Class 11 Mathematics Sets MCQs Set 04

Answer: (c) ∈, ∉, ∉

Explanation:
1. 4 is listed in A, so 4 ∈ A.
2. −4 is not listed in A, so −4 ∉ A.
3. 12 is not listed in A, so 12 ∉ A. Hence the symbols are ∈, ∉, ∉.

Teacher's Note:
1. The symbol ∈ is used only between an element and a set, never between two sets.
2. Read the roster form carefully; a negative number or a number beyond the last listed value is not an element.

Answer: (a) {1, 3, 9}

Explanation:
1. The positive integers that divide 9 exactly are 1, 3 and 9.
2. Hence the set in roster form is {1, 3, 9}.

Teacher's Note:
1. The divisors of a number always include 1 and the number itself.
2. Do not confuse divisors (which divide 9) with multiples (which 9 divides).

Answer: (c) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Explanation:
1. 4x + 9 < 50 ⇒ 4x < 41 ⇒ x < 10.25
2. Natural numbers less than 10.25 are 1, 2, 3, …, 10.
3. So the set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Teacher's Note:
1. Solve the inequality first, then pick only the values allowed by the type of number (here natural numbers).
2. A strict inequality x < 10.25 still allows 10 because 10 is less than 10.25.

Answer: (d) \( \left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}\right\} \)

Explanation:
1. n takes the values 1, 2, 3.
2. n = 1: \( \frac{1}{1+1} = \frac{1}{2} \); n = 2: \( \frac{2}{1+4} = \frac{2}{5} \); n = 3: \( \frac{3}{1+9} = \frac{3}{10} \)
3. Hence the set is \( \left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}\right\} \).

Teacher's Note:
1. Substitute each allowed value of n one by one and simplify.
2. Check the denominator 1 + n² carefully; for n = 3 it is 10, not 5.

Answer: (b) A = {x : x = 7n, n ∈ N and 2 ≤ n ≤ 14}

Explanation:
1. Every element is a multiple of 7: 14 = 7 × 2 and 98 = 7 × 14.
2. So x = 7n where n runs from 2 to 14.
3. Hence A = {x : x = 7n, n ∈ N and 2 ≤ n ≤ 14}.

Teacher's Note:
1. Divide the first and last elements by the common factor to get the range of n.
2. Always check both ends of the range; 7 × 15 = 105 is not in A.

Answer: (a) 0

Explanation:
1. φ denotes the empty (null) set.
2. The empty set has no element, so n(A) = 0.

Teacher's Note:
1. φ and {φ} are different: {φ} has one element, namely φ.
2. The empty set is a finite set with zero elements.

Answer: (a) singleton set

Explanation:
1. B has exactly one element, 5.
2. A set with only one element is called a singleton set.

Teacher's Note:
1. Singleton sets are written as {a}; they are not the same as the element a itself.
2. An empty set has no elements, so {5} cannot be empty.

Answer: (b) finite

Explanation:
1. A year has exactly 12 months.
2. A set whose elements can be counted completely is a finite set, so X = finite.

Teacher's Note:
1. A set is finite if counting its elements comes to an end.
2. Finite does not mean small; it only means the counting stops.

Answer: (a) 4

Explanation:
1. A = {1, 4, 9, 16, 25, 36, 49, 64, 81}
2. The even elements of A are 4, 16, 36, 64.
3. So A ∩ B = {4, 16, 36, 64} and its cardinality is 4.

Teacher's Note:
1. Cardinality means the number of elements in a finite set, written n(A ∩ B).
2. The square of a number is even only when the number itself is even, which gives a quick count.

Answer: (c) infinite

Explanation:
1. The set {1, 2, 3, …} is the set of all natural numbers.
2. Its elements never end, so it is an infinite set.

Teacher's Note:
1. Three dots at the end of a roster usually signal an infinite set.
2. The set of natural numbers is the most common example of an infinite set.

Answer: (c) Sets A, C and B, D

Explanation:
1. n(A) = 3, n(B) = 5, n(C) = 3, n(D) = 5
2. Equivalent sets have the same number of elements.
3. So A, C are equivalent and B, D are equivalent.

