Class 11 Mathematics Relations and Functions MCQs Set 12

Answer: (c) {a, b}, {1, 3, 2}

Explanation:
1. First elements: a, b, so A = {a, b}.
2. Second elements: 1, 3, 2, so B = {1, 3, 2}.

Teacher's Note:
1. The order of listing in a set does not matter, so {1, 3, 2} = {1, 2, 3}.
2. Check: 2 × 3 = 6 pairs, as given.

Answer: (d) 12

Explanation:
1. n(A) = 3 and n(B) = 4
2. n(A × B) = n(A) × n(B) = 3 × 4 = 12

Teacher's Note:
1. Multiply, do not add, the numbers of elements.
2. 7 = 3 + 4 would be the largest possible size of A ∪ B, not of A × B.

Answer: (a) {1, 2, 3}, {3, 5}

Explanation:
1. First elements 1, 2, 3 are in A and second elements 3, 5 are in B.
2. Then n(A) ≥ 3 and n(B) ≥ 2, and n(A) × n(B) = 6 forces n(A) = 3, n(B) = 2.
3. So A = {1, 2, 3} and B = {3, 5}.

Teacher's Note:
1. Use n(A × B) = 6 to confirm that no other elements can exist.
2. The second elements decide B, so B cannot be {5} alone.

Answer: (b) (1, 5), (2, 3), (3, 5)

Explanation:
1. A = {1, 2, 3}, B = {3, 5}, so A × B = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)}.
2. Removing the three given pairs leaves (1, 5), (2, 3), (3, 5).

Teacher's Note:
1. Every pair of A × B must have its first element in A and second element in B, so pairs like (5, 1) are impossible.
2. List A × B completely before removing the known pairs.

Answer: (c) {a, b, 1, 2}

Explanation:
1. n(P × P) = n(P)² = 16 ⇒ n(P) = 4.
2. From (a, 1) and (b, 2), the elements a, b, 1, 2 all belong to P.
3. These are already 4 elements, so P = {a, b, 1, 2}.

Teacher's Note:
1. In P × P, both components of every pair come from P.
2. n(P × P) = n(P)² gives the exact size of P.

Answer: (a) 6, 0

Explanation:
1. a − 3 = 3 ⇒ a = 6
2. b + 7 = 7 ⇒ b = 0

Teacher's Note:
1. Equate corresponding components separately.
2. Substitute back to check: (6 − 3, 0 + 7) = (3, 7).

Answer: (c) −6, 8

Explanation:
1. x + 6 = 0 ⇒ x = −6
2. y − 2 = 6 ⇒ y = 8

Teacher's Note:
1. Watch the signs while transposing terms.
2. Substitute back: (−6 + 6, 8 − 2) = (0, 6).

Answer: (d) 3, −2

Explanation:
1. x + 2 = 5 ⇒ x = 3
2. 4 = 2x + y = 6 + y ⇒ y = −2

Teacher's Note:
1. Solve the equation with one unknown first, then substitute it into the other.
2. Check: (3 + 2, 4) = (5, 6 − 2) = (5, 4).

Answer: (b) A × B ≠ B × A

Explanation:
1. A ⊇ {1, 2, 3}, B ⊇ {4, 6}, and 3 × 2 = 6, so A = {1, 2, 3}, B = {4, 6}.
2. (1, 4) ∈ A × B but (1, 4) ∉ B × A, since 1 ∉ B.
3. So A × B ≠ B × A.

Teacher's Note:
1. A × B = B × A only when A = B (or one of the sets is empty).
2. Option (c) lists only 3 pairs, but A × B has 6.

Answer: (b) 17

Explanation:
1. n(A) × n(B) = 45, so n(A) must be a factor of 45.
2. Factors of 45 are 1, 3, 5, 9, 15, 45.
3. 17 is not a factor of 45, so n(A) cannot be 17.

Teacher's Note:
1. Both n(A) and n(B) are natural numbers whose product is n(A × B).
2. Listing the factors of the given product answers such questions instantly.

Answer: (c) R = {(0, 5), (1, 3), (2, 1)}

Explanation:
1. b = 5 − 2a must be a whole number, so a can be 0, 1, 2.
2. a = 0 ⇒ b = 5; a = 1 ⇒ b = 3; a = 2 ⇒ b = 1 (a = 3 gives b = −1, not allowed).
3. So R = {(0, 5), (1, 3), (2, 1)}.

