Ques. 1JEE Main 2026, 22 January (Evening Shift)
Let S be the set of the first 11 natural numbers. Then the number of elements in A = \B S : n(B) 2 and the product of all elements of B is even\ is .
Previous year questions with solutions · 2025 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 2 | 19 | 0.11 | 0.42% ▼ 50% vs last year |
| 2025 | 4 | 19 | 0.21 | 0.84% |
Let S be the set of the first 11 natural numbers. Then the number of elements in A = \B S : n(B) 2 and the product of all elements of B is even\ is .
Let A = \x : |x2 - 10| 6\ and B = \x : |x - 2| > 1\. Then
Let A = \( , ) R R : | - 1| 4 and | - 5| 6\ and B = \( , ) R R : 16( - 2)2 + 9( - 6)2 144\. Then
Let A = \1, 6, 11, 16, \ and B = \9, 16, 23, 30, \ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A B) is
Let A=\(x,y) :|x+y| 3\ and B=\(x,y) :|x|+|y| 3\. If C=\(x,y) :x=0 or y=0\, then (x,y) |x+y| is :
Let A=\1,2,3, ,10\ and B= \ mn:m,n A, m<n and (m,n)=1 \. Then n(B) is equal to :