Teacher's Note:
1. Equivalent sets need only the same number of elements; the elements themselves can be different.
2. Equal sets are always equivalent, but equivalent sets need not be equal.

Answer: (a) equal

Explanation:
1. A = {A, L, O, Y} (repeated L is written once).
2. B = {L, O, Y, A} = {A, L, O, Y}
3. Both sets have exactly the same elements, so A = B.

Teacher's Note:
1. In roster form a repeated letter is listed only once.
2. The order of elements does not matter when comparing sets.

Answer: (a) X ⊂ Y

Explanation:
1. X = {0, 49, 490, …} and Y = {0, 49, 98, 147, …}, the multiples of 49 starting from 0.
2. By the binomial theorem, 8ⁿ = (1 + 7)ⁿ = 1 + 7n + 49k for some whole number k, so 8ⁿ − 7n − 1 is always a multiple of 49.
3. Hence every element of X is in Y, but 98 ∈ Y is not in X. So X ⊂ Y.

Teacher's Note:
1. Writing the first few elements of both sets often reveals the relationship quickly.
2. To show X ⊂ Y, every element of X must be in Y; one element of Y missing from X shows X ≠ Y.

Answer: (c) 11

Explanation:
1. The set is the interval (−5, 6].
2. Length of an interval = upper end − lower end = 6 − (−5) = 11.

Teacher's Note:
1. The length of an interval does not depend on whether the end points are included or not.
2. Use ( for an excluded end and ] for an included end.

Answer: (d) {0, 1, 2, 3, 4, 5, 6, 7}

Explanation:
1. A universal set must contain every element of all the given sets.
2. A ∪ B ∪ C = {0, 1, 2, 3, 4, 5, 6, 7}
3. Among the options only (d) contains all of them.

Teacher's Note:
1. Every set under discussion must be a subset of the universal set.
2. If even one element of a given set is missing, the option cannot be the universal set.

Answer: (a) A

Explanation:
1. A ⊂ A ∪ B, since every element of A is in A ∪ B.
2. If P ⊂ Q, then P ∩ Q = P.
3. So A ∩ (A ∪ B) = A.

Teacher's Note:
1. This is called the absorption law; similarly A ∪ (A ∩ B) = A.
2. A quick check with A = {1}, B = {2} gives {1} ∩ {1, 2} = {1} = A.

Answer: (d) {2, 4, 6, 8, 10, 12}

Explanation:
1. A ∪ B contains every element that is in A or in B (or both).
2. A ∪ B = {2, 4, 6, 8, 10, 12}; the common elements 6, 8 are written once.

Teacher's Note:
1. Union collects elements from both sets; intersection keeps only the common ones.
2. Do not repeat common elements while writing the union.

Answer: (d) {1, 2, 3, 4, 5, 6, 9}

Explanation:
1. A ∪ B contains all the elements of A and of B.
2. A ∪ B = {1, 2, 3, 4, 5, 6, 9}

Teacher's Note:
1. Count check: A and B are disjoint here, so A ∪ B must have 3 + 4 = 7 elements.
2. Options with fewer elements must be missing something.

Answer: (a) A

Explanation:
1. Every element of B (a, i, u) is already in A, so B ⊂ A.
2. If B ⊂ A, then A ∪ B = A.

Teacher's Note:
1. When one set is a subset of the other, the union is the bigger set and the intersection is the smaller set.
2. Here A ∩ B = B and A ∪ B = A.

Answer: (d) four points

Explanation:
1. Points of A ∩ B satisfy both x² + y² = 25 and x² + 9y² = 144.
2. Subtracting, 8y² = 119, so \( y^2 = \frac{119}{8} \) and \( x^2 = 25 - \frac{119}{8} = \frac{81}{8} \).
3. Both x and y take two non-zero values each (±), giving 4 points \( \left(\pm\frac{9}{2\sqrt2}, \pm\sqrt{\frac{119}{8}}\right) \).

