JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q4

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If $\alpha=1$ and $\beta=1+i\sqrt2$ , where $i=\sqrt{-1}$ are two roots of the equation $x^3+ax^2+bx+c=0$ , $a,b,c\in\mathbb R$ , then $\displaystyle\int_{-1}^1\left(x^3+ax^2+bx+c\right)dx$ is equal to:

-2

-4

-8

-10

Answer and worked solution

Official answer: C

Cubic. Real coefficients put the conjugate $1-i\sqrt2$ among the roots too:
$(x-1)(x^2-2x+3)=x^3-3x^2+5x-3.$

Integral. On $[-1,1]$ the odd terms vanish:
$\int_{-1}^1(-3x^2-3)\,dx=-2-6=-8.$

(C)

Topics:Theory of equations (higher degree)Properties of definite integrals

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JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 4