Exercise 1.2
1) If U={1,2,3,4,……9,10},
A={1,2,3,4}, B={3,5,7,8} and C={1,2,7,8}, Find:
Solution:
a) A U B = {1, 2,
3, 4} U {3, 5, 7, 8} = {1, 2, 3, 4, 5, 7, 8}
b. A ∩ C = {1, 2,
3, 4} ∩ {1, 2, 7, 8} {1, 2}
c. (A – B) ∩ C =
({1, 2, 3, 4} – {3, 5, 7, 8}) ∩ {1, 2, 7, 8}= {1, 2, 4} ∩ {1, 2, 7, 8} = {1, 2}
d. A U C = {1, 2,
3, 4} U {1, 2, 7, 8}
So, A U C = {1, 2, 3, 4, 7, 8}
Now, $\overline
{{\rm{AUC}}} $ = U – (AUC)
So, $\overline {{\rm{AUC}}} $ = {5, 6, 9, 10}
e. ${\rm{\bar B}}$
= U – B = {1, 2, 3, …9, 10} – {3, 5, 7, 8}
= {1, 2, 4, 6, 9, 10}
Or, ${\rm{\bar C}}$ = U – C = {1, 2, 3, ….9, 10} – {1, 2, 7,
8}
Or, ${\rm{\bar B\: }}$U ${\rm{\bar C}}$ = {1, 2, 4, 6, 9,
10} U {3, 4, 5, 6, 9, 10}
Or, ${\rm{\bar B\: }}$U ${\rm{\bar C}}$ = {1, 2, 3, 4, 5, 6,
9, 10}
f. (AUB) – C =
{1, 2, 3, 4, 5, 7, 8} – {1, 2, 7,
8} [from1(a)]
= {3, 4, 5}
2) If U={a,b,c,d,…….i,j,k},
A={b,c,d,e}, B={d,e,f,g,h,i}, C={a,e,I,o,u}, D={b,d,j,k}.
Solution:
a. AUB
= {b, c, d, e} U
{d, e, f, g, h, i}
= {b, c, d, e, f, g, h, i}
b. A∩C
= {b, c, d, e} ∩
{a, e, i, o, u}
= {e}
c. A – B
= {b, c, d, e} –
{d, e, f, g, h, i}
= {b, c}
d. A – C
= {b, c, d, e} –
{a, e, i, o, u}
= {b, c, d}
e. (A – C) ∩ C
= {b, c, d} ∩ {a,
e, i, o, u} [from 2(d)]
= ɸ
f. A $\Delta $ D
= (A – D) U (D – A)
=({b, c, d, e} – {b, d, j, k}) U ({b, d, j, k} – {b, c, d,
e})
= {c, e} U {j, k}
= {c, e, j, k}
g) (AUB) – C
= {b, c, d, e, f, g, h, i} – {a, e, i, o, u} [from 2(a)]
= {b, c, d, f, g, h}
h) A∩ ${\rm{\bar B}}$
=A∩(U – B)
= {b,c,d,e}∩[{a,
b, c, d, ….i, j, k} – {d, e, f, g, h, i}]
={b,c,d,e}∩ {a,
b, c}={b,c}
i) ${\rm{\bar A-B}}$
= U – (A – B)
= {a, b, c, d, ….i, j, k} – {b, c}
= {a, d, e, f, g, h, i, j, k}
3. a)Given the sets
U = {x:x is a positive integer les than 12}, A={3,5,7,9},
B={1,2,3,8,9}, C={1,4,7,10}, find $\overline {{\rm{AUB}}} $, (A-B) U C, (A-C) ∩
B.
Solution:
Here, U = {x:x is a positive integer les than 12}
So, U = {1, 2, 3, ….11}
Now, A U B = {3, 5, 7, 9} U {1, 2, 3, 8, 9} = {1, 2, 3, 5,
7, 8, 9}
So,
$\overline
{{\rm{AUB}}} $
= U – (AUB)
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} – P1, 2, 3, 5, 7, 8,
9}
= {4, 6, 10, 11}
Again, A – B = {3, 5, 7, 9} – {1, 2, 3, 8, 9} = {5, 7}
So,
(A – B) U C
= {5, 7} U {1, 4, 5, 7, 10} = {1, 4, 5, 7, 10}
And,
A – C = {3, 5, 7, 9} – {1, 4, 7, 10} = {3, 5, 9}
So,
(A – C) ∩ B
= {3, 5, 9} ∩ {1, 2, 3, 8, 9} {3, 9}
b) Given U = {x:x is
a natural number upto 20}, A={ x:x≥ 6}, B = {x:x ≤ 8} and C = {x:10< x <
15, find B U C, A∩B, A-C and $\overline {{\rm{AUB}}} $.
Solution:
Here, U = {x:x is a natural number upto 20}
So, U = {1, 2, 3,...20}
A = {x:x≥ 6} = {6, 7, 8, ….20}
B = {x:x ≤ 8} = {1, 2, 3, ….8}
C = {x:10< x < 15} = {11, 12, 13, 14}
Now, B U C = {1, 2, 3, 4, 5, 6, 7, 8} U {11, 12, 13, 14}
So,
B U C
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
Again,
A∩B
= {6, 7, 8, …20} ∩ {1, 2, 3, ….8} = {6, 7, 8}
And,
A – C
= {6, 7, 8….20} – {11, 12, 13, 14}
= {6, 7, 8, 9, 10, 15, 16, 17, 18, 19, 20}
And A U B = {6, 7, 8, …., 20} U {1, 2, 3….8}
= {1, 2, 3, …..20} = U.
