Question:

Step 1: Evaluate the discriminant $D$. The problem begins by evaluating a discriminant $D$ given by the expression $(2n-1)^2 - 4n(n-1)$. The calculation provided in the solution is: $$D = (2n-1)^2 - 4n(n-1)$$ Expanding the terms as presented: $$D = (4n^2 - 4n + 1) - (4n^2 - 4n + 1)$$ $$D = 0$$ Step 2: Analyze the condition for the product of roots. The problem then states a condition that the product of roots is less than zero. The expression for the product of roots is given as $\frac{n}{n(n-1)}$. So, we establish the inequality: $$\frac{n}{n(n-1)} < 0$$ For this expression to be defined, the denominator $n(n-1)$ cannot be zero, which means $n \neq 0$ and $n \neq 1$. Since $n \neq 0$, we can simplify the fraction by cancelling $n$ from the numerator and the denominator: $$\frac{1}{n-1} < 0$$ For a fraction with a positive numerator ($1 > 0