Maths Q4 – JEE-MAIN 2026

Home Maths Complex & QuadraticQuestion #74

α = 1

and

β = 1 + i \sqrt{2}

, where

i = \sqrt{- 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:

Clue 1 of 3: Properties of Polynomial Roots

For a polynomial with real coefficients, complex roots always appear in conjugate pairs.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify all roots and coefficients✦ Active

Since the polynomial $x^{3} + a x^{2} + b x + c = 0$ has real coefficients ( $a, b, c \in R$ ) and $1 + i \sqrt{2}$ is a root, its conjugate $1 - i \sqrt{2}$ must also be a root. The three roots are $r_{1} = 1$ , $r_{2} = 1 + i \sqrt{2}$ , and $r_{3} = 1 - i \sqrt{2}$ .

Using Vieta's formulas:

r_{1} + r_{2} + r_{3} = - a ⟹ 1 + (1 + i \sqrt{2}) + (1 - i \sqrt{2}) = 3 ⟹ - a = 3 ⟹ a = - 3

r_{1} r_{2} r_{3} = - c ⟹ 1 \cdot (1 + i \sqrt{2}) \cdot (1 - i \sqrt{2}) = 1 \cdot (1^{2} - (i \sqrt{2})^{2}) = 1 \cdot (1 - (- 2)) = 3 ⟹ - c = 3 ⟹ c = - 3

The polynomial is $x^{3} - 3 x^{2} + b x - 3 = 0$ .

Step 2: Evaluate the definite integral using properties of odd/even functions○ Expand

The integral to evaluate is $\int_{- 1}^{1} (x^{3} + a x^{2} + b x + c) d x$ . Substitute the values of $a = - 3$ and $c = - 3$ :

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

Separate the integrand into odd and even functions. For symmetric limits $[- L, L]$ , $\int_{- L}^{L} f (x) d x = 0$ if $f (x)$ is odd, and $\int_{- L}^{L} f (x) d x = 2 \int_{0}^{L} f (x) d x$ if $f (x)$ is even.

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

The term $(x^{3} + b x)$ is an odd function, so $\int_{- 1}^{1} (x^{3} + b x) d x = 0$ . The term $(- 3 x^{2} - 3)$ is an even function, so $\int_{- 1}^{1} (- 3 x^{2} - 3) d x = 2 \int_{0}^{1} (- 3 x^{2} - 3) d x$ .

💡 Teacher's Secret Hint

Remember that the coefficient 'b' does not affect the integral over symmetric limits for the odd term.

Step 3: Calculate the remaining integral○ Expand

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

= 2 {[- x^{3} - 3 x]}_{0}^{1}

= 2 [(- 1^{3} - 3 \cdot 1) - (0^{3} - 3 \cdot 0)]

= 2 [(- 1 - 3) - 0]

= 2 [- 4]

= - 8

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Maths Question 4 – JEE-MAIN 2026

Clue 1 of 3: Properties of Polynomial Roots

🧠 Full Solution Path