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

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Definite Integrals

α = 1

and

β = 1 + i 2

, where

i = - 1

are two roots of the equation

x^{3} + ax^{2} + bx + c = 0, a, b, c \in R

, then

\int_{- 1}^{1} (x^{3} + a x^{2} + b x + c) d x

is equal to :

Answer: (C)

With real coefficients, the non-real root

1 + i 2

is accompanied by its conjugate

1 - i 2

, so the three roots are

1, 1 + i 2, 1 - i 2

. Their sum is

3

, so

a = - 3

; the product of the two conjugate roots is

(1)^{2} + (2)^{2} = 3

, so the product of all three is

3

and

c = - 3

. Then

\int_{- 1}^{1} (x^{3} - 3 x^{2} + b x - 3) d x

: the odd terms

x^{3}

and

b x

integrate to

0

over

[- 1, 1]

, leaving

\int_{- 1}^{1} (- 3 x^{2} - 3) d x = 2 \int_{0}^{1} (- 3 x^{2} - 3) d x = 2 [- x^{3} - 3 x]_{0}^{1} = 2 (- 4) = - 8

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option