Question

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

• A. \(-2\)
• B. \(-4\)
• C. \(-8\)
• D. \(-10\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Because the polynomial has real coefficients, complex roots must appear in conjugate pairs. Finding the third root allows us to deduce the polynomial's coefficients using Vieta's formulas. Finally, we integrate the polynomial over symmetric limits.
Step 2: Key Formula or Approach:
Complex conjugate root theorem: If \(1 + i\sqrt{2}\) is a root, so is \(\gamma = 1 - i\sqrt{2}\).
Vieta's formulas:
Sum of roots \(= -a\)
Product of roots taken two at a time \(= b\)
Product of all roots \(= -c\)
Step 3: Detailed Explanation:
The roots are \(1\), \(1 + i\sqrt{2}\), and \(1 - i\sqrt{2}\).
Find \(a\):
\(-a = 1 + (1 + i\sqrt{2}) + (1 - i\sqrt{2}) = 3 \implies a = -3\)
Find \(c\):
\(-c = (1)(1 + i\sqrt{2})(1 - i\sqrt{2}) = 1^2 - (i\sqrt{2})^2 = 1 - (-2) = 3 \implies c = -3\)
Find \(b\):
\(b = (1)(1+i\sqrt{2}) + (1)(1-i\sqrt{2}) + (1+i\sqrt{2})(1-i\sqrt{2}) = 2 + 3 = 5\)
The polynomial is \(P(x) = x^3 - 3x^2 + 5x - 3\).
Evaluate the definite integral:
\[ I = \int_{-1}^1 (x^3 - 3x^2 + 5x - 3) dx \] Using the symmetric limits property (\(\int_{-A}^A f(x) dx = 0\) for odd functions):
The odd terms \(x^3\) and \(5x\) integrate to 0.
\[ I = \int_{-1}^1 (-3x^2 - 3) dx = 2 \int_{0}^1 (-3x^2 - 3) dx \] \[ I = 2 \left[ -x^3 - 3x \right]_0^1 = 2(-1 - 3) = -8 \] Step 4: Final Answer:
The integral evaluates to \(-8\).