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For a non-zero complex number z, let (z) denote the principal argument of z, with - < (z) . Let be the cube root of unity for which 0< ( )< . Let = ( n=12025(- )n ). Then the value of 3 is .
JEE Advanced 2025 Paper 2
Previous year questions with solutions · 2024 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 2 | 2 | 1 | 5.88% ↓37.3% |
| 2025 | 3 | 2 | 1.5 | 9.38% ↑218.7% |
| 2024 | 1 | 2 | 0.5 | 2.94% |
For a non-zero complex number z, let (z) denote the principal argument of z, with - < (z) . Let be the cube root of unity for which 0< ( )< . Let = ( n=12025(- )n ). Then the value of 3 is .
Let f(x)=x4+ax3+bx2+c be a polynomial with real coefficients such that f(1)=-9. Suppose that i 3 is a root of the equation 4x3+3ax2+2bx=0, where i= -1. If 1, 2, 3, and 4 are all the roots of the equation f(x)=0, then | 1|2+| 2|2+| 3|2+| 4|2 is equal to .