Previous year questions: Relations and Functions

Relations and Functions28 min readLesson 5 of 5

Q1 (CBSE Board 2023, 2 marks) The map f(x) = 2x takes A = {1, 2, 3, 4} onto a set B and is stated to be bijective. Work out the set B. Answer: Apply the rule to each element of A: the images are 2, 4, 6 and 8, all distinct, so injectivity is visible. Since the map is surjective, B cannot contain anything else, hence B = {2, 4, 6, 8}. Writing an extra element in B would break surjectivity, and reading the images in order confirms that each input of A has exactly one image.

Q2 (CBSE Board 2024, 3 marks) On the set {1, 2, 3, 4, 5} a relation collects the pairs (x, y) with |x² - y²| < 8. Test the three defining properties of an equivalence relation. Answer: Reflexive: put y = x, then |x² - x²| = 0 < 8, so every diagonal pair qualifies. Symmetric: the expression |x² - y²| is symmetric in x and y, so a pair qualifies exactly when its reverse does. Transitive: the pairs (2, 3) and (3, 4) qualify because |4 - 9| = 5 and |9 - 16| = 7, but (2, 4) fails because |4 - 16| = 12. The relation is reflexive and symmetric, yet transitivity breaks. Note that a single failing triple is enough to reject transitivity, so no exhaustive check is needed. The relation is also not universal on the set, because pairs such as (1, 5) are missing, so the pair list stays a proper subset of the product.

Q3 (CBSE Board 2026, 3 marks) Take A = R - {3} and B = R - {1}. The rule f(x) = (x - 2)/(x - 3) maps A into B. Decide whether this rule is injective and surjective. Answer: Injectivity: from (a - 2)/(a - 3) = (b - 2)/(b - 3), cross multiplication gives (a - 2)(b - 3) = (b - 2)(a - 3). Expanding, ab - 3a - 2b + 6 = ab - 3b - 2a + 6, so -3a - 2b = -3b - 2a and hence a = b. Surjectivity: solve y = (x - 2)/(x - 3). Then y(x - 3) = x - 2, so x(y - 1) = 3y - 2 and x = (3y - 2)/(y - 1). For y in B the denominator is non-zero, and x = 3 is impossible because 3y - 2 = 3y - 3 has no solution. So x lies in A and the rule is onto B. The hole at x = 3 in the domain and the hole at y = 1 in the codomain are exactly what make the two conditions work.

Q4 (CBSE SQP 2021-22, 1 mark) On A = {1, 2, 3}, let R = {(1, 1), (1, 2), (2, 2), (3, 3)}. Deleting exactly one ordered pair turns R into an equivalence relation on A. Name that pair. Answer: The diagonal pairs (1, 1), (2, 2) and (3, 3) already make R reflexive. The pair (1, 2) lacks its reverse (2, 1), so symmetry fails. Deleting (1, 2) leaves only diagonal pairs, and that relation passes all three tests. The pair to delete is (1, 2). After the deletion the relation contains only diagonal pairs, which are symmetric because each pair equals its reverse.

Q5 (CBSE Board 2023, 5 marks) On N \times N, put (a, b) R (c, d) exactly when ad(b + c) = bc(a + d). Prove that R is an equivalence relation. Answer: Reflexive: for (a, b) the statement becomes ab(b + a) = ba(a + b), which is an identity, so the pair is related to itself. Symmetric: start from ad(b + c) = bc(a + d). Rewriting the two products as ab(d - c) = cd(b - a) shows the condition is unchanged when the two pairs trade places. Transitive: assume (a, b) R (c, d) and (c, d) R (e, f). Dividing the first relation by acbd and the second by cedf gives 1/b - 1/a = 1/d - 1/c and 1/d - 1/c = 1/f - 1/e. Adding those equations yields 1/b - 1/a = 1/f - 1/e. Write it as (a - b)/(ab) = (e - f)/(ef). Cross multiplying gives ef(a - b) = ab(e - f), so aef - bef = abe - abf, which rearranges to af(b + e) = be(a + f). That is exactly the condition for (a, b) R (e, f), so transitivity holds.