So,
$\overline
{{\rm{AUB}}} $
= U – (AUB) = U – U = ɸ
4. a) If A = {x:x =
2n + 1, n≤5, nԑN} and B= {x:x = 3n – 2, n≤4, nԑN}, find A U B, A ∩ B and B – A.
Solution:
A = {x:x = 2n + 1, n≤5, nԑN}
So, A = {3, 5, 7, 9, 11}
B = {x:x = 3n – 2, n≤4, nԑN}
So, B = {1, 4, 7, 10}
Now,
A U B
= {3, 5, 7, 9, 11} U {1, 4, 7, 10}
= {1, 3, 4, 5, 7, 9, 10, 11}
A ∩ B
= {3, 5, 7, 9, 11} ∩ {1, 4, 7, 10}
= {7}
And,
B – A =
{1, 4, 7, 10} – {3, 5, 7, 9, 11} = {1, 4, 10}
b) If A is set of
multiples of 3 less than 20, B is the set of multiples of 6 less or equal to 30
and U is the set of all natural numbers, find A ∩ B and A-B.
Solution:
here, U = {1, 2, 3, ….},
A = {3, 6, 9, 12, 15, 18}
B = {6, 12, 18, 24, 30}
So,
A ∩ B =
{3, 6, 9, 12, 15, 18} ∩ {6, 12, 18, 24, 30}.
= {6, 12, 18}
And
A – B
= {3, 6, 9, 12, 15, 18} – {6, 12, 18, 24, 30}
= {3, 9, 15}
c) If U ={x: –1≤x – 2
≤7}, A ={x:x is a prime number} and B = {x:x is a odd number}, find
Solution
Here, U = {x: –1≤x – 2 ≤7} = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A ={x:x is a prime number} = {2, 3, 5, 7}
B = {x:x is a odd number} = {1, 3, 5, 7, 9}
Now,
(i) The sets if elements
which are either prime or odd
A U B = {2, 3, 5, 7} U {1, 3, 5, 7, 9} = {1, 2, 3, 5, 7, 9}
(ii) The sets if
elements which are prime as well as odd
A U B = {2, 3, 5, 7} ∩ {1, 3, 5, 7, 9} = {3, 5, 7}
(iii) The sets if
elements which are prime but not odd
A – B = {2, 3, 5, 7}
– {1, 3, 5, 7, 9} = {2}
5. Let U= {a, b, c, d,
e, f, g, h}, A= {a, b, c, d}, B= {c, d, e, f} and C= {d, e, f, g, h}. Verify
the following relations:
Solution:
A U B = {a, b, c, d} U {c, d, e, f} = {a, b, c, d, e, f}
A U C = {a, b, c, d} U {d, e, f, g, h} = {a, b, c, d, e, f,
g, h}
B ∩ C = {c, d, e, f} U {d, e, f, g, h} = {d, e, f}
a) A U (B ∩ C) = (A U
B) ∩ (A U C)
L.H.S. = A U (B ∩ C)
= {a, b, c, d, e, f} U {d, e, f} = {a, b, c, d, e, f}
R.H.S. = (A U B) ∩ (A U C)
= {a, b, c, d, e, f} ∩ {a, b, c, d, e, f, g, h} = {a, b, c,
d, e, f}
Hence, A U (B ∩ C) = (A U B) ∩ (A U C)
b) A ∩ (B U C) =
(A∩B) U (A ∩ C)
A ∩ B = {a, b, c, d} ∩ {c, d, e, f} = {c, d}
A ∩ C = {a, b, c, d} ∩ {d, e, f, g, h} = {d}
B U C = {c, d, e, f} U {d, e, f, g, h} = {c, d, e, f, g, h}
Now, LHS = A ∩ (B U C) = {a, b, c, d} ∩ {c, d, e, f, g, h} =
{c, d}
RHS = (A∩B) U (A ∩ C) = {c, d} U {d} = {c, d}
Hence, A ∩ (B U C) =
(A∩B) U (A ∩ C)
c) $\overline
{{\rm{AUB}}} = {\rm{\bar A}}\mathop\cap\nolimits^ {\rm{\bar B}}$
Solution:
${\rm{\bar A}}$ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c, d} = {e, f, g,
h}
${\rm{\bar B}}$ = U – B = {a, b, c, d, e, f, g, h} – {c, d,
e, f} = {a, b, g, h}
Now,
LHS = $\overline
{{\rm{AUB}}} $
= U – (AUB)
= {a, b, c, d, e, f, g, h} – {a, b, c, d, e, f} = {g, h}
R.H.S. =
${\rm{\bar A}}\mathop\cap\nolimits^ {\rm{\bar B}}$
= {e, f, g, h} ∩ {a, b, g, h}
= {g, h}
Hence, $\overline
{{\rm{AUB}}} = {\rm{\bar A}}\mathop\cap\nolimits^ {\rm{\bar B}}$
d) A – (B U C)= (A –
B) ∩ (A – C)
Here,
A – B = {a, b, c, d} – {c, d, e, f} = {a, b}
A – C = {a, b, c, d} – {d, e, f, g, h} = {a, b, c}
Now,
LHS = A – (B U C)
= {a, b, c, d} – {c, d, e, f, g, h}
= {a, b}
RHS = (A – B) ∩ (A – C)
= {a, b} ∩ {a, b, c}
= {a, b}
Hence, A – (B U C)=
(A – B) ∩ (A – C)