Chapter 1

Class 11 Mathematics · 61 questions · 0 with answers

Solved examples

example-1Multiple choice

Write the following sets in the roaster form.

(i)A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}
(ii)C = {x : x2 + 7x – 8 = 0, x ∈ R} Solution (i) 2x – 1 is always an odd number for all positive integral values of x. In particular, 2x – 1 is an odd number for x = 1, 2, ... , 9. Thus, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}. (ii) x2 + 7x – 8 = 0 or (x + 8) (x – 1) = 0 giving x = – 8 or x = 1 Thus, C = {– 8, 1}

example-2Multiple choice

State which of the following statements are true and which are false. Justify your answer.

(i)37 ∉ {x | x has exactly two positive factors}
(ii)28 ∈ {y | the sum of the all positive factors of y is 2y}
(iii)7,747 ∈ {t | t is a multiple of 37} Solution (i) False Since, 37 has exactly two positive factors, 1 and 37, 37 belongs to the set. (ii) True Since, the sum of positive factors of 28 = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2(28) (iii) False 7,747 is not a multiple of 37.

example-3Multiple choice

If X and Y are subsets of the universal set U, then show that

(i)Y ⊂ X ∪ Y
(ii)X ∩ Y ⊂ X
(iii)X ⊂ Y ⇒ X ∩ Y = X Solution (i) X ∪ Y = {x | x ∈ X or x ∈ Y} Thus x∈Y⇒x∈X∪Y Hence, Y⊂X∪Y SETS 5 (ii) X ∩ Y = {x | x ∈ X and x ∈ Y} Thus x∈X∩Y⇒x∈X Hence X∩Y⊂X (iii) Note that x∈X∩Y⇒x∈X Thus X∩Y⊂X Also, since X ⊂ Y, x∈X⇒x∈Y⇒x∈X∩Y so that X⊂X∩Y Hence the result X = X ∩ Y follows.

example-4Multiple choice

Given that N = {1, 2, 3, ..., 100}, then

(i)Write the subset A of N, whose element are odd numbers.
(ii)Write the subset B of N, whose element are represented by x + 2, where x ∈ N. Solution (i) A = {x | x ∈ N and x is odd}= {1, 3, 5, 7, ..., 99} (ii) B = {y | y = x + 2, x ∈ N} So, for 1 ∈ N, y = 1 + 2 = 3

example-5Multiple choice

Given that E = {2, 4, 6, 8, 10}. If n represents any member of E, then, write the following sets containing all numbers represented by

(i)n + 1
(ii)n 2 Solution Given E = {2, 4, 6, 8, 10} (i) Let A = {x | x = n + 1, n ∈ E} Thus, for 2 ∈ E, x = 3

example-6Multiple choice

Let X = {1, 2, 3, 4, 5, 6}. If n represent any member of X, express the following as sets :

(i)n ∈ X but 2n ∉ X
(ii)n + 5 = 8
(iii)n is greater than 4. Solution (i) For X = {1, 2, 3, 4, 5, 6}, it is the given that n ∈ X, but 2n ∉ X. Let, A = {x | x ∈ X and 2x ∉ X} Now, 1∉A as 2.1 = 2 ∈ X 2∉A as 2.2 = 4 ∈ X 3∉A as 2.3 = 6 ∈ X But 4∈A as 2.4 = 8 ∉ X 5∈A as 2.5 = 10 ∉ X 6∈A as 2.6 = 12 ∉ X So, A = {4, 5, 6} (ii) Let B = {x | x ∈ X and x + 5 = 8} Here, B = {3} as x = 3 ∈ X and 3 + 5 = 8 and there is no other element belonging to X such that x + 5 = 8. (iii) Let C = {x | x ∈ X, x > 4} Therefore, C = {5, 6}

example-7Multiple choice

Draw the Venn diagrams to illustrate the followoing relationship among sets E, M and U, where E is the set of students studying English in a school, M is the set of students studying Mathematics in the same school, U is the set of all students in that school.

