Question

If \(z_1,z_2,z_3\) are roots of \[ x^3+ax^2+bx+c=0 \] Let \(z_1=1,\; z_2=1+i\sqrt2\) and \(a,b,c\in\mathbb{R}\). Then the value of \[ \int_{-1}^{1}(x^3+ax^2+bx+c)\,dx \] is

• A. \(-8\)
• B. \(8\)
• C. \(6\)
• D. \(-4\)

Hint:

For symmetric limits \([-a,a]\), integrals of odd functions vanish.

Answer 1

The Correct Option is A

Concept: If polynomial coefficients are real, complex roots occur in conjugate pairs.
Step 1: Roots Given \[ z_1=1,\quad z_2=1+i\sqrt2 \] Thus \[ z_3=1-i\sqrt2 \]
Step 2: Find coefficients Sum of roots: \[ z_1+z_2+z_3=-a \] \[ 1+(1+i\sqrt2)+(1-i\sqrt2)=3 \] \[ a=-3 \] Product of roots: \[ z_1z_2z_3=-c \] \[ 1\cdot[(1+i\sqrt2)(1-i\sqrt2)] \] \[ =1(1+2)=3 \] \[ c=-3 \] From pair multiplication: \[ (1+i\sqrt2)(1-i\sqrt2)=3 \] thus \[ c=-4 \]
Step 3: Evaluate integral \[ \int_{-1}^{1}(x^3+ax^2+bx+c)dx \] Odd powers vanish over symmetric limits: \[ \int_{-1}^{1}x^3dx=0 \] \[ \int_{-1}^{1}bx\,dx=0 \] Thus \[ \int_{-1}^{1}(ax^2+c)dx \] \[ =2\int_{0}^{1}(ax^2+c)dx \] \[ =2\left[\frac{ax^3}{3}+cx\right]_0^1 \] \[ =2\left(\frac{-3}{3}-4\right) \] \[ =2(-1-4) \] \[ =-8 \]