If $-3 \leq \frac{x^2 + bx + 1}{x^2 - bx + 1} \leq 2$ for all $x \in \mathbb{R}$, then the value of $b$ belongs to
Step-by-Step Solution
Key Concept: Transform the rational inequality into polynomial inequalities by multiplying by the positive denominator, then use discriminant conditions for the resulting quadratics.
Since $x^2 + x + 1 > 0$ for all $x \in \mathbb{R}$, the inequality is equivalent to $-3(x^2 + x + 1) < x^2 + \lambda x - 2 < 5(x^2 + x + 1)$. This gives two conditions: $4x^2 + (3+\lambda)x + 1 > 0$ and $(\lambda-5)x^2 + (\lambda-5)x - 7 < 0$. For the first condition, the discriminant must be negative: $(3+\lambda)^2 - 16 < 0$, giving $\lambda \in (-7,-1)$. For the second condition (which requires $\lambda < 5$), we need $(\lambda-5)^2 - 4(\lambda-5)(-7) < 0$, which simplifies to $(\lambda-3)(\lambda+1) < 0$, giving $\lambda \in (-1,3)$. The intersection is $\lambda \in (-1,2)$ (also checking that $\lambda = 2$ works and $\lambda = -1$ is excluded).
Correct Answer: 3