(i)All the students who study Mathematics study English, but some students who study English do not study Mathematics.
(ii)There is no student who studies both Mathematics and English.
(iii)Some of the students study Mathematics but do not study English, some study English but do not study Mathematics, and some study both.
(iv)Not all students study Mathematics, but every students studying English studies Mathematics. Solution (i) Since all of the students who study mathematics study English, but some students who study English do not study Mathematics. Therefore, M⊂E⊂U Thus the Venn Diagram is Fig 1.4 SETS 7 (ii) Since there is no student who study both English and Mathematics Hence, E ∩ M = φ. Fig 1.5 (iii) Since there are some students who study both English and Mathematics, some English only and some Mathematics only. Thus, the Venn Diagram is Fig 1.6 (iv) Since every student studying English studiesMathematics. Hence, E⊂M⊂U Fig 1.7

example-8Short answer

For all sets A, B and C Is (A ∩ B) ∪ C = A ∩ (B ∪ C)? Justify your statement. Solution No. consider the following sets A, B and C : A = {1, 2, 3} B = {2, 3, 5} C = {4, 5, 6} Now (A ∩ B) ∪ C = ({1, 2, 3} ∩ {2, 3, 5}) ∪ {4, 5, 6} = {2, 3} ∪ {4, 5, 6} = {2, 3, 4, 5, 6} And A ∩ (B ∪ C) = {1, 2, 3} ∩ [{2, 3, 5} ∪ {4, 5, 6} = {1, 2, 3} ∩ {2, 3, 4, 5, 6} = {2, 3} Therefore, (A ∩ B) ∪ C ≠ A ∩ (B ∪ C)

example-9Short answer

Use the properties of sets to prove that for all the sets A and B A – (A ∩ B) = A – B Solution We have A – (A ∩ B) = A ∩ (A ∩ B)′ (since A – B = A ∩ B′) = A ∩ (A′ ∪ B′) [by De Morgan’s law) = (A ∩ A′) ∪ (A ∩ B′) [by distributive law] = φ ∪ (A ∩ B′) = A ∩ B′ = A – B

example-10Long answer

For all sets A, B and C Is (A – B) ∩ (C – B) = (A ∩ C) – B? Justify your answer. Solution Yes Let x ∈ (A – B) ∩ (C – B) ⇒ x ∈ A – B and x ∈ C – B ⇒ (x ∈ A and x ∉ B) and (x ∈ C and x ∉ B) ⇒ (x ∈ A and x ∈ C) and x ∉ B ⇒ (x ∈ A ∩ C) and x ∉ B ⇒ x ∈ (A ∩ C) – B So (A – B) ∩ (C – B) ⊂ (A ∩ C) – B ... (1) Now, conversely SETS 9 Let y ∈ (A ∩ C) – B ⇒ y ∈ (A ∩ C) and y ∉ B ⇒ (y ∈ A and y ∈ C) and (y ∉ B) ⇒ (y ∈ A and y ∉ B) and (y ∈ C and y ∉ B) ⇒ y ∈ (A – B) and y ∈ (C – B) ⇒ y ∈ (A – B) ∩ (C – B) So (A ∩ C) – B ⊂ (A – B) ∩ (C – B) ... (2) From (1) and (2), (A – B) ∩ (C – B) = (A ∩ C) – B

example-11Long answer

Let A, B and C be sets. Then show that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Solution We first show that A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C) Let x ∈ A ∪ (B ∩ C). Then x∈A or x∈B∩C ⇒ x∈A or (x ∈ B and x ∈ C) ⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C) ⇒ (x ∈ A ∪ B) and (x ∈ A ∪ C) ⇒ x ∈ (A ∪ B) ∩ (A ∪ C) Thus, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C) ... (1) Now we will show that (A ∪ B) ∩ (A ∪ C) ⊂ (A ∪ C) Let x ∈ (A ∪ B) ∩ (A ∪ C) ⇒ x ∈ A ∪ B and x ∈ A ∪ C ⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C) ⇒ x ∈ A or (x ∈ B and x ∈ C) ⇒ x ∈ A or (x ∈ B ∩ C) ⇒ x ∈ A ∪ (B ∩ C) Thus, (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C) ... (2) So, from (1) and (2), we have A ∩ (B ∪ C) = (A ∪ B) ∩ (A ∪ C)