Teacher's Note:
1. W = {0, 1, 2, …} includes 0, so start checking from a = 0.
2. Stop when b becomes negative.

Answer: (a) R = {(9, 3), (9, −3), (4, 2), (4, −2), (25, 5), (25, −5)}

Explanation:
1. Each arrow gives an ordered pair (start, end).
2. Arrows: 9 → ±3, 4 → ±2, 25 → ±5.
3. So R = {(9, 3), (9, −3), (4, 2), (4, −2), (25, 5), (25, −5)}.

Teacher's Note:
1. The number of ordered pairs equals the number of arrows (here 6).
2. Elements of Q with no arrow do not appear in R.

Answer: (a) R = {(x, y) : x ∈ P, y ∈ Q and x is the square of y}

Explanation:
1. In each pair, x = y²: 9 = (±3)², 4 = (±2)², 25 = (±5)².
2. So R = {(x, y) : x ∈ P, y ∈ Q and x is the square of y}.

Teacher's Note:
1. Test the rule on every pair, including the negative second elements.
2. ‘y is the square of x’ would need y = 81 for x = 9, which is not in Q.

Answer: (d) 16

Explanation:
1. n(A × B) = 2 × 2 = 4
2. Number of relations = number of subsets of A × B = 2⁴ = 16.

Teacher's Note:
1. The count 2^pq includes the empty relation and A × B itself.
2. 4 is the number of ordered pairs, not the number of relations.

Answer: (c) 2¹⁸

Explanation:
1. n(B) = 6, so n(A × B) = 3 × 6 = 18.
2. Number of relations = 2¹⁸.

Teacher's Note:
1. Count the elements of B carefully before multiplying.
2. Relations from A to B are exactly the subsets of A × B.

Answer: (b) Relation

Explanation:
1. The curve opens to the right, so a vertical line x = k (k > 0) cuts it at two points.
2. So one x has two images, which is not allowed for a function.
3. Hence the curve represents a relation but not a function.

Teacher's Note:
1. Vertical line test: a curve is the graph of a function if no vertical line meets it more than once.
2. Every function is a relation, but a relation is a function only if it passes the vertical line test.

Answer: (c) f(3)

Explanation:
1. f(1) = 1 + 2 + 3 = 6
2. f(2) = 4 + 4 + 3 = 11
3. f(3) = 9 + 6 + 3 = 18, which is the maximum.

Teacher's Note:
1. Substitute carefully and compare the numbers.
2. For x > 0, x² + 2x + 3 increases as x increases, so the largest input gives the largest value.

Answer: (a) h² − 2h + 2

Explanation:
1. Put 1 + x = 2 − h, i.e. x = 1 − h.
2. f(2 − h) = (1 − h)² + 1 = 1 − 2h + h² + 1 = h² − 2h + 2

Teacher's Note:
1. Match the input: choose x so that 1 + x equals the required argument.
2. Alternatively write f(t) = (t − 1)² + 1 and substitute t = 2 − h.

Answer: (c) \( \left[\frac{1}{3}, 1\right] \)

Explanation:
1. −1 ≤ sin 3x ≤ 1 ⇒ 1 ≤ 2 − sin 3x ≤ 3
2. Taking reciprocals (all positive) reverses the inequality: \( \frac{1}{3} \le \frac{1}{2 - \sin 3x} \le 1 \)
3. So the range is \( \left[\frac{1}{3}, 1\right] \).

Teacher's Note:
1. sin of any angle lies in [−1, 1], whatever the multiple of x.
2. Reciprocals of positive numbers reverse the order of an inequality.

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

Explanation:
1. 4x + 3 = 3x + 5 ⇒ x = 2 and y = −2, so A is true.
2. A × A must have 3 × 3 = 9 pairs, but only 6 are listed (for example (3, 3) and (4, 4) are missing), so R is false.
3. Hence A is true and R is false.

Teacher's Note:
1. n(A × A) = n(A)², so a set of 3 elements gives 9 pairs.
2. A × A always contains the pairs (a, a) for every a ∈ A.

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

Explanation:
1. First elements x, y, z belong to A and n(A) = 3, so A = {x, y, z}; second elements 1, 2 belong to B and n(B) = 2, so B = {1, 2}. A is true.
2. n(A × B) = 3 × 2 = 6, so R is true.
3. R gives the size of A × B, but A is found from the first and second elements; R does not explain A.

Teacher's Note:
1. First components of the pairs belong to A and second components belong to B.
2. Knowing n(A) and n(B) tells us when we have found all the elements.

