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Let f be a polynomial function such that 2(f(x))= ( 2 (2+ 23+ 29+ ) ) 3 (1+ f(x)f ( 1x ) ), x>0 and f(6)=37. Then n=110f(n) is equal to .
JEE Main 2026 (Online) 8th April Evening Shift
Previous year questions with solutions · 2025 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 20 | 19 | 1.1 | 4.21% ↓21.1% |
| 2025 | 16 | 12 | 1.3 | 5.33% |
Let f be a polynomial function such that 2(f(x))= ( 2 (2+ 23+ 29+ ) ) 3 (1+ f(x)f ( 1x ) ), x>0 and f(6)=37. Then n=110f(n) is equal to .
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Let A=\1, 2, 3, 4, 5, 6\. The number of one-one functions f : A A such that f(1) 3, f(3) 4 and f(2)+f(3)=5, is .
The number of relations on the set A = \1, 2, 3\, containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetric, is .