Students can download Maths Chapter 1 Set Language Ex 1.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.3

Question 1.

Using the given venn diagram, write the elements of

(i) A

(ii) B

(iii) A∪B

(iv) A∩B

(v) A – B

(vi) B – A

(vii) A’

(viii) B’

(ix) U

Solution:

(i) A = {2, 4, 7, 8, 10}

(ii) B = {3, 4, 6, 7, 9, 11}

(iii) A∪B = {2, 3, 4, 6, 7, 8, 9, 10, 11}

(iv) A∩B = {4, 7}

(v) A – B = {2, 8, 10}

(vi) B – A = {3, 6, 9, 11}

(vii) A’ = {1, 3, 6, 9, 11, 12}

(viii) B’ = {1,2, 8, 10, 12}

(ix) U = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}

Question 2.

Find A∪B, A∩B, A – B and B – A for the following sets

(i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16}

Solution:

A∪B = {2, 6, 10, 14} ∪ {2, 5, 14, 16}

= {2, 5, 6, 10, 14, 16}

A∩B = {2, 6, 10, 14} ∩ {2, 5, 14, 16}

= {2, 14}

A – B = {2, 6, 10, 14} – {2, 5, 14, 16}

= {6, 10}

B – A = {2, 5, 14, 16} – {2, 6, 10, 14}

= {5, 16}

(ii) A = {a, b, c, e, u} and B = {a, e, i, o, u}

Solution:

A∪B = {a, b, c, e, u} ∪ {a, e, i, o, u}

= {a, b, c, e, i, o, u}

A∩B = {a, b, c, e, u} ∩ {a, e, i, o, u}

= {a, e, u}

A – B = {a, b, c, e, u} – {a, e, i, o, u}

= {b, c}

B – A = {a, e, i, o, u} – {a, b, c, e, u}

= {i, o}

(iii) A = {x : x ∈ N, x ≤ 10} and B = {x : x ∈ W, x < 6}

Solution:

A = {1, 2, 3, 4, 5, 6,7, 8, 9, 10} and B = {0, 1, 2, 3, 4, 5}

A∪B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∪ {0, 1, 2, 3, 4, 5}

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A∩B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 3, 4, 5}

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

A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {0, 1, 2, 3, 4, 5}

= {6, 7, 8, 9, 10}

B – A = {0, 1, 2, 3, 4, 5} – {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

= {0}

(iv) A = Set of all letters in the word “mathematics” and B = Set of all letters in the word “geometry”

Solution:

A = {m, a, t, h, e, i, c, s}

B = {g, e, o, m, t, r, y}

A∪B = {m, a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y}

= {a, c, e, g, h, i, m, o, r, s, t, y}

A∩B= {m, a, t, h, e, i, c, s} ∩ {g, e, o, m, t, r, y}

= {e, m, t}

A – B = {m, a, t, h, e, i, c, 5} – {g, e, o, m, t, r, y}

= {a, c, h, i, s}

B -A = {g, e, o, m, t, r, y} – {m, a, t, h, e, i, c, s}

= {g, o, r, y}

Question 3.

If U = {a, b, c, d, e, f, g, h), A = {b, d, f, h} and B = {a, d, e, h}, find the following sets.

(i) A’

Solution:

A’ = U – A

= {a, b, c, d, e, f, g, h} – {b, d, f, h)

= {a, c, e, g}

(ii) B’

Solution:

B’ = U – B

= {a, b, c, d, e, f, g, h} – {a, d, e, h}

= {b, c, f, g}

(iii) A’∪B’

Solution:

A’∪B’ = {a, c, e, g} ∪ {b, c,f, g}

= {a, b, c, e, f, g}

(iv) A’∩B’

Solution:

A’∩B’ = {a, c, e, g} ∩ {b, c, f, g}

= {c, g}

(v) (A∪B)’

Solution:

A∪B = {b, d, f, h} ∪ {a, d, e, h}

= {a, b, d, e, f, h}

(A∪B)’ = U – (A∪B)

= {a, b, c, d, e, f, g, h} – {a, b, d, e, f, h}

= {c, g}

(vi) (A∩B)’

Solution:

(A∩B) = {b, d, f, h) ∩ {a, d, e, h)

= {d, h}

(A∩B)’ = U – (A∩B)

= {a, b, c, d, e, f, g, h} – {d, h}

= {a, b, c, e, f, g}

(vii) (A’)’

Solution:

A’ = {a, c, e, g}

(A’)’ = U – A’

= {a, b, c, d, e, f, g, h} – {a, c, e, g}

= {b, d, f, h}

(viii) (B’)’

Solution:

B’ = {b, c, f, g}

(B’)’ = U – B’

= {a, b, c, d, e, f, g, h) – {b, c, f, g)

= {a, d, e, h}

Question 4.

Let U = {0, 1, 2, 3, 4, 5, 6, 7} A = {1, 3, 5, 7} and B = {0, 2, 3, 5, 7}, find the following sets.

