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Unit 17: SETS AND FUNCTIONS
Solved Exercise 17.2
1. Which of the following sets are disjoint, overlapping, exhaustive and cells.(i) {1,2,3,5,7} and {4,6,8,9,10}
(ii) {1,2,3,6} and {1,2,4,8}
(iii) E and 0 when U = Z (U denotes universal set)
(iv) N and W whenU= {x|x ∈ Z ⋀ x ≥ 0}
(v) Q and Q' when U =R
SOLUTION:
(i) {1,2,3,5,7} and {4,6,8,9,10}
Ans: Disjoint set
(ii) {1,2,3,6} and {1,2,4,8}
Ans: Overlapping set
(iii) E and 0 when U = Z (U denotes universal set)
Ans: Cell
(iv) N and W whenU= {x|x ∈ Z ⋀ x ≥ 0}
Ans: Exhaustive set
(v) Q and Q' when U =R
Ans: Cell
2. If A ={1,2,3,4,5,6} and B ={2,4,6,8,10} then find:
" (i) AUB
(ii) B∩A
(iii) A — B
(iv) B — A
(v) AΔB
(i) AUB
SOLUTION:
AUB = {1,2,3,4,5,6}U{2,4,6,8,10}
AUB = {1, 2, 3, 4, 5, 6, 8, 10} Ans.
(ii) B∩A
SOLUTION:
B∩A = {2,4,6,8,10}∩{1,2,3,4,5,6}
B∩A = {2, 4, 6} Ans.
(iii) A — B
SOLUTION:
A — B = {1,2,3,4,5,6} — {2,4,6,8,10}
A — B = {1, 3, 5} Ans.
(iv) B — A
SOLUTION:
B — A = {2,4,6,8,10} — {1,2,3,4,5,6}
B — A = {8, 10} Ans.
(v) AΔB
SOLUTION:
AΔB = (AUB) — (A∩B)
AΔB = {(1,2,3,4,5,6)U(2,4,6,8,10)} — {(1,2,3,4,5,6)∩(2,4,6,8,10)}
AΔB = {1, 2, 3, 4, 5, 6, 8, 10} — {2, 4, 6}
AΔB = {1, 3, 5, 8, 10} Ans.
3. If U ={1,2,3,...,10}, A ={1,2,3,4,5} and B ={1,3,5,7,9} then find: (i) A'
(ii) B'
(iii) A' U B'
(iv) A' ∩ B'
(v) (AUB)'
(vi) (A∩B)'
(vii) A'ΔB'
(viii) (AΔB)'
(ix) A — B'
(x) A' — B
Note: U = {1,2,3,...,10} = {1,2,3,4,5,6,7,8,9,10}
(i) A'
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10} Ans.
(ii) B'
SOLUTION:
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10} Ans.
(iii) A' U B'
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A' U B' = {6, 7, 8, 9,10} U {2, 4, 6, 8, 10}
A' U B' = {2, 4, 6, 7, 8, 9, 10} Ans.
(iv) A' ∩ B'
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A' ∩ B' = {6, 7, 8, 9,10} ∩ {2, 4, 6, 8, 10}
A' ∩ B' = {6, 8, 10} Ans.
(v) (AUB)'
SOLUTION:
AUB = {1,2,3,4,5} U {1,3,5,7,9}
AUB = {1, 2, 3, 4, 5, 7, 9}
(AUB)' = U — (AUB)
(AUB)' = {1,2,3,4,5,6,7,8,9,10} — {1, 2, 3, 4, 5, 7, 9}
(AUB)' = {6, 8, 10} Ans.
(vi) (A∩B)'
SOLUTION:
A∩B = {1,2,3,4,5} ∩ {1,3,5,7,9}
A∩B = {1, 3, 5}
(A∩B)' = U — (A∩B)
(A∩B)' = {1,2,3,4,5,6,7,8,9,10} — {1, 3, 5}
(A∩B)' = {2, 4, 6, 7, 8, 9, 10} Ans.
