TOPICS

Unit-1(Examples)

Sets

**Example-1 :-** Write the solution set of the equation x² + x – 2 = 0 in roster form.

The given equation can be written as x² + x – 2 = 0 x² + 2x – x - 2 = 0 x (x + 2) - 1 (x + 2) = 0 (x – 1) (x + 2) = 0 i.e., x = 1, – 2 Therefore, the solution set of the given equation can be written in roster form as {1, – 2}.

**Example-2 :-** Write the set {x : x is a positive integer and x² < 40} in the roster form.

The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form is {1, 2, 3, 4, 5, 6}.

**Example-3 :-** Write the set A = {1, 4, 9, 16, 25, . . . } in set-builder form.

The set A = {1, 4, 9, 16, 25, . . . }, We may write as A = {x : x is the square of a natural number}. Alternatively, we can write A = {x : x = n², where n ∈ N}

**Example-4 :-** Write the set {1/2, 2/3, 3/4, 4/5, 5/6, 6/7} in the set-builder form

We see that each member in the given set has the numerator one less than the denominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the set-builder form the given set is {x : x = n/n+1, where is a natural number and 1 ≤ n ≤ 6}.

**Example-5 :-** Match each of the set on the left described in the roster form with the same set on the right described in the set-builder form :

(i) {P, R, I, N, C, A, L} (a) { x : x is a positive integer and is a divisor of 18}

(ii) { 0 } (b) { x : x is an integer and x² – 9 = 0}

(iii) {1, 2, 3, 6, 9, 18} (c) {x : x is an integer and x + 1= 1}

(iv) {3, –3} (d) {x : x is a letter of the word PRINCIPAL}

Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I are repeated, so (i) matches (d). Similarly, (ii) matches (c) as x + 1 = 1 implies x = 0. Also, 1, 2 ,3, 6, 9, 18 are all divisors of 18 and so (iii) matches (a). Finally, x² – 9 = 0 implies x = 3, –3 and so (iv) matches (b).

**Example-6 :-** State which of the following sets are finite or infinite :

(i) {x : x ∈ N and (x – 1) (x –2) = 0}

(ii) {x : x ∈ N and x² = 4}

(iii) {x : x ∈ N and 2x –1 = 0}

(iv) {x : x ∈ N and x is prime}

(v) {x : x ∈ N and x is odd}

(i) Given set = {1, 2}. Hence, it is finite. (ii) Given set = {2}. Hence, it is finite. (iii) Given set = φ. Hence, it is finite. (iv) The given set is the set of all prime numbers and since set of prime numbers is infinite. Hence the given set is infinite (v) Since there are infinite number of odd numbers, hence, the given set is infinite.

**Example-7 :-** Find the pairs of equal sets, if any, give reasons:

A = {0},

B = {x : x > 15 and x < 5},

C = {x : x – 5 = 0 },

D = {x: x² = 25},

E = {x : x is an integral positive root of the equation x² – 2x –15 = 0}.

Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E, it follows that, A ≠ B, A ≠ C, A ≠ D, A ≠ E. Since B = φ but none of the other sets are empty. Therefore B ≠ C, B ≠ D and B ≠ E. Also C = {5} but –5 ∈ D, hence C ≠ D. Since E = {5}, C = E. Further, D = {–5, 5} and E = {5}, we find that, D ≠ E. Thus, the only pair of equal sets is C and E.

**Example-8 :-** Which of the following pairs of sets are equal? Justify your answer.

(i) X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.

(ii) A = {n : n ∈ Z and n² ≤ 4} and B = {x : x ∈ R and x² – 3x + 2 = 0}.

(i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are equal sets as repetition of elements in a set do not change a set. Thus, X = {A, L, O, Y} = B. (ii) A = {–2, –1, 0, 1, 2}, B = {1, 2}. Since 0 ∈ A and 0 ∉ B, A and B are not equal sets.

**Example-9 :-** Consider the sets φ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol ⊂ or ⊄ between each of the following pair of sets:

(i) φ . . . B, (ii) A . . . B, (iii) A . . . C, (iv) B . . . C

(i) φ ⊂ B as φ is a subset of every set. (ii) A ⊄ B as 3 ∈ A and 3 ∉ B (iii) A ⊂ C as 1, 3 ∈ A also belongs to C (iv) B ⊂ C as each element of B is also an element of C.

**Example-10 :-** Let A = { a, e, i, o, u} and B = { a, b, c, d}. Is A a subset of B ? No. (Why?). Is B a subset of A? No. (Why?)

Given that A = { a, e, i, o, u} and B = { a, b, c, d}. No, A is not subset of B, because every element of set A is not present in set B. No, B is not subset of A, because every element of set B is not present in set A.

**Example-11 :-** Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that A ⊂ C?. If not, give an example.

No, Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}. Here A ∈ B as A = {1} and B ⊂ C. But A ⊄ C as 1 ∈ A and 1 ∉ C. Note that an element of a set can never be a subset of itself.

**Example-12 :-** Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A ∪ B.

Given that A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. A ∪ B = { 2, 4, 6, 8} ∪ { 6, 8, 10, 12} A ∪ B = {2, 4, 6, 8, 10, 12}

**Example-13 :-** Let A = { a, e, i, o, u } and B = { a, i, u }. Show that A ∪ B = A

We have, A ∪ B = { a, e, i, o, u } = A. This example illustrates that union of sets A and its subset B is the set A itself, i.e., if B ⊂ A, then A ∪ B = A.

**Example-14 :-** Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from Class XI who are in the school football team. Find X ∪ Y and interpret the set.