(i) A’

Solution:

A’ = U – A

= {0, 1, 2, 3, 4, 5, 6, 7} – {1, 3, 5, 7}

= {0, 2, 4, 6}

(ii) B’ = U – B

Solution:

= {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 3, 5, 7}

= {1, 4, 6}

(iii) A’∪B’

Solution:

A’∪B’ = {0, 2, 4, 6,}∪{1, 4, 6}

{0, 1, 2, 4, 6}

(iv) A’∩B’

Solution:

A’∩B’ = {0, 2, 4, 6,}∩{1, 4, 6}

{4, 6}

(v) (A∪B)’

Solution:

A∪B = {1, 3, 5, 7}∪{0, 2, 3, 5, 7}

= {0, 1, 2, 3, 5, 7}

(A∪B)’ = U – (A∪B)

{0, 1, 2, 3, 4, 5, 6, 7} – {0, 1, 2, 3, 5, 7}

{4, 6}

(vi) (A∩B)’

Solution:

(A∩B)= {1, 3, 5, 7}∩{0, 2, 3, 5, 7}

= {3, 5, 7}

(A∩B)’ = U – (A∩B)

= {0, 1, 2, 3, 4, 5, 6, 7} – {3, 5, 7}

= {0, 1, 2, 4, 6}

(vii) (A’)’

A’ = {0, 2, 4, 6}

(A’)’ = U – A’

= {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 4, 6}

= {1, 3, 5, 7}

(viii) (B’)’

B’ = {1, 4, 6}

(B’)’ = {0, 1, 2, 3, 4, 5, 6, 7} – {1, 4, 6}

= {0, 2, 3, 5, 7}

Question 5.

Find the symmetric difference between the following sets.

(i) P = {2, 3, 5, 7, 11} and Q = {1, 3, 5, 11}

Solution:

Use anyone of the formula to find A & B

AΔB = (A – B)∪(B – A) or AΔB = (A∪B) – (A∩B)

P∪Q = {2, 3, 5, 7, 11} ∪ {1,3,5, 11}

= {1, 2, 3, 5, 7, 11}

P∩Q = {2, 3, 5, 7, 11}∩{1, 3, 5,11}

= {3, 5, 11}

PΔQ = (P∪Q) – (P∩Q)

= {1, 2, 3, 5, 7, 11} – {3, 5, 11}

= {1, 2, 7}

(OR)

P – Q = {2, 3, 5, 7, 11} – {1, 3, 5, 11}

= {2,7}

Q – P = {1, 3, 5, 11} – {2, 3, 5, 7, 11}

= {1}

PΔQ = (P – Q)∪(Q – P)

= {2, 7} ∪ {1}

= {1, 2, 7}

(ii) R = {l, m, n, o, p} and S = {j, l, n, q}

Solution:

R- S = {l, m, n, o, p} – {j, l, n, q}

= {m, o, p}

S – R = {j, l, n, q} – {l, m, n, o, p}

= {j, q}

RΔS = (R – S)∪(S – R)

= {m, o, p} – {j, q} = {j, m, o, p, q)

(OR)

R∪S = {l, m, n, o, p} ∪ {j,l,n,q}

= {l, m, n, o, p, j, q}

R∩S = {l, m, n, o,p} ∩ {j, l, n, q}

= {l, n}

RΔS = (R∪S) – (R∩S)

= {l, m, n, o, p, j, q} – { l, n}

= {m, o, p, j, q}

(iii) X = {5, 6, 7} and Y = {5, 7, 9, 10}

Solution:

X∪Y = {5, 6, 7} ∪ {5, 7, 9, 10}

= {5 ,6, 7, 9, 10}

X∩Y = {5, 6, 7} ∩ {5, 7, 9, 10}

= {5, 7}

XΔY = (X∪Y) – (X∩Y)

= {5, 6, 7, 9, 10} – {5, 7}

= {6, 9, 10}

OR

X – Y = {5, 6, 7} – {5, 7, 9, 10} = {6}

Y – X = {5, 7, 9, 10} – {5, 6, 7} = {9, 10}

XΔY = (X – Y) ∪ (Y – X)

= {6}∪{9, 10}

= {6, 9, 10}

Question 6.

Using the set symbols, write down the expressions for the shaded region in the following

Solution:

Question 7.

Let A and B be two overlapping sets and the universal set U. Draw appropriate Venn diagram for each of the following,

(i) A∪B

(ii) A∩B

(iii) (A∩B)’

(iv) (B – A)’

(v) A’∪B’

(Vi) A’∩B’

(vii) What do you observe from the Venn diagram (iii) and (v)?

Solution:

(i) A∪B

(ii) A∩B

(iii) (A∩B)’

(iv) (B – A)’

(v) A’∪B’

(Vi) A’∩B’

(vii) What do you observe from the diagram (iii) and (v)?

From the diagram (iii) and (v) we get (A∩B)’ = A’∪B’