JEE Main 2026 — Mathematics Algebra

Since $a, b, c \in \mathbb{R}$a,b,c∈R, the complex roots of the equation $x^3+ax^2+bx+c=0$x3+ax2+bx+c=0 must occur in conjugate pairs.

Given roots are $\alpha = 1$α=1 and $\beta = 1+i\sqrt{2}$β=1+i2​.

The third root must be $\gamma = 1-i\sqrt{2}$γ=1−i2​.

The polynomial is given by:

$$x^3+ax^2+bx+c = (x-1)(x-(1+i\sqrt{2}))(x-(1-i\sqrt{2}))$$x3+ax2+bx+c=(x−1)(x−(1+i2​))(x−(1−i2​))

$$= (x-1)((x-1)^2 - (i\sqrt{2})^2)$$=(x−1)((x−1)2−(i2​)2)

$$= (x-1)(x^2 - 2x + 1 + 2)$$=(x−1)(x2−2x+1+2)

$$= (x-1)(x^2 - 2x + 3)$$=(x−1)(x2−2x+3)

$$= x^3 - 3x^2 + 5x - 3$$=x3−3x2+5x−3

We need to evaluate the integral:

$$I = \int_{-1}^{1} (x^3 - 3x^2 + 5x - 3) dx$$I=∫−11​(x3−3x2+5x−3)dx

Using the property of definite integrals $\displaystyle \int_{-a}^{a} f(x) dx = 0$∫−aa​f(x)dx=0 for odd functions and $\displaystyle 2\int_{0}^{a} f(x) dx$2∫0a​f(x)dx for even functions:

$$I = \int_{-1}^{1} x^3 dx - 3\int_{-1}^{1} x^2 dx + 5\int_{-1}^{1} x dx - 3\int_{-1}^{1} 1 dx$$I=∫−11​x3dx−3∫−11​x2dx+5∫−11​xdx−3∫−11​1dx

$$I = 0 - 3(2)\int_{0}^{1} x^2 dx + 0 - 3(2)\int_{0}^{1} 1 dx$$I=0−3(2)∫01​x2dx+0−3(2)∫01​1dx

$$I = -6 \left[ \dfrac{x^3}{3} \right]_{0}^{1} - 6 \left[ x \right]_{0}^{1}$$I=−6[3x3​]01​−6[x]01​

$$I = -6 \left( \dfrac{1}{3} \right) - 6(1)$$I=−6(31​)−6(1)

$$I = -2 - 6 = -8$$I=−2−6=−8

Answer: $-8$−8