Class 11 Mathematics Relations and Functions MCQs Set 11

Answer: (a) (p, q), p ∈ P and q ∈ Q

Explanation:
1. An ordered pair is written with round brackets, as (p, q), with first element from P and second from Q.
2. [p, q] denotes an interval and {p, q} denotes a set, in which order does not matter.
3. So only (p, q) is an ordered pair.

Teacher's Note:
1. In an ordered pair the order matters: (p, q) ≠ (q, p) unless p = q.
2. In a set the order does not matter: {p, q} = {q, p}.

Answer: (d) 5, −2

Explanation:
1. Equal ordered pairs have equal first elements and equal second elements.
2. 2a − 5 = 5 ⇒ a = 5
3. 4 = b + 6 ⇒ b = −2. Hence a = 5, b = −2.

Teacher's Note:
1. Compare first with first and second with second; never cross-compare.
2. Check: (2 × 5 − 5, 4) = (5, 4) and (5, −2 + 6) = (5, 4).

Answer: (d) {(a₁, b₁), (a₁, b₂), (a₁, b₃), (a₁, b₄), (a₂, b₁), (a₂, b₂), (a₂, b₃), (a₂, b₄)}

Explanation:
1. A × B = {(a, b) : a ∈ A, b ∈ B}, so each element of A is paired with every element of B.
2. a₁ pairs with b₁, b₂, b₃, b₄ and a₂ pairs with b₁, b₂, b₃, b₄.
3. So A × B has 2 × 4 = 8 ordered pairs, which is option (d).

Teacher's Note:
1. n(A × B) = n(A) × n(B); use this to reject options with the wrong number of pairs.
2. The first element of every pair in A × B must come from A.

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

Explanation:
1. A pair (x, y) is in both A × B and B × A only if x and y are both in A ∩ B.
2. A ∩ B = {1, 2}
3. So (A × B) ∩ (B × A) = {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}.

Teacher's Note:
1. The result (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B) saves a lot of listing.
2. Check any option pair: (6, 1) is not in B × A because 6 ∉ B.

Answer: (c) A = {a, b} and B = {1, 2, 3}

Explanation:
1. A is the set of first elements: {a, b}.
2. B is the set of second elements: {1, 2, 3}.
3. Check: 2 × 3 = 6 pairs, which matches.

Teacher's Note:
1. First components give A and second components give B.
2. Always confirm with n(A × B) = n(A) × n(B).

Answer: (d) 2^mn − 1

Explanation:
1. A relation from A to B is any subset of A × B, and n(A × B) = mn.
2. A set with mn elements has 2^mn subsets.
3. Leaving out the empty relation, the number of non-empty relations is 2^mn − 1.

Teacher's Note:
1. Number of relations from A to B = 2^{n(A) × n(B)}, including the empty relation.
2. Read the question carefully: ‘non-empty’ means subtract 1.

Answer: (c) 12

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

Teacher's Note:
1. The number of elements in a Cartesian product is the product, not the sum, of the sizes.
2. 64 = 4³ would be the number of functions from B to A, which is a different idea.

Answer: (a) Arrows drawn from each of 1, 3, 6 (in A) to both x and y (in B)

Explanation:
1. A × B = {(1, x), (1, y), (3, x), (3, y), (6, x), (6, y)}
2. Each pair (a, b) is shown by an arrow from a in A to b in B.
3. So every element of A has arrows to both x and y, which is option (a).

Teacher's Note:
1. In an arrow diagram of A × B, arrows always start in A and end in B.
2. The number of arrows equals n(A × B) = 3 × 2 = 6.

Answer: (a) R₁ only

Explanation:
1. A relation from A to B is a subset of A × B, so each pair must have first element in A and second in B.
2. R₁: all first elements 1, 2, 3 ∈ A and second elements 5, 7, 8 ∈ B, so R₁ is a relation.
3. R₂ contains (5, 2) and R₃ contains (6, 2); 5, 6 ∉ A and 2 ∉ B, so they are not relations from A to B.

Teacher's Note:
1. Check every pair; one wrong pair is enough to reject a relation.
2. A pair (b, a) with b ∈ B, a ∈ A belongs to B × A, not A × B.

Answer: (c) {(9, −3), (9, 3), (4, −2), (4, 2), (25, −5), (25, 5)}

Explanation:
1. Read each arrow as an ordered pair (start, end).
2. 9 → 3, 9 → −3, 4 → 2, 4 → −2, 25 → 5, 25 → −5
3. So R = {(9, −3), (9, 3), (4, −2), (4, 2), (25, −5), (25, 5)}.

Teacher's Note:
1. Count the arrows first; here 6 arrows means R has 6 pairs.
2. Elements of Q with no arrow (such as 1) do not appear in R.