example-12Long answer

Let P be the set of prime numbers and let S = {t | 2t – 1 is a prime}. Prove that S ⊂ P. Solution Now the equivalent contrapositive statement of x ∈ S ⇒ x ∈ P is x ∉ P ⇒ x ∉ S. Now, we will prove the above contrapositive statement by contradiction method Let x∉P ⇒ x is a composite number Let us now assume that x ∈ S ⇒ 2x – 1 = m (where m is a prime number) ⇒ 2 =m+1 Which is not true for all composite number, say for x = 4 because

example-13Long answer

From 50 students taking examinations in Mathematics, Physics and Chemistry, each of the student has passed in at least one of the subject, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry. What is the largest possible number that could have passed all three examination? Solution Let M be the set of students passing in Mathematics P be the set of students passing in Physics C be the set of students passing in Chemistry Now, n(M ∪ P ∪ C) = 50, n(M) = 37, n(P) = 24, n(C) = 43 n(M ∩ P) ≤ 19, n(M ∩ C) ≤ 29, n(P ∩ C) ≤ 20 (Given) n(M ∪ P ∪ C) = n(M) + n(P) + n(C) – n(M ∩ P) – n(M ∩ C) – n(P ∩ C) + n(M ∩ P ∩ C) ≤ 50 ⇒ 37 + 24 + 43 – 19 – 29 – 20 + n(M ∩ P ∩ C) ≤ 50 ⇒ n(M ∩ P ∩ C) ≤ 50 – 36 ⇒ n(M ∩ P ∩ C) ≤ 14 Thus, the largest possible number that could have passed all the three examinations is 14.

example-14Multiple choice

Each set Xr contains 5 elements and each set Yr contains 2 elements 20 n and ∪ X r = S = ∪ Yr . If each element of S belong to exactly 10 of the X ’s and to r =1 r =1 exactly 4 of the Yr’s, then n is

(A)10
(B)20
(C)100
(D)50 Solution The correct answer is (B) Since, n(Xr) = 5, ∪ X r = S , we get n(S) = 100 r =1 But each element of S belong to exactly 10 of the Xr’s So, = 10 are the number of distinct elements in S. Also each element of S belong to exactly 4 of the Yr’ s and each Yr contain 2 elements. If S has n number of Yr in it. Then 2n = 10 which gives n = 20

example-15Multiple choice

Two finite sets have m and n elements respectively. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The values of m and n respectively are.

(A)7, 6
(B)5, 1
(C)6, 3
(D)8, 7 Solution The correct answer is (C). Since, let A and B be such sets, i.e., n (A) = m, n (B) = n So n (P(A)) = 2m, n (P(B)) = 2n Thus n (P(A)) – n (P(B)) = 56, i.e., 2m – 2n = 56 ⇒ 2n (2m – n – 1) = 23 7 ⇒ n = 3 , 2m – n – 1 = 7 ⇒ m=6

example-16Multiple choice

The set (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∩ C′ is equal to

(A)B ∩ C′
(B)A ∩ C
(C)B ∪ C′
(D)A ∩ C′ Solution The correct choice is (A). Since (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∩ C′ = (A ∪ (B ∪ C)) ∩ (A′ ∪ (B ∪ C)) ∩ C′ = (A ∩ A′) ∪ (B ∪ C) ∩ C′ = φ ∪ (B ∪ C) ∩ C′ = B ∩ C′ ∪ φ = B ∩ C′ Fill in the blanks in Examples 17 and 18 :

example-17Multiple choice

If A and B are two finite sets, then n(A) + n(B) is equal to ___________ Solution Since n(A ∪ B) = n