We have, X ∪ Y = {Ram, Geeta, Akbar, David, Ashok}. This is the set of students from Class XI who are in the hockey team or the football team or both.

**Example-15 :-** Consider the sets A and B of Example 12. Find A ∩ B.

Given that A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. A ∩ B = { 2, 4, 6, 8} ∩ { 6, 8, 10, 12} A ∩ B = {6, 8}

**Example-16 :-** Consider the sets X and Y of Example 14. Find X ∩ Y.

We see that element ‘Geeta’ is the only element common to both. Hence, X ∩ Y = {Geeta}.

**Example-17 :-** Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. Find A ∩ B and hence show that A ∩ B = B.

Given that A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = { 2, 3, 5, 7 }. A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ { 2, 3, 5, 7 } A ∩ B = {2, 3, 5, 7} = B

**Example-18 :-** Let A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find A – B and B – A.

Given that A = { 1, 2, 3, 4, 5, 6 } and B = { 2, 4, 6, 8 }. A - B = { 1, 2, 3, 4, 5, 6 } - { 2, 4, 6, 8 } A - B = {1, 3, 5} B - A = { 2, 4, 6, 8 } - { 1, 2, 3, 4, 5, 6 } B - A = {8}

**Example-19 :-** Let V = { a, e, i, o, u } and B = { a, i, k, u}. Find V – B and B – V.

Given that V = { a, e, i, o, u } and B = { a, i, k, u }. V - B = { a, e, i, o, u } - { a, i, k, u } V - B = {e, o} B - V = { a, i, k, u} - { a, e, i, o, u } B - V = {k}

**Example-20 :-** Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.

Given that U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {1, 3, 5, 7, 9} A' = {2, 4, 6, 8, 10}

**Example-21 :-** Let U be universal set of all the students of Class XI of a coeducational school and A be the set of all girls in Class XI. Find A′

Since A is the set of all girls, A′ is clearly the set of all boys in the class. Note : If A is a subset of the universal set U, then its complement A′ is also a subset of U.

**Example-22 :-** Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′.

Given that U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. A′ = {1, 4, 5, 6}, B′ = { 1, 2, 6 }. Hence A′ ∩ B′ = { 1, 6 } Also A ∪ B = { 2, 3, 4, 5 }, so that (A ∪ B)′ = { 1, 6 } ( A ∪ B )′ = { 1, 6 } = A′ ∩ B′

**Example-23 :-** If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X ∩ Y have ?

Given that n ( X ∪ Y ) = 50, n ( X ) = 28, n ( Y ) = 32, n (X ∩ Y) = ? By using the formula n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y ), we find that n ( X ∩ Y ) = n ( X ) + n ( Y ) – n ( X ∪ Y ) = 28 + 32 – 50 = 10

**Example-24 :-** In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teach both physics and mathematics. How many teach physics ?

Let M denote the set of teachers who teach mathematics and P denote the set of teachers who teach physics. So, n ( M ∪ P ) = 20 , n ( M ) = 12 and n ( M ∩ P ) = 4 We wish to determine n ( P ) = ? Using the result n ( M ∪ P ) = n ( M ) + n ( P ) – n ( M ∩ P ), we obtain 20 = 12 + n ( P ) – 4 Thus n ( P ) = 12 Hence 12 teachers teach physics

**Example-25 :-** In a class of 35 students, 24 like to play cricket and 16 like to play football.
Also, each student likes to play at least one of the two games. How many students like to play both cricket and football ?

Let X be the set of students who like to play cricket and Y be the set of students who like to play football. So, Given that n ( X) = 24, n ( Y ) = 16, n ( X ∪ Y ) = 35, n (X ∩ Y) = ? Using the formula n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y ), we get 35 = 24 + 16 – n (X ∩ Y) Thus, n (X ∩ Y) = 5 i.e., 5 students like to play both games.

**Example-26 :-** In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.

Let U denote the set of surveyed students and A denote the set of students taking apple juice and B denote the set of students taking orange juice. Then n (U) = 400, n (A) = 100, n (B) = 150 and n (A ∩ B) = 75. Now n (A′ ∩ B′) = n (A ∪ B)′ = n (U) – n (A ∪ B) = n (U) – n (A) – n (B) + n (A ∩ B) = 400 – 100 – 150 + 75 = 225 Hence 225 students were taking neither apple juice nor orange juice.

**Example-27 :-** There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C1, 50 to chemical C2, and 30 to both the chemicals C1 and C2.
Find the number of individuals exposed to

(i) Chemical C1 but not chemical C2

(ii) Chemical C2 but not chemical C1

(iii) Chemical C1 or chemical C2

Let U denote the universal set consisting of individuals suffering from the skin disorder, A denote the set of individuals exposed to the chemical C1 and B denote the set of individuals exposed to the chemical C2. Here n ( U) = 200, n ( A ) = 120, n ( B ) = 50 and n ( A ∩ B ) = 30 (i) we have A =( A – B ) ∪ ( A ∩ B ). n (A) = n( A – B ) + n( A ∩ B ) n ( A – B ) = n ( A ) – n ( A ∩ B ) = 120 –30 = 90 Hence, the number of individuals exposed to chemical C1 but not to chemical C2 is 90. (ii) we have B = ( B – A) ∪ ( A ∩ B). and so, n (B) = n (B – A) + n ( A ∩ B) n ( B – A ) = n ( B ) – n ( A ∩ B ) = 50 – 30 = 20 Thus, the number of individuals exposed to chemical C2 and not to chemical C1 is 20. (iii) The number of individuals exposed either to chemical C1 or to chemical C2, i.e., n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B ) = 120 + 50 – 30 = 140.

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