1. **Problem statement:** Verify if there is a mistake in question 2 of the first part regarding the equation $g(x) = 0$ having a unique solution $\alpha$ such that $1.89 < \alpha < 1.90$. 2. **Recall the function:** $$g(x) = 2\ln x - 1 - \frac{1}{x^2}$$ 3. **Check the behavior of $g(x)$ near the interval:** - Evaluate $g(1.89)$: $$g(1.89) = 2\ln(1.89) - 1 - \frac{1}{(1.89)^2}$$ Calculate approximate values: $\ln(1.89) \approx 0.6386$ $$g(1.89) \approx 2 \times 0.6386 - 1 - \frac{1}{3.5721} = 1.2772 - 1 - 0.2799 = -0.0027$$ - Evaluate $g(1.90)$: $$g(1.90) = 2\ln(1.90) - 1 - \frac{1}{(1.90)^2}$$ $\ln(1.90) \approx 0.6419$ $$g(1.90) \approx 2 \times 0.6419 - 1 - \frac{1}{3.61} = 1.2838 - 1 - 0.2770 = 0.0068$$ 4. **Interpretation:** Since $g(1.89) < 0$ and $g(1.90) > 0$, by the Intermediate Value Theorem, there is a unique root $\alpha$ in $(1.89, 1.90)$. 5. **Conclusion:** There is no mistake in question 2; the interval for the unique solution $\alpha$ is correctly stated. **Final answer:** The statement in question 2 is correct and consistent with the function behavior.