(A)+ n
(B)– n (A ∩ B) So n(A) + n (B) = n (A ∪ B) + n (A ∩ B)

example-18Fill in the blanks

If A is a finite set containing n element, then number of subsets of A is __________ Solution 2n State true or false for the following statements given in Examples 19 and 20.

example-19Short answer

Let R and S be the sets defined as follows: R = {x ∈ Z | x is divisible by 2} S = {y ∈ Z | y is divisible by 3} then R∩S=φ Solution False Since 6 is divisible by both 3 and 2. Thus R∩S ≠φ

example-20Short answer

Q ∩ R = Q, where Q is the set of rational numbers and R is the set of real numbers. Solution True Since Q⊂R So Q∩R=Q

Questions

Q2Short answer

∈ N, y = 2 + 2 = 4, and so on. Therefore, B = {3, 4, 5, 6, ... , 100}

Q4Short answer

∈ E, x = 5, and so on. Therefore, A = {3, 5, 7, 9, 11}. (ii) Let B = {x | x = n2, n ∈ E} So, for 2 ∈ E, x = (2)2 = 4, 4 ∈ E, x = (4)2 = 16, 6 ∈ E, x = (6)2 = 36, and so on. Hence, B = {4, 16, 36, 64, 100}

Q24Long answer

= 16 which can not be equal to the sum of any prime number m and 1. Thus, we arrive at a contradiction ⇒ x ∉ S. Thus, when x ∉ P, we arrive at x ∉ S So S ⊂ P.

Q1Multiple choice

Write the following sets in the roaster from

(i)A = {x : x ∈ R, 2x + 11 = 15}
(ii)B = {x | x2 = x, x ∈ R}
(iii)C = {x | x is a positive factor of a prime number p} SETS 13

Q2Multiple choice

Write the following sets in the roaster form : w−2

(i)D = {t | t3 = t, t ∈ R}
(ii)E = {w | = 3 , w ∈ R} w+3
(iii)F = {x | x4 – 5x2 + 6 = 0, x ∈ R}

Q3Short answer

If Y = {x | x is a positive factor of the number 2p – 1 (2p – 1), where 2p – 1 is a prime number}.Write Y in the roaster form.

Q4Multiple choice

State which of the following statements are true and which are false. Justify your answer.

(i)35 ∈ {x | x has exactly four positive factors}.
(ii)128 ∈ {y | the sum of all the positive factors of y is 2y}
(iii)3 ∉ {x | x4 – 5x3 + 2x2 – 112x + 6 = 0}
(iv)496 ∉ {y | the sum of all the positive factors of y is 2y}.

Q5Short answer

Given L = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5} Verify that L – (M ∪ N) = (L – M) ∩ (L – N)

Q6Multiple choice

If A and B are subsets of the universal set U, then show that

(i)A ⊂ A ∪ B
(ii)A ⊂ B ⇔ A ∪ B = B
(iii)(A ∩ B) ⊂ A

Q7Multiple choice

Given that N = {1, 2, 3, ... , 100}. Then write

(i)the subset of N whose elements are even numbers.
(ii)the subset of N whose element are perfect square numbers.

Q8Multiple choice

If X = {1, 2, 3}, if n represents any member of X, write the following sets containing all numbers represented by

(i)4n
(ii)n + 6
(iii)
(iv)n – 1

Q9Multiple choice

If Y = {1, 2, 3, ... 10}, and a represents any element of Y, write the following sets, containing all the elements satisfying the given conditions.

(i)a ∈ Y but a2 ∉ Y
(ii)a + 1 = 6, a ∈ Y
(iii)a is less than 6 and a ∈ Y

Q10Short answer

A, B and C are subsets of Universal Set U. If A = {2, 4, 6, 8, 12, 20} B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.