(vii) A'ΔB'
Method I:
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A'ΔB' = (A'UB') — (A'∩B')
A'ΔB' = {(6,7,8,9,10)U(2,4,6,8,10) — (6,7,8,9,10)∩(2,4,6,8,10)} A'ΔB' = {(2,4,6,7,8,9,10) — (6,8,10)} A'ΔB' = {2, 4, 7, 9} Ans.
Method II:
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A'ΔB' = {6,7,8,9,10)Δ(2,4,6,8,10}
(Hint: select uncommon elements from both sets)
A'ΔB' = {2, 4, 7, 9} Ans.
(viii) (AΔB)'
Method I:
SOLUTION:
AΔB = (AUB) — (A∩B)
AΔB = {(1,2,3,4,5) U (1,3,5,7,9)} — {(1,2,3,4,5) ∩ (1,3,5,7,9)} AΔB = {(1,2,3,4,5,7,9) — (1,3,5)}
AΔB = {2,4,7,9}
(AΔB)' = U — AΔB
(AΔB)' = {1,2,3,4,5,6,7,8,9,10} — {2,4,7,9}
(AΔB)' = {1, 3, 5, 6, 8, 10} Ans.
Method II:
SOLUTION:
AΔB = {1,2,3,4,5} Δ {1,3,5,7,9}
(Hint: select uncommon elements from both sets) AΔB = {2, 4, 7, 9}
(AΔB)' = U — AΔB
(AΔB)' = {1,2,3,4,5,6,7,8,9,10} — {2,4,7,9}
(AΔB)' = {1, 3, 5, 6, 8, 10} Ans.
(ix) A — B'
SOLUTION:
B' = U — B
B' = {1,2,3,4,5,6,7,8,9,10} — {1,3,5,7,9}
B' = {2, 4, 6, 8, 10}
A — B' = {1,2,3,4,5} — {2, 4, 6, 8, 10}
A — B' = {1, 3, 5} Ans.
(x) A' — B
SOLUTION:
A' = U — A
A' = {1,2,3,4,5,6,7,8,9,10} — {1,2,3,4,5}
A' = {6, 7, 8, 9,10}
A' — B = {6, 7, 8, 9,10} — {1,3,5,7,9}
A' — B = {6, 8, 10} Ans.
4. If U = {x|x ∈ Z ⋀ -4 ＜ x ＜ 6}, P = {p|p ∈ E ⋀ -4 ＜ p ＜ 6} Q = {q|q ∈ P ⋀ q ＜ 6} then show that:
(i) P — Q = P∩Q'
(ii) Q — P = Q∩P'
(iii) (PUQ)' = P'∩Q'
(iv) (P∩Q)' = P'UQ'
Note:
U = {x|x ∈ Z ⋀ -4 ＜ x ＜ 6} = {-3,-2,-1,0,1,2,3,4,5}
P = {p|p ∈ E ⋀ -4 ＜ p ＜ 6} = {-2,0,2,4}
Q = {q|q ∈ P ⋀ q ＜ 6} = {2,3,5}
(i) P — Q = P∩Q'
SOLUTION:
Taking L.H.S:
P — Q = {-2,0,2,4} — {2,3,5}
P — Q = {-2,0,4}
Taking R.H.S
P∩Q'
First we find Q'
Q'= U — Q
Q' = {-3,-2,-1,0,1,2,3,4,5} — {2,3,5}
Q' = {-3, -2, -1, 0, 1, 4}
Now we find P∩Q'
P∩Q' = {-2,0,2,4} ∩ {-3,-2,-1,0,1,4}
P∩Q' = {-2, 0, 4}
P — Q = P∩Q'
L.H.S = R.H.S (Hence Proved)
(ii) Q — P = Q∩P'
SOLUTION:
Taking L.H.S:
Q — P = {2,3,5} — {-2,0,2,4}
Q — P = {3, 5}
Taking R.H.S
Q∩P'
First we find P'
P'= U — P
P' = {-3,-2,-1,0,1,2,3,4,5} — {-2,0,2,4}
