JEE Main 2026 — Apr 4, Shift 2 — Mathematics solution

JEE Main2026Apr 4, Shift 2Mathematics

If $α = 1$ and $β = 1 + i 2$ , where $i = - 1$ , are two roots of the equation $x^{3} + a x^{2} + b x + c = 0$ , $a, b, c \in R$ , then $\int_{- 1}^{1} (x^{3} + a x^{2} + b x + c) d x$ is equal to:

A
$- 2$
B
$- 4$
C
$- 8$
D
$- 10$

Solution

Real-coefficient polynomial $\Rightarrow$ complex roots come in conjugate pairs. So the three roots are $1$ , $1 + i 2$ , and $1 - i 2$ .

By Vieta: sum of roots $= - a$ . Sum $= 1 + (1 + i 2) + (1 - i 2) = 3$ , so $a = - 3$ .

Product of roots $= - c$ : $1 \cdot (1 + i 2) (1 - i 2) = 1 \cdot (1 + 2) = 3$ , so $c = - 3$ .

Now integrate. By symmetry, odd-power terms vanish over $[- 1, 1]$ :

$I = \int_{- 1}^{1} (x^{3} - 3 x^{2} + b x - 3) d x = 2 \int_{0}^{1} (- 3 x^{2} - 3) d x = 2 [- x^{3} - 3 x]_{0}^{1} = 2 (- 1 - 3) = - 8$ .

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