JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Since a, b, c R , the complex roots of the equation x^3+ax^2+bx+c=0 must occur in conjugate pairs. Given roots are = 1 and = 1+i 2 . The third root must be = 1-i 2 . The polynomial is given by: x^3+ax^2+bx+c = (x-1)(x-(1+i 2 ))(x-(1-i 2 )) = (x-1)((x-1)^2 - (i 2 )^2) = (x-1)(x^2 - 2x + 1 + 2) = (x-1)(x^2 - 2x + 3) = x^3 - 3x^2 + 5x - 3 We need to evaluate the integral: I = _ -1 ^ 1 (x^3 - 3x^2 + 5x - 3) dx Using the property of definite integrals _ -a ^ a f(x) dx = 0 for odd functions and 2 _ 0 ^ a f(x) dx for even functions: I = _ -1 ^ 1 x^3 dx - 3 _ -1 ^ 1 x^2 dx + 5 _ -1 ^ 1 x dx - 3 _ -1 ^ 1 1 dx I = 0 - 3(2) _ 0 ^ 1 x^2 dx + 0 - 3(2) _ 0 ^ 1 1 dx I = -6 [ x^3 3 ]_ 0 ^ 1 - 6 [ x ]_ 0 ^ 1 I = -6 ( 1 3 ) - 6(1) I = -2 - 6 = -8 Answer: -8