Sets, Relations and Functions: JEE Main PYQ, 2020–2025

Answer: (A)

Test each property. Reflexive: every triangle has the same angles as itself, so T R T. Symmetric: if T₁ has the same angles as T₂, then T₂ has the same angles as T₁. Transitive: if T₁ and T₂ have equal angles, and T₂ and T₃ have equal angles, then T₁ and T₃ do too. R satisfies all three, so it is an equivalence relation.

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)

A function is one-one if f(x₁) = f(x₂) implies x₁ = x₂. It is onto if every element of the codomain has a preimage in the domain. One-one check: 2x₁ = 2x₂ gives x₁ = x₂, so f is one-one. Onto check: for f(x) = 1 we need x = 1/2, which is not an integer, so 1 has no preimage in ℤ. The range is the set of even integers, which is a proper subset of the codomain ℤ. Hence f is one-one but not onto.

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: (B)

A set does not count repeated elements. MATHEMATICS has 11 letters, and M, A and T each appear twice. The distinct letters are M, A, T, H, E, I, C, S, so there are 11 − 3 = 8 elements.

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: (C)

A has 2⁶ = 64 subsets. Those with no even number are the subsets of {1, 3, 5}, and there are 2³ = 8 of them. So 64 − 8 = 56 subsets contain at least one even number.