Q11Short answer

Let U be the set of all boys and girls in a school, G be the set of all girls in the school, B be the set of all boys in the school, and S be the set of all students in the school who take swimming. Some, but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationship among sets U, G, B and S.

Q12True / False

For all sets A, B and C, show that (A – B) ∩ (C – B) = A – (B ∪ C) Determine whether each of the statement in Exercises 13 – 17 is true or false. Justify your answer.

Q13Short answer

For all sets A and B, (A – B) ∪ (A ∩ B) = A

Q14Short answer

For all sets A, B and C, A – (B – C) = (A – B) – C

Q15Short answer

For all sets A, B and C, if A ⊂ B, then A ∩ C ⊂ B ∩ C

Q16Short answer

For all sets A, B and C, if A ⊂ B, then A ∪ C ⊂ B ∪ C

Q17Short answer

For all sets A, B and C, if A ⊂ C and B ⊂ C, then A ∪ B ⊂ C. Using properties of sets prove the statements given in Exercises 18 to 22

Q18Short answer

For all sets A and B, A ∪ (B – A) = A ∪ B

Q19Short answer

For all sets A and B, A – (A – B) = A ∩ B

Q20Short answer

For all sets A and B, A – (A ∩ B) = A – B

Q21Short answer

For all sets A and B, (A ∪ B) – B = A – B x+5 4 x − 40

Q22Short answer

Let T = x| −5= . Is T an empty set? Justify your answer. x−7 13 − x

Q23Long answer

Let A, B and C be sets. Then show that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Q24Multiple choice

Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science; 4 in English and Science; 4 in all the three. Find how many passed

(i)in English and Mathematics but not in Science
(ii)in Mathematics and Science but not in English
(iii)in Mathematics only
(iv)in more than one subject only

Q25Long answer

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither? SETS 15

Q26Long answer

In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.

Q27Multiple choice

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find

(a)The number of families which buy newspaper A only.
(b)The number of families which buy none of A, B and C

Q28Multiple choice

In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows: French = 17, English = 13, Sanskrit = 15 French and English = 09, English and Sanskrit = 4 French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study

(i)French only (v) French and Sanskrit but not English
(ii)English only (vi) French and English but not Sanskrit
(iii)Sanskrit only (vii) at least one of the three languages
(iv)English and Sanskrit (viii) none of the three languages but not French

Q29Multiple choice

Suppose A1, A2, ..., A30 are thirty sets each having 5 elements and B1, B2, ..., Bn 30 n are n sets each with 3 elements, let ∪ Ai = ∪ B j = S and each element of S i =1 j =1 belongs to exactly 10 of the Ai’s and exactly 9 of the B,’S. then n is equal to

(A)15
(B)3
(C)45
(D)35

Q30Multiple choice

Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. The values of m and n are, respectively,

(A)4, 7
(B)7, 4
(C)4, 4
(D)7, 7

Q31Multiple choice

The set (A ∩ B′)′ ∪ (B ∩ C) is equal to

(A)A′ ∪ B ∪ C
(B)A′ ∪ B
(C)A′ ∪ C′
(D)A′ ∩ B

Q32Multiple choice

Let F1 be the set of parallelograms, F2 the set of rectangles, F3 the set of rhombuses, F4 the set of squares and F5 the set of trapeziums in a plane. Then F1 may be equal to

(A)F2 ∩ F3
(B)F3 ∩ F4
(C)F2 ∪ F5
(D)F2 ∪ F3 ∪ F4 ∪ F1

Q33Multiple choice

Let S = set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then

(A)S ∩ T ∩ C = φ
(B)S ∪ T ∪ C = C
(C)S ∪ T ∪ C = S
(D)S ∪ T = S ∩ C

Q34Multiple choice

Let R be set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of x-axis and y-axis. Then

(A)R = {(x, y) : 0 ≤ x ≤ a, 0 ≤ y ≤ b}
(B)R = {(x, y) : 0 ≤ x < a, 0 ≤ y ≤ b}
(C)R = {(x, y) : 0 ≤ x ≤ a, 0 < y < b}
(D)R = {(x, y) : 0 < x < a, 0 < y < b}