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

Explanation:
1. n(A) = 2, n(B) = 2, so n(A × B) = 4.
2. By R, the number of relations = 2⁴ = 16, so A is true.
3. R is true and directly gives A.

Teacher's Note:
1. Every subset of A × B, including φ and A × B itself, is a relation from A to B.
2. Do not confuse 2^pq (relations) with pq (pairs in A × B).

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

Explanation:
1. In (6, 3), b = 3 is not divisible by a = 6, so the roster in A is wrong; A is false.
2. R = {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)}.
3. Both the first and the second elements cover 1, 2, 3, 4, 6, so R is true. Hence (d).

Teacher's Note:
1. ‘b is exactly divisible by a’ means a is a factor of b, so a ≤ b for positive numbers.
2. Since (a, a) ∈ R for every a, the domain and range are the whole set.

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

Explanation:
1. The arrows give (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), so R is true.
2. The rule is y = x², but x takes the values −2, −1, 0, which are not natural numbers.
3. So the condition x, y ∈ N is wrong (it should be x ∈ Z); A is false. Hence (d).

Teacher's Note:
1. In set-builder form, the stated set of x-values must include every first element shown.
2. Natural numbers start from 1, so 0 and negative numbers are not in N.

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

Explanation:
1. R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
2. Domain = {0, 1, 2, 3, 4, 5}, so A is true.
3. Range = {5, 6, 7, 8, 9, 10}, so R is false.

Teacher's Note:
1. Write the relation in roster form first; domain and range can then be read off directly.
2. The range is obtained by applying the rule x + 5 to each element of the domain.

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

Explanation:
1. x = 1, 2, …, 7, so R = {(3, 5), (4, 6), (5, 7), (6, 8), (7, 9), (8, 10), (9, 11)}.
2. Domain = {3, 4, 5, 6, 7, 8, 9}, so A is true.
3. Range = {5, 6, 7, 8, 9, 10, 11}, so R is false.

Teacher's Note:
1. Find the allowed values of x first (here 1 to 7), then compute both components.
2. The set {1, …, 7} is the set of x-values, not the range.

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

Explanation:
1. Each of p, q, r, s has exactly one arrow (p → 1, q → 4, r → 1, s → 4), so the diagram represents a function; A is true.
2. For x ≠ 2, \( f(x) = \frac{(x-2)(x+2)}{x-2} = x + 2 \), which differs from g(x) = x + 3; also the domains differ.
3. So f ≠ g and R is false.

Teacher's Note:
1. In an arrow diagram, a function needs exactly one arrow from every element of the first set; elements of the second set may receive none or many.
2. For equality of functions, both the rules and the domains must match.

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

Explanation:
1. For x > 0, −3x < 0, so 2 − 3x < 2. The range is (−∞, 2), not R; A is false.
2. x² ≥ 0 ⇒ x² + 2 ≥ 2, with every value ≥ 2 attained; the range is [2, ∞), so R is true.

Teacher's Note:
1. A restricted domain (such as x > 0) usually restricts the range too.
2. The minimum of x² + c is c, reached at x = 0.

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

Explanation:
1. R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}; every element 1, 2, 3, 5 of A appears as a first element. A is true.
2. |0| = 0, which is not positive; |x| is only non-negative. So R is false.
3. Hence A is true and R is false.

Teacher's Note:
1. |x − y| is odd exactly when one of x, y is odd and the other is even.
2. The correct statement is |x| ≥ 0 for all real x, with equality only at x = 0.

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

Explanation:
1. √(x − 1) is real only when x − 1 ≥ 0, i.e. x ≥ 1. So the domain is [1, ∞), not R − {1}; A is false.
2. √(x − 1) takes every value from 0 upwards, so the range is [0, ∞); R is true.

Teacher's Note:
1. For √(expression), solve expression ≥ 0 to get the domain.
2. A square root is never negative, so its range starts at 0.

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

Explanation:
1. \( [f(x)]^3 = \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) = f(x^3) + 3f(x) \)
2. Since \( f\left(\frac{1}{x}\right) = \frac{1}{x} + x = f(x) \), the statement in A is true.
3. f(a + b) = (a + b − a)²(a + b − b)² = b²a², which is not 0 in general; R is false.

Teacher's Note:
1. Use (p + q)³ = p³ + q³ + 3pq(p + q) with pq = x · (1/x) = 1.
2. For x + 1/x, replacing x by 1/x leaves the function unchanged.