Answer: (c) {(x, y) : x is the square of y, x ∈ P, y ∈ Q}

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

Teacher's Note:
1. To write set-builder form, find one rule that links every first element to its second element.
2. Check the direction: x = y² here, not y = x².

Answer: (c) {0, ±3, ±4, ±5}

Explanation:
1. Integer solutions of a² + b² = 25: (0, ±5), (±5, 0), (±3, ±4), (±4, ±3).
2. The first elements a are 0, ±3, ±4, ±5.
3. So domain (R) = {0, ±3, ±4, ±5}.

Teacher's Note:
1. Since the relation is on Z, negative values must be included as well.
2. The domain is the set of all first elements of the ordered pairs in R.

Answer: (c) {1, 2, 6}

Explanation:
1. R = {(1, 1), (1, 2), (1, 6), (2, 2), (2, 6), (6, 6)}
2. The second elements are 1, 2, 6.
3. So range of R = {1, 2, 6}.

Teacher's Note:
1. Every number divides itself, so (a, a) ∈ R for each a; this alone puts every element in the range.
2. The range is the set of second elements of the pairs.

Answer: (a) {1, 2, 3, 4} and {1, 4, 9, 16, 25}

Explanation:
1. R = {(1, 1), (2, 4), (3, 9), (4, 16)}
2. Domain = set of first elements = {1, 2, 3, 4}.
3. Codomain = the whole set B = {1, 4, 9, 16, 25}.

Teacher's Note:
1. The codomain of a relation from A to B is always B, while the range is only the images actually used.
2. Here the range is {1, 4, 9, 16}, which is a proper subset of the codomain.

Answer: (a) Yes

Explanation:
1. The first elements 3, 4, 5, 6, 7 are all different.
2. So each first element has exactly one image, and the relation is a function.

Teacher's Note:
1. A relation fails to be a function only when some first element is repeated with different second elements.
2. Repeated second elements are allowed in a function.

Answer: (c) R₁ and R₃ are functions

Explanation:
1. R₁: first elements 2, 3, 4 are distinct, so R₁ is a function.
2. R₂: 2 is paired with both 2 and 4, so R₂ is not a function.
3. R₃: first elements 1 to 6 are distinct, so R₃ is a function.

Teacher's Note:
1. Look only at the first elements: any repetition with a different image rules out a function.
2. In R₁, two elements having the same image 1 is perfectly allowed.

Answer: (b) [−a, a]

Explanation:
1. The square root is real only when a² − x² ≥ 0.
2. x² ≤ a² ⇒ −a ≤ x ≤ a
3. So the domain is [−a, a].

Teacher's Note:
1. The expression under a square root must be non-negative, so the end points where it is 0 are included.
2. Use square brackets for end points that are included.

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

Explanation:
1. −1 ≤ cos x ≤ 1 ⇒ −2 ≤ −2cos x ≤ 2 ⇒ −1 ≤ 1 − 2cos x ≤ 3
2. When 1 − 2cos x is in (0, 3], its reciprocal is in \( \left[\frac{1}{3}, \infty\right) \).
3. When 1 − 2cos x is in [−1, 0), its reciprocal is in (−∞, −1]. So the range is \( (-\infty, -1] \cup \left[\frac{1}{3}, \infty\right) \).

Teacher's Note:
1. When the denominator can take both negative and positive values, take reciprocals of the two parts separately.
2. The value 0 of the denominator is excluded, which is why the range splits into two pieces.

Answer: (c) not equal (domains are not same)

Explanation:
1. For x ≠ 3, f(x) = (x − 3)(x + 3)/(x − 3) = x + 3, so the rules agree.
2. But the domain of f is R − {3} while the domain of g is R.
3. Two functions are equal only if their domains are equal, so f ≠ g.

Teacher's Note:
1. Equal functions need the same domain, the same codomain and the same rule.
2. Cancelling a factor such as (x − 3) does not remove the restriction x ≠ 3.

Answer: (c) Domain = R − {4}, Range = {−1}

Explanation:
1. f is not defined when x − 4 = 0, i.e. x = 4. So domain = R − {4}.
2. For x ≠ 4, \( f(x) = \frac{-(x - 4)}{x - 4} = -1 \).
3. So the range is {−1}.

Teacher's Note:
1. Always remove the zeros of the denominator from the domain first.
2. If the function simplifies to a constant, the range is a single value.

Answer: (b) Domain = R, Range = (−∞, 2]

Explanation:
1. f(x) is defined for every real x, so domain = R.
2. |x − 5| ≥ 0 ⇒ 2 − |x − 5| ≤ 2, with equality at x = 5.
3. So range = (−∞, 2].

Teacher's Note:
1. The modulus is never negative, so subtracting it can only lower the value.
2. Include the end point 2 because it is actually reached at x = 5.

