' = [U – A] ∩ [U∪ B]

=  [ {1, 2, 3, 4, 5, 6, 7, 8, 9}– {2, 4, 6, 8}] ∩  [ {1, 2, 3, 4, 5, 6, 7, 8, 9}– {2, 3, 5, 7}]

= {1,3,5,7,9} ∩ {1,4,6,8,9}

= {1,9}

? L.H.S. = R.H.S.

(A B) = A ∩ B.

 

(ii) L.H.S. = (A ∩ B)' = U – (A ∩ B)

= {1,2,3,4,5,6

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New answer posted•a year ago
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a year ago

3. (i) {x : x is an odd natural number}

(ii) {x : x is an even natural number}

(iii) {x : x is not a multiple of 3}

(iv) {x : x is a positive composite number and x = 1}

(v) {x : x is a natural number not divisible by 3 and 5}.

(vi) {x : x is not a perfect square}

(vii) {x : x is not a perfect c

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New answer posted•a year ago
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a year ago

35. (i) A' = U – A = {a, b, c, d, e, f, g, h} – {a, b, c}

= {d, e, f, g, h}

(ii) B' = U – B = {a, b, c, d, e, f, g, h} – {d, e, f, g}

= {a, b, c, h}.

(iii) C' = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g}.

= {b, d, f, h}

(iv) D' = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a}

= {b, c, d, e}

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New answer posted•a year ago
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a year ago

34. (i) A' = U – A = {1,2,3,4,5,6,7,8,9} – {1,2,3,4}

= {5,6,7,8,9}

(ii) B' = U – B = {1,2,3,4,5,6,7,8,9} – {2,4,6,8}

= {1,3,5,7,9}.

(iii) (A ∪C)' = A' ∩ C'

= {5,6,7,8,9} ∩ [U – C] [? (i)]

= {5,6,7,8,9} ∩ [ {1,2,3,4,5,6,1,8,9} – {3,4,5,6}]

= {5,6,7,8,9} ∩ {1,2,7,8,9}

= {7,8,9}

(iv) A∪ B)' = A' ∩ B' [By demorgan

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New answer posted•a year ago
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a year ago

33. (i) False, as {2,3,4,5} ∩ {3,6} = {3} ≠ ?  .Hence sets are not disjoint.

(ii) False as {a, e, i, o, u} ∩ {a, b, c, d} = {a} ≠ ?  Hence sets are not disjoint.

(iii) True as {2,6,10,14} ∩ {3,7,11,15} = ?  . Hence sets are disjoint.

(iv) True as {2,6,10} ∩ {3,7,11}= ?  .

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New answer posted•a year ago
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a year ago

32. R – Q = {x: x is a real number but not rational number}

= {x: x is an irrational number}

Since real number = rational number + irrational number

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New answer posted•a year ago
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a year ago

31. (i) X – Y = {a, b, c, d} – (f, b, d, g}

= {a, c}

(ii) Y – X = {f, b, d, g} – {a, b, c, d}

= {f, g}

(iii) X ∩ Y = {a, b, c, d} ∩ {f, b, d, g}

= {b, d}.

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New answer posted•a year ago
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a year ago

30. (i) A – B = {3,6,9,12,15,18,21} – {4,8,12,16,20}

= {3,6,9,15,18,21}

(ii) A – C = {3,6,9,12,15,18,21} – {2,4,6,8,10,12,14,16}

= {3,9,15,18,21}

(iii) A – D = {3,6,9,12,15,18,21} – {5,10,15,20}

= {3,6,9,12,18,21}

(iv) B – A = {4,8,12,16,20} – {3,6,9,12,15,18,21}

= {4,8,16,20}

(v) C – A = {2,4,6,8,10,1

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New answer posted•a year ago
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a year ago

29. (i) {1,2,3,4} ∩ {x : x is a natural number and 4 ≤ x ≤ 6}

{1, 2, 3, 4} ∩ {4, 5, 6}

{4} ≠∅

Hence, the given pair of set is not disjoint.

(ii) {a, e, i, o, u} ∩ {c, d, e, f}

{e} ≠∅

Hence, the given pair of set is not disjoint.

(iii) {x: x is an even integer} ∩ {x: x is are odd integer}.

=∅

As

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New answer posted•a year ago
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a year ago

28. A = {1,2,3,4,5, 6, …}

B = {2,4,6, …}

C = {1,3,5, …}

D = {2,3,5, …, }

(i) A ∩ B = {1,2,3,4 …} ∩ {2,4,6, …} = {2,4,6 …} = B.

(ii) A ∩ C = {1,2,3,4, …} ∩ {1,3,5 …} = {1,3,5, …} = C.

(iii) B ∩ C = {2,4,6, …} ∩ {1,3,5, …} = ?  .

(iv) B ∩ D = {2,4,6 …} ∩ {2,3,5 …} = {2}

(v) C ∩ D = {1,3,5, …} ∩

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