JEE Main 2023 · 11 Apr 2023 Shift 2

Let A = {1, 3, 4, 6, 9} and B = {2, 4, 5, 8, 10}. Let R be a relation defined on A × B such that R = {((a₁, b₁), (a₂, b₂)) : a₁ ≤ b₂ and b₁ ≤ a₂}. Then the number of elements in the set R is

1. A26
2. B52
3. C160
4. D180

The domain of the function f(x) = 1/√([x]² − 3[x] − 10) is (where [x] denotes the greatest integer less than or equal to x)

1. A(−∞, −2) ∪ [6, ∞)
2. B(−∞, −3] ∪ [6, ∞)
3. C(−∞, −2) ∪ (5, ∞)
4. D(−∞, −3] ∪ (5, ∞)

For a ∈ C, let A = {z ∈ C : Re(a + z̄) > Im(ā + z)} and B = {z ∈ C : Re(a + z̄) < Im(ā + z)}. Then among the two statements: (S1) : If Re(a), Im(a) > 0, then the set A contains all the real numbers (S2) : If Re(a), Im(a) < 0, then the set B contains all the real numbers,

1. Aboth are true
2. Bboth are false
3. Conly (S1) is true
4. Donly (S2) is true

If the system of linear equations 7x + 11y + αz = 13 5x + 4y + 7z = β 175x + 194y + 57z = 361 has infinitely many solutions, then α + β + 2 is equal to:

1. A3
2. B4
3. C5
4. D6

If |(x + 1 x x; x x + λ x; x x x + λ²)| = 9/8(103x + 81), then λ, λ/3 are the roots of the equation

1. A4x² + 24x − 27 = 0
2. B4x² − 24x − 27 = 0
3. C4x² + 24x + 27 = 0
4. D4x² − 24x + 27 = 0

If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial numbers, then the serial number of the word THAMS is

1. A101
2. B102
3. C103
4. D104

The sum of the coefficients of three consecutive terms in the binomial expansion of (1 + x)ⁿ⁺², which are in the ratio 1:3:5, is equal to

1. A25
2. B41
3. C63
4. D92