Previous year questions: Relations and Functions

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Six items below come from the official JEE Main 2026 papers of the National Testing Agency and four come from the NCERT Exemplar for Class 11 Mathematics. The heading shows the source and the marks. Every question was matched against the official PDF page and every final value against the official answer key before it was rewritten here. Attempt each one first; the worked answer follows the same language as the class notes.

Q1 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

The function f maps the interval from 1 to infinity into itself by f(x) equal to (x minus 1) to the fourth power plus 1. Consider two statements. Statement I: the set of x with f(x) equal to the inverse of f at x contains exactly two elements. Statement II: the set of x with f(x) equal to the inverse of f at x plus 1 is empty. Decide which statement holds.

Answer: The function is strictly increasing on its domain and its range is the same interval, so it has an inverse, and the inverse rule is 1 plus the fourth root of (x minus 1). For statement I, put t equal to the fourth root of (x minus 1), so x is t to the fourth plus 1. The equation f(x) equal to the inverse at x becomes t to the sixteenth plus 1 equal to t plus 1, that is t to the sixteenth equal to t. So t is 0 or 1, giving x equal to 1 or 2. Exactly two elements appear, so statement I is true. For statement II, the equation becomes (x minus 1) to the fourth equal to the fourth root of x. At x equal to 1 the left side is 0 while the right side is 1, and at x equal to 2 the left side is 1 while the right side is about 1.19; at x equal to 3 the left side is 16 and the right side is about 1.32. By continuity there is a solution between 2 and 3, so the set is not empty and statement II is false. Only statement I is true.

Q2 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

A function f from the reals to the reals satisfies f(x + y) equal to f(x) plus 2y squared plus y plus axy for every real x and y, where a is a constant. Also f(0) equals minus 1 and f(1) equals 2. Find the sum from n equal to 1 to 5 of (a plus f(n)).

Answer: Set x equal to 0 in the rule. Then f(y) equals f(0) plus 2y squared plus y, that is 2y squared plus y minus 1. This formula must be consistent with the general rule for every x and y. Expand the left side using the formula: f(x + y) is 2(x + y) squared plus (x + y) minus 1, which is 2x squared plus 4xy plus 2y squared plus x plus y minus 1. The right side is 2x squared plus x minus 1 plus 2y squared plus y plus axy. Comparing the cross terms gives a equal to 4. The required sum is 5 times 4 plus the sum of f(n) from 1 to 5. Now f(n) is 2n squared plus n minus 1, and summing gives 2 times 55 plus 15 minus 5, that is 120. Adding the 5 times 4 gives 140, which is the official answer.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

On the set A containing the integers from 1 to 6, count the one-one functions from A to A that satisfy f(1) at least 3, f(3) at most 4, and f(2) plus f(3) equal to 5.

Answer: Since f(3) is at most 4 and f(2) is 5 minus f(3), the second value must stay inside A and must differ from f(3) because the function is one-one. Test the four possible values of f(3). If f(3) is 1, then f(2) is 4 and f(1) can be any of 3, 5 or 6, which is three choices; the remaining three domain values fill the remaining three images in 6 ways, giving 18. The same count appears for f(3) equal to 2, 3 and 4: in each case f(2) is fixed, f(1) has three allowable values and the rest give 6 arrangements, so each case contributes 18. The total is 4 times 18, that is 72. The official answer is 72.

Q4 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

The function f from the reals to the reals is given by (2x squared minus 3x plus 2) divided by (3x squared plus x plus 3). Classify it as one-one and onto, or otherwise.

Answer: First check one-one. Differentiate the quotient. The derivative numerator simplifies to 11(x squared minus 1) times a positive denominator factor, so the derivative vanishes at x equal to 1 and x equal to minus 1 and changes sign at both points. The function rises, then falls, then rises, so two different inputs give the same output and the function is not one-one.

Now check onto by finding the range. Write y equal to f(x) and rearrange into a quadratic in x: (3y minus 2)x squared plus (y plus 3)x plus (3y minus 2) equal to 0. For a real x the discriminant must be non-negative, and that condition is (y plus 3) squared at least 4(3y minus 2) squared. This reduces to (7 minus 5y)(7y minus 1) at least 0, so y lies between 1 by 7 and 7 by 5. The range is a closed interval around zero and does not cover the reals, so the function is not onto either. The correct description is neither one-one nor onto, matching the official answer.

Q5 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

On the set of ordered pairs of natural numbers, take R as the set of pairs (x, y) with the natural logarithm of (x + y) at most 2. Find the minimum number of ordered pairs that must be added to R so that it becomes transitive.

Answer: Since the logarithm is at most 2 exactly when x + y is at most e squared, and e squared is about 7.39, the relation contains the pairs of positive integers with sum at most 7. That is 21 pairs. A relation is transitive when (a, b) and (b, c) in the relation force (a, c) to be present. Take (a, b) and (b, c) with a + b at most 7 and b + c at most 7. Then a + c equals (a + b) + (b + c) minus 2b, which is at most 12 minus 2b. A missing pair (a, c) can therefore be forced only when a + c lies between 8 and 12 while both entries are at most 6. Conversely, for any such pair the value b equal to 7 minus the larger of a and c works, because then a + b and b + c are both at most 7. So the closure keeps the original 21 pairs and adds every pair of numbers from 1 to 6 whose sum is 8 or more. There are 36 pairs from 1 to 6 in all, and 21 have sum at most 7, so 15 remain. The answer is 15.

Q6 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Count the functions from the set 1, 2, 3, 4 to the set a, b, c that are not onto.

Answer: Each of the four inputs can choose any of the three outputs, so there are 3 to the fourth power, that is 81, functions in total. An onto function must use all three outputs, and the count by inclusion and exclusion is 3 to the fourth minus 3 times 2 to the fourth plus 3 times 1 to the fourth, because we subtract the functions missing one output and add back those missing two. That gives 81 minus 48 plus 3, which is 36 onto functions. The number that are not onto is therefore 81 minus 36 equal to 45. The official answer is 45.

Q7 (NCERT Exemplar, 3 marks)

With A equal to the set containing minus 1, 2 and 3, and B equal to the set containing 1 and 3, write down the four products A cross B, B cross A, B cross B and A cross A.

Answer: Pair every element of the first set with every element of the second. A cross B contains (minus 1, 1), (minus 1, 3), (2, 1), (2, 3), (3, 1) and (3, 3). B cross A contains (1, minus 1), (1, 2), (1, 3), (3, minus 1), (3, 2) and (3, 3). B cross B contains (1, 1), (1, 3), (3, 1) and (3, 3). A cross A contains nine pairs, starting with (minus 1, minus 1) and ending with (3, 3). The sizes are 6, 6, 4 and 9, which follow the rule that the product size is the product of set sizes. Source: NCERT Exemplar Class 11 Mathematics, Chapter 2, Exercise 2.3, Short Answer Type, Q1.

Q8 (NCERT Exemplar, 3 marks)

Let R be the relation from the whole numbers to the whole numbers given by pairs (x, y) with x squared plus y squared equal to 25. Find its domain and range.

Answer: List the whole number pairs whose squares add to 25. The options are (0, 5), (5, 0), (3, 4) and (4, 3), because 9 plus 16 is 25. So R contains exactly those four ordered pairs. The domain collects the first entries, giving the set 0, 3, 4 and 5, and the range collects the second entries, giving the same set. The symmetry is expected because the equation treats x and y alike. Source: NCERT Exemplar Class 11 Mathematics, Chapter 2, Exercise 2.3, Short Answer Type, Q6.

Q9 (NCERT Exemplar, 3 marks)

A relation R1 consists of the pairs (x, y) with y equal to 2x plus 7, where x is real and lies between minus 5 and 5. Find its domain and range.

Answer: The domain is the set of allowed first entries, so it is the closed interval from minus 5 to 5. The rule is increasing, so the smallest output comes from the smallest input and the largest from the largest. At x equal to minus 5 the value is minus 3, and at x equal to 5 it is 17. The range is therefore the closed interval from minus 3 to 17. Both endpoints are included because the domain endpoints are included. Source: NCERT Exemplar Class 11 Mathematics, Chapter 2, Exercise 2.3, Short Answer Type, Q7.

Q10 (NCERT Exemplar, 3 marks)

Decide whether g equal to the set of pairs (1, 1), (2, 3), (3, 5), (4, 7) is a function, and if it is described by g(x) equal to ax plus b, find a and b.

Answer: The first entries are 1, 2, 3 and 4, all different, so every input has exactly one output and g is a function. To find the rule, substitute two pairs. From (1, 1) we get a plus b equal to 1, and from (2, 3) we get 2a plus b equal to 3. Subtracting gives a equal to 2, and then b equals minus 1. Check the remaining pairs: 2 times 3 minus 1 is 5 and 2 times 4 minus 1 is 7, so the rule fits all four pairs. The values are a equal to 2 and b equal to minus 1. Source: NCERT Exemplar Class 11 Mathematics, Chapter 2, Exercise 2.3, Long Answer Type, Q16.