Teacher's Note:
1. A ∩ B for sets of points means the common points, found by solving the two equations together.
2. When both x² and y² come out positive, each gives two signs, so there are four points.

Answer: (c) {Geeta}

Explanation:
1. X ∩ Y contains students who are in both teams.
2. Only Geeta appears in both X and Y, so X ∩ Y = {Geeta}.

Teacher's Note:
1. Intersection answers the question ‘who belongs to both groups?’.
2. Union would answer ‘who belongs to at least one group?’.

Answer: (b) A ∩ B = B

Explanation:
1. Every element of B (2, 3, 5, 7) is in A, so B ⊂ A.
2. Hence A ∩ B = {2, 3, 5, 7} = B.

Teacher's Note:
1. If B ⊂ A, then A ∩ B = B always.
2. A ∩ B is always a subset of both A and B, so option (c) can never be true.

Answer: (d) 41

Explanation:
1. S = {3, 6, 9, …, 99}; since 99 = 3 × 33, n(S) = 33.
2. P = {2, 3, 5, 7, 11, 13, 17, 19}, so n(P) = 8.
3. n(S) + n(P) = 33 + 8 = 41.

Teacher's Note:
1. To count multiples of k up to N, divide the largest multiple by k.
2. Remember that 1 is not a prime number and 2 is the only even prime.

Answer: (a) Yes

Explanation:
1. No student can be both a boy and a girl.
2. So A ∩ B = φ, which means A and B are disjoint.

Teacher's Note:
1. Two sets are disjoint when they have no common element, i.e. A ∩ B = φ.
2. Disjoint sets are shown in a Venn diagram by two non-overlapping circles.

Answer: (b) φ

Explanation:
1. A contains only odd numbers and B contains only even numbers.
2. They have no common element, so A ∩ B = φ.

Teacher's Note:
1. An odd number and an even number can never be equal, so such sets are always disjoint.
2. Write φ (not {φ}) for the empty intersection.

Answer: (a) {3, 5, 7}

Explanation:
1. Compare each element of A with C: 3, 5, 7 are in both; 1 and 9 are not in C.
2. So A ∩ C = {3, 5, 7}.

Teacher's Note:
1. Go through the smaller set and keep only those elements that also appear in the other set.
2. The element 11 of C is not in A, so it is not in the intersection.

Answer: (a) A ∩ B = φ

Explanation:
1. Two sets are disjoint if they have no element in common.
2. This is written as A ∩ B = φ.

Teacher's Note:
1. Disjointness is about the intersection, not the union.
2. A ∪ B = φ happens only when both sets are empty.

Answer: (b) A ∩ C ≠ φ

Explanation:
1. Two sets are intersecting (overlapping) if they have at least one common element.
2. So A ∩ C ≠ φ; here A ∩ C = {3, 5, 7}.

Teacher's Note:
1. Intersecting sets are the opposite of disjoint sets.
2. In a Venn diagram, intersecting sets are drawn as overlapping circles.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For any object we can decide definitely whether it is a natural number less than 100, so the collection is well defined.
2. Hence A is true, and it is true because of the definition stated in R.
3. So both are true and R explains A.

Teacher's Note:
1. A collection is a set only if membership can be decided without any doubt.
2. Collections like ‘good students’ or ‘tall boys’ are not sets because they are not well defined.

Answer: (d) A is false, but R is true.

Explanation:
1. 60 = 2² × 3 × 5, so its prime divisors are 2, 3, 5; D = {2, 3, 5}. 1 and 4 are not prime, so A is false.
2. Letters of TRIGONOMETRY without repetition are T, R, I, G, O, N, M, E, Y, so R is true.
3. Hence A is false and R is true.

Teacher's Note:
1. 1 is neither prime nor composite, so it is never listed as a prime divisor.
2. While listing letters of a word, write each repeated letter only once.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The elements 1, 4, 9, …, 100 are 1², 2², …, 10², so A is true.
2. R is a true property of roster form.
3. R does not explain why the set-builder form in A is correct, so (b).

