Mathematics · Definite Integration

JEE Main 2026 — 4 April, Evening Shift — Question 28

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

Option A:
-2
Option B:
-4
Option C:
-8
Correct
Option D:
-10

Answer: C

Step-by-step solution

Roots are $1 \pm \mathrm{i} \sqrt{2}$ & $1$ $-\mathrm{a}=$ sum of roots $\mathrm{a}=-3$ $-\mathrm{c}=1(1+\sqrt{2} \mathrm{i})(1-\mathrm{i} \sqrt{2})$ $\mathrm{c}=-3$ $I=\int_{-1}^{1}\left(x^{3}-3 x^{2}+b x-3\right) d x$ $=2 \int_{0}^{1}\left(-3 x^{2}-3\right) d x$ $=2\left(-x^{3}-3 x\right)_{0}^{1}=-8$

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Definite Integration
Topic: Introduction to Definite Integration

Answer: C

Step-by-step solution

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