Q6 (CBSE Board 2025, 5 marks) Let A be the set of positive integers and define R on A \times A by (a, b) R (c, d) when ad = bc. Show that R is an equivalence relation. Answer: Reflexive: ab = ba for every pair, so each (a, b) is related to itself. Symmetric: if ad = bc then cb = da, which is the reverse condition. Transitive: assume ad = bc and cf = de. Multiplying the first equation by f and the second by b gives adf = bcf and bcf = bde, hence adf = bde. Since d is a positive integer, cancel d to get af = be, which is exactly (a, b) R (e, f). All three conditions hold. The reflexive step is ab = ba, the symmetric step swaps the two products, and the transitive step cancels the positive integer d before comparing the outer products.

Q7 (CBSE Board 2024, 5 marks) For f(x) = 2x/(1 + x²) on the real line, show that the map is neither injective nor surjective, then give a codomain that makes it surjective. Answer: Take the two inputs 2 and 1/2. Both give the same output, because f(2) = 4/5 and f(1/2) = 2(1/2)/(1 + 1/4) = 1/(5/4) = 4/5. Two different inputs share one output, so injectivity fails. For surjectivity, note that the output always stays inside [-1, 1]. Trying to hit 3 is impossible: 3 = 2x/(1 + x²) leads to 3x² - 2x + 3 = 0 with discriminant 4 - 36 < 0. So the range is [-1, 1] and the map misses every number outside it. Taking A = [-1, 1] as the codomain makes the same rule surjective.

Q8 (CBSE Board 2026, 5 marks) Fix a prime p and define R on Z by x R y when p divides |x - y|. Determine whether R is reflexive, symmetric and transitive. Answer: Reflexive: |x - x| = 0 and p divides 0 because 0 = p times 0, so every integer is related to itself. Symmetric: |x - y| and |y - x| are equal, so the condition is mutual. Transitive: suppose p divides |x - y| and p divides |y - z|. Then p divides the sum (x - y) + (y - z) = x - z, so p divides |x - z| and x R z. R is an equivalence relation. The transitive step works because the two differences add to x - z, and a divisor of both numbers divides their sum.

Q9 (CBSE Board 2023, 5 marks) The rule f(x) = 16 - x² is considered on the domain [-4, 4] with codomain [0, 4]. Show surjectivity and failure of injectivity, then locate every a with f(a) = 7. Answer: Surjectivity: pick y in [0, 4] and set x = square root of (16 - y). Since y lies between 0 and 4, 16 - y lies between 12 and 16, so x lies between 2 and 4, inside the domain. The rule sends this x to y. Injectivity fails because f(2) = f(-2) = 12, two different inputs sharing one output. Finally 16 - a² = 7 gives a² = 9, so a = 3 or a = -3. The same test works whenever a quadratic equation has a negative discriminant, which is the quickest way to show that a value is outside the range.

Q10 (CBSE Board 2026, 5 marks) On the non-negative reals, f(x) = 4x² + 4x - 5 maps into [-5, infinity). Prove that this map is bijective onto that interval and find p at least 0 with f(p) = 3. Answer: Write 4x² + 4x = (2x + 1)² - 1, so f(x) = (2x + 1)² - 6. Injectivity: if (2a + 1)² = (2b + 1)² with a and b non-negative, both bracketed numbers are positive, so 2a + 1 = 2b + 1 and a = b. Surjectivity: solve y = 4x² + 4x - 5 for x at least 0. The quadratic formula gives x = (-1 + square root of (y + 6))/2, which is non-negative because y is at least -5. So every y in [-5, infinity) is attained. For f(p) = 3, solve p² + p - 2 = 0, giving p = 1 or p = -2, and p at least 0 selects p = 1. A quick check gives f(1) = 4 + 4 - 5 = 3, which confirms the value.