Teacher's Note:
1. To convert a roster to set-builder form, find the rule connecting each element with its position.
2. Order and repetition do not change a set, but they do not help in finding the rule.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Months with fewer than 31 days are February, April, June, September and November, so A is true.
2. Consonants before k are b, c, d, f, g, h, j (a, e, i are vowels), so R is true.
3. The two statements are about different sets, so R is not the explanation of A.

Teacher's Note:
1. Remember the months with 31 days: January, March, May, July, August, October, December.
2. When listing consonants, skip the vowels a, e, i, o, u.

Answer: (c) A is true, but R is false.

Explanation:
1. The only even prime is 2, so there is no even prime greater than 2. A is true.
2. x² = 4 gives x = 2 or −2, both even, so no odd x works and B = φ. R is false.
3. Hence A is true and R is false.

Teacher's Note:
1. To show a set is empty, show that no object can satisfy its condition.
2. Check every solution of the equation against the extra condition (here ‘x is odd’).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A has 6 elements, so it is finite; A is true.
2. The number of men living today is very large but countable, so B is finite; R is true.
3. R is about a different set and does not explain A.

Teacher's Note:
1. A set can be very large and still finite; what matters is that counting would come to an end.
2. Infinite sets, like N, have no last element.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. 101, 102, 103, … never end, so the set in A is infinite; A is true.
2. There are only 25 primes less than 99 (counting stops), so R is true.
3. R does not explain A.

Teacher's Note:
1. A condition with only a lower bound on integers usually gives an infinite set.
2. A condition like ‘less than 99’ on natural numbers or primes always gives a finite set.

Answer: (c) A is true, but R is false.

Explanation:
1. A = {2, 3, 5} and 30 = 2 × 3 × 5, so B = {2, 3, 5}. Hence A = B; A is true.
2. Q = {2, 2, 1, 3, 3} = {1, 2, 3}, since repeated elements are counted once, so P = Q. R is false.
3. Hence A is true and R is false.

Teacher's Note:
1. Repeating an element in roster form does not create a new element.
2. Two sets are equal when they have exactly the same elements, regardless of order or repetition.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 1, 2, 3 are all elements of B, so A ⊂ B; A is true.
2. R is the definition of a subset, and it is exactly why A holds.
3. Both are true and R explains A.

Teacher's Note:
1. To check A ⊂ B, test each element of A one by one for membership in B.
2. Every set is a subset of itself, and φ is a subset of every set.

Answer: (c) A is true, but R is false.

Explanation:
1. −4 is excluded and 6 is included, so the interval is (−4, 6]; A is true.
2. −12 < x < −10 excludes both ends, so it is (−12, −10), not [−12, −10]; R is false.
3. Hence A is true and R is false.

Teacher's Note:
1. Use a round bracket for < or > and a square bracket for ≤ or ≥.
2. [a, b] is a closed interval; (a, b) is an open interval.

Answer: (c) A is true, but R is false.

Explanation:
1. Every vowel is a letter of the English alphabet, so the alphabet can serve as the universal set; A is true.
2. No vowel is a consonant, so the set of vowels is not a subset of the set of consonants; R is false.
3. Hence A is true and R is false.

Teacher's Note:
1. A universal set must contain all the elements of every set under discussion.
2. The vowels and consonants are disjoint sets whose union is the whole alphabet.

Answer: (d) A is false, but R is true.

Explanation:
1. The diagram shows A and B as two separate (non-overlapping) circles, which represents disjoint sets.
2. If A ⊂ B, the circle of A must lie completely inside the circle of B, so A is false.
3. R is the definition of a subset and is true. Hence (d).

Teacher's Note:
1. In a Venn diagram, A ⊂ B is shown by drawing the circle A inside the circle B.
2. Separate circles mean the sets have no common element.

Answer: (d) A is false, but R is true.

Explanation:
1. Both a and b are in B, so A ⊂ B. The statement A ⊄ B is false.
2. If A ⊂ B, every element of A is already in B, so A ∪ B = B. R is true.
3. Hence A is false and R is true.

Teacher's Note:
1. The symbol ⊄ means ‘is not a subset of’.
2. When A ⊂ B, the union is B and the intersection is A.