Q35Multiple choice

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Then, the number of students who play neither is

(A)0
(B)25
(C)35
(D)45

Q36Multiple choice

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is

(A)210
(B)290
(C)180
(D)260

Q37Multiple choice

If X = {8 – 7n – 1 | n ∈ N} and Y = {49n – 49 | n ∈ N}. Then

(A)X ⊂ Y
(B)Y ⊂ X
(C)X = Y
(D)X ∩ Y = φ

Q38Multiple choice

A survey shows that 63% of the people watch a News Channel whereas 76% watch another channel. If x% of the people watch both channel, then

(A)x = 35
(B)x = 63
(C)39 ≤ x ≤ 63
(D)x = 39

Q39Multiple choice

If sets A and B are defined as A = {(x, y) | y = , 0 ≠ x ∈ R} B = {(x, y) | y = – x, x ∈ R}, then

(A)A ∩ B = A
(B)A ∩ B = B
(C)A ∩ B = φ
(D)A ∪ B = A SETS 17

Q40Multiple choice

If A and B are two sets, then A ∩ (A ∪ B) equals

(A)A
(B)B
(C)φ
(D)A ∩ B

Q41Multiple choice

IfA = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ... , 18} and N the set of natural numbers is the universal set, then A′ ∪ (A ∪ B) ∩ B′) is

(A)φ
(B)N
(C)A
(D)B

Q42Multiple choice

Let S = {x | x is a positive multiple of 3 less than 100} P = {x | x is a prime number less than 20}. Then n(S) + n(P) is

(A)34
(B)31
(C)33
(D)30

Q43Multiple choice

If X and Y are two sets and X′ denotes the complement of X, then X ∩ (X ∪ Y)′ is equal to

(A)X
(B)Y
(C)φ
(D)X ∩ Y Fill in the blanks in each of the Exercises from 44 to 51 :

Q44Fill in the blanks

The set {x ∈ R : 1 ≤ x < 2} can be written as ______________.

Q45Fill in the blanks

When A = φ, then number of elements in P(A) is ______________.

Q46Fill in the blanks

If A and B are finite sets such that A ⊂ B, then n (A ∪ B) = ______________.

Q47Fill in the blanks

If A and B are any two sets, then A – B is equal to ______________.

Q48Fill in the blanks

Power set of the set A = {1, 2} is ______________.

Q49Fill in the blanks

Given the sets A = {1, 3, 5}. B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Then the universal set of all the three sets A, B and C can be ______________.

Q50Multiple choice

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}. Then

(i)(B ∪ C)′ is ______________.
(ii)(C – A)′ is ______________.

Q51Fill in the blanks

For all sets A and B, A – (A ∩ B) is equal to ______________.

Q52Multiple choice

Match the following sets for all sets A, B and C (i) ((A′ ∪ B′) – A)′

(a)A – B (ii) [B′ ∪ (B′ – A)]′
(b)A (iii) (A – B) – (B – C)
(c)B (iv) (A – B) ∩ (C – B)
(d)(A × B) ∩ (A × C) (v) A × (B ∩ C) (e) (A × B) ∪ (A × C) (vi) A × (B ∪ C) (f) (A ∩ C) – B State True or False for the following statements in each of the Exercises from 53 to 58 :

Q53Multiple choice

If A is any set, then A ⊂ A

Q54Multiple choice

Given that M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and if B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then B ⊄ M

Q55Multiple choice

The sets {1, 2, 3, 4} and {3, 4, 5, 6} are equal.

Q56Multiple choice

Q ∪ Z = Q, where Q is the set of rational numbers and Z is the set of integers.

Q57Multiple choice

Let sets R and T be defined as R = {x ∈ Z | x is divisible by 2} T = {x ∈ Z | x is divisible by 6}. Then T ⊂ R

Q58Multiple choice

Given A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2}. Then A = B.