Mathematics

The key insight is that the denominator $(x-1)(x^2+bx+a)$ must have exactly one real root: $x=1$. This means the quadratic $x^2+bx+a$ cannot have any real roots, so its discriminant must be negative. Additionally, since $x=1$ appears in the denominator, the numerator must not also equal zero there—otherwise the factor would cancel and the domain would exclude fewer points. Set up these two conditions: $b^2-4a<0$ (no real roots in the quadratic) and $1+a+b\neq 0$ (numerator nonzero at $x=1$), then test the given pairs.

The key insight is that the denominator $(x-1)(x^2+bx+a)$ must have exactly one real root: $x=1$. This means the quadratic $x^2+bx+a$ cannot have any real roots, so its discriminant must be negative. Additionally, since $x=1$ appears in the denominator, the numerator must not also equal zero there—otherwise the factor would cancel and the domain would exclude fewer points. Set up these two conditions: $b^2-4a<0$ (no real roots in the quadratic) and $1+a+b\neq 0$ (numerator nonzero at $x=1$), then test the given pairs.