P' = {-3, -1, 1, 3, 5}
Now we find Q∩P'
Q∩P' = {2,3,5} ∩ {-3, -1, 1, 3, 5}
Q∩P' = {3, 5}
Q — P = Q∩P'
L.H.S = R.H.S (Hence Proved)
(iii) (PUQ)' = P'∩Q'
SOLUTION:
Taking L.H.S:
(PUQ)'
PUQ = {-2,0,2,4} U {2,3,5}
PUQ = {-2, 0, 2, 3, 4, 5}
(PUQ)' = U — (PUQ)
(PUQ)' = {-3,-2,-1,0,1,2,3,4,5} — {-2,0,2,3,4,5}
(PUQ)' = {-3, -1, 1}
Taking R.H.S
First we find P' and Q'
P'= U — P
P' = {-3,-2,-1,0,1,2,3,4,5} — {-2,0,2,4}
P' = {-3, -1, 1, 3, 5}
Q'= U — Q
Q' = {-3,-2,-1,0,1,2,3,4,5} — {2,3,5}
Q' = {-3, -2, -1, 0, 1, 4}
Now we find P'∩Q'
P'∩Q' = {-3, -1, 1, 3, 5} ∩ {-3, -2, -1, 0, 1, 4}
P'∩Q' = {-3, -1, 1}
(PUQ)' = P'∩Q'
L.H.S = R.H.S (Hence Proved)
(iv) (P∩Q)' = P'UQ'
SOLUTION:
Taking L.H.S:
(P∩Q)' =
P∩Q = {-2,0,2,4} U {2,3,5}
P∩Q = {2}
(P∩Q)' = U — (P∩Q)
(P∩Q)' = {-3,-2,-1,0,1,2,3,4,5} — {2}
(P∩Q)' = {-3, -2, -1, 0, 1, 3, 4, 5}
Taking R.H.S
First we find P' and Q'
P'= U — P
P' = {-3,-2,-1,0,1,2,3,4,5} — {-2,0,2,4}
P' = {-3, -1, 1, 3, 5}
Q'= U — Q
Q' = {-3,-2,-1,0,1,2,3,4,5} — {2,3,5}
Q' = {-3, -2, -1, 0, 1, 4}
Now we find P'UQ'
P'UQ' = {-3, -1, 1, 3, 5} U {-3, -2, -1, 0, 1, 4}
P'UQ' = {-3, -2, -1, 0, 1, 3, 4, 5}
(P∩Q)' = P'UQ'
L.H.S = R.H.S (Hence Proved)
5. If A = {2n | n ∈ N}, B = {3n | n ∈ N} and C = {4n | n ∈ N} then find:
i) A∩B
ii) AUC
(iii) B∩C
Note:
N = {1,2,3,4,5,6,.....}
A = {2n | n ∈ N} = {2,4,6,8,10,12,.....}
B = {3n | n ∈ N} = {3,6,9,12,15,18,.....}
C = {4n | n ∈ N} = {4,8,12,16,20,24}
i) A∩B
SOLUTION:
A∩B = {2,4,6,8,10,12,.....} ∩ {3,6,9,12,15,18,.....}
A∩B = {6, 12, 18, 24........}
(Hint: First common number is 6, therefore its common are multiple of 6)
A∩B = {6n | n ∈ N} Ans.
ii) AUC
SOLUTION:
AUC = {2,4,6,8,10,12,.....} U {4,8,12,16,20,24}
AUC = {2,4,6,8,10,.....}
(Hint: AUC is multiple of 2)
AUC = {2n | n ∈ N} Ans.
(iii) B∩C
B∩C = {3,6,9,12,15,18,.....} ∩ {4,8,12,16,20,24}
B∩C = {12, 24, 36, .......}
(Hint: First common number is 12, therefore its common are multiple of 12)
B∩C = {2n | n ∈ N} Ans.
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