Answer: (d) None of these

Explanation:
1. For x ≥ 0, x − |x| = x − x = 0, and division by √0 is not defined.
2. For x < 0, x − |x| = x + x = 2x < 0, and its square root is not real.
3. So f is not defined for any real x; the domain is the empty set, i.e. none of the given options.

Teacher's Note:
1. For a square root in the denominator, the expression inside must be strictly positive.
2. Splitting into the cases x ≥ 0 and x < 0 is the standard way to handle |x|.

Answer: (c) \( \left[-\frac{1}{2}, \frac{1}{2}\right] \)

Explanation:
1. Let y = x/(1 + x²). Then yx² − x + y = 0.
2. For real x (when y ≠ 0) the discriminant must be ≥ 0: 1 − 4y² ≥ 0 ⇒ −1/2 ≤ y ≤ 1/2. Also y = 0 at x = 0.
3. So the range is \( \left[-\frac{1}{2}, \frac{1}{2}\right] \); the extreme values occur at x = ±1.

Teacher's Note:
1. Writing y = f(x) and asking for which y the equation has a real solution x is a reliable way to find a range.
2. Check the end points: f(1) = 1/2 and f(−1) = −1/2.

Answer: (a) the straight line y = x passing through the origin

Explanation:
1. The identity function is f(x) = x for every real x.
2. Its graph is the set of points (x, x), i.e. the line y = x through the origin making 45° with the x-axis.

Teacher's Note:
1. A line parallel to the x-axis is the graph of a constant function, not the identity function.
2. A vertical line is not the graph of any function, since one x would have many images.

Answer: (b) constant function

Explanation:
1. The graph is a horizontal line parallel to the x-axis.
2. So every x has the same image, i.e. f(x) = c for all x. This is a constant function.

Teacher's Note:
1. The graph of a constant function f(x) = c is a horizontal line at height c.
2. Its range is the single value {c}.

Answer: (a) a V-shaped graph with its vertex at (2, 0), opening upwards

Explanation:
1. |x − 2| ≥ 0 and equals 0 only at x = 2.
2. f(x) = x − 2 for x ≥ 2 and f(x) = 2 − x for x < 2, two straight lines meeting at (2, 0).
3. So the graph is a V with vertex (2, 0).

Teacher's Note:
1. The graph of |x − a| is the graph of |x| shifted a units to the right.
2. The vertex is the point where the expression inside the modulus becomes zero.

Answer: (c) a set consisting of two elements

Explanation:
1. For x > 0, |x| = x, so f(x) = 1.
2. For x < 0, |x| = −x, so f(x) = −1.
3. So the range is {−1, 1}, a set with two elements.

Teacher's Note:
1. x/|x| is the signum function for x ≠ 0.
2. The signum function also takes the value 0 at x = 0, but here x = 0 is excluded.

Answer: (b) Greatest integer function

Explanation:
1. The graph is made of horizontal steps: the value stays at an integer n for n ≤ x < n + 1 and jumps at each integer.
2. Each step has a filled dot at its left end and an open dot at its right end.
3. This is the graph of the greatest integer function f(x) = [x].

Teacher's Note:
1. [x] is the greatest integer less than or equal to x, e.g. [2.7] = 2 and [−1.2] = −2.
2. The signum function has only three values (−1, 0, 1), while [x] takes every integer value.

Answer: (c) \( \frac{1}{2} \)

Explanation:
1. x = 1/2 lies in 0 ≤ x < 1, so f(x) = x is used.
2. \( f\left(\frac{1}{2}\right) = \frac{1}{2} \)

Teacher's Note:
1. For a piecewise function, first decide which interval the input belongs to.
2. Using the wrong piece (x² or 1/x) gives the distractor answers 1/4 and 2.

Answer: (a) 0

Explanation:
1. \( f\left(\frac{1}{x}\right) = \frac{1}{x^3} - x^3 \)
2. \( f(x) + f\left(\frac{1}{x}\right) = x^3 - \frac{1}{x^3} + \frac{1}{x^3} - x^3 = 0 \)

Teacher's Note:
1. To find f(1/x), replace every x in the rule by 1/x.
2. Functions with f(1/x) = −f(x) always give this sum as 0.

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

Explanation:
1. By R, x + 1 = 3 and y − 2 = 1.
2. So x = 2 and y = 3; A is true.
3. A follows directly from R, so R is the correct explanation.

Teacher's Note:
1. Equality of ordered pairs gives two separate equations.
2. Always substitute back: (2 + 1, 3 − 2) = (3, 1).

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

Explanation:
1. A is the definition of the Cartesian product, so it is true.
2. Pairing red and blue with each of b, c, s gives exactly the 6 pairs listed, so R is true.
3. R is only an example of the definition, not a reason for it, so (b).

Teacher's Note:
1. An example that agrees with a definition does not explain the definition.
2. Check n(A × B) = 2 × 3 = 6 to confirm the listing.