Sets, Relations and Functions: JEE Advanced PYQ, 2007–2026

Answer: (D)

log₂[x(x − 2)] = 3 gives x² − 2x = 8, so (x − 4)(x + 2) = 0. Both logarithms need x > 2, so x = 4.

Answer: (A)

Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B) to get n(A ∩ B) = 6. The elements in exactly one set are those in the union but not in the intersection, so n(A Δ B) = n(A ∪ B) − n(A ∩ B) = 15 − 6 = 9. Equivalently, n(A − B) + n(B − A) = 6 + 3 = 9.

Answer: (A)

For x ≥ 0, x² + p is strictly increasing with range [p, ∞). For x < 0, qx + 3 is one-one if q ≠ 0, and the range is (−∞, 3) if q > 0 or (3, ∞) if q < 0. If q < 0, the ranges (3, ∞) and [p, ∞) always overlap, so f is not one-one. If q > 0, the ranges (−∞, 3) and [p, ∞) must be disjoint for one-one, which needs p ≥ 3, and must together cover ℝ for onto, which needs p ≤ 3. So p = 3 and q > 0. Because the left piece is open at 3, the value 3 is attained only at x = 0 and is not repeated.

Answer: (A)

De Morgan's laws give A′ ∪ B′ = (A ∩ B)′ and A′ ∩ B′ = (A ∪ B)′. So n(A ∩ B) = 80 − 55 = 25. Then n(A ∪ B) = 50 + 45 − 25 = 70. Therefore n(A′ ∩ B′) = 80 − 70 = 10. As a check, n(A′) + n(B′) − n(A′ ∪ B′) = 30 + 35 − 55 = 10.

Answer: (D)

The condition f(f(x)) = f(x) says that f(y) = y for every y in the range of f. So if the image is a set S with k elements, f fixes each element of S, and each of the 3 − k elements outside S can be sent to any element of S. This gives kᵏ⁻ᵏ… precisely k^(3−k) functions for each choice of S. For k = 1, there are C(3,1) · 1² = 3 functions (the constants). For k = 2, there are C(3,2) · 2¹ = 6 functions. For k = 3, there is C(3,3) · 3⁰ = 1 function (the identity). The total is 3 + 6 + 1 = 10.

Answer: (A)

Split the elements by remainder on division by 3: {3, 6} (remainder 0), {1, 4} (remainder 1) and {2, 5} (remainder 2). The two multiples of 3 do not change the remainder of the sum, so they give 2² = 4 free choices. Suppose i elements come from {1, 4} and j from {2, 5}. The sum has remainder i + 2j ≡ i − j (mod 3). Since 0 ≤ i, j ≤ 2, this is 0 only when i = j. The number of ways is C(2,0)² + C(2,1)² + C(2,2)² = 1 + 4 + 1 = 6. The total is 4 × 6 = 24.

Answer: (A)

For x ∈ [n, n+1) with n an integer, [x] = n, so f(x) = x + n takes each value in [2n, 2n+1) exactly once. Distinct integers n give disjoint intervals that are ordered, so f is strictly increasing and therefore one-one. The range is the union of all [2n, 2n+1), which misses every value in [2n+1, 2n+2). For example, f(x) = 1 has no solution, because the interval for n = 0 reaches only values below 1 and the one for n = 1 starts at 2. So f is not onto.

Answer: (A)

A continuous cubic is one-one on ℝ exactly when it is monotonic. Here f′(x) = 3x² + 6x + k = 3(x + 1)² + (k − 3). This is ≥ 0 for all x iff k − 3 ≥ 0. When k = 3, f′ = 3(x + 1)² is zero only at x = −1, so f is still strictly increasing. For k < 3, f′ is negative on an interval and f is not one-one. So k ≥ 3.

Answer: (A)

Let a_n be the number of such subsets of {1,...,n}. If n is not in the subset, the rest is any valid subset of {1,...,n−1}, giving a_{n−1} choices. If n is in the subset, then n−1 is not, and the rest is a valid subset of {1,...,n−2}, giving a_{n−2} choices. So a_n = a_{n−1} + a_{n−2}, with a_0 = 1 (only the empty set) and a_1 = 2. The sequence is 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, so a_10 = 144.

Answer: (C)

Reflexive needs (1,1), (2,2) and (3,3). (3,3) is absent, so R is not reflexive. Symmetric: (1,2) has (2,1), and the diagonal pairs are their own reverses, so R is symmetric. Transitive: the chains (1,2),(2,1) give (1,1); (2,1),(1,2) give (2,2); (1,2),(2,2) give (1,2); and so on. All the resulting pairs are in R, so R is transitive. Hence R is symmetric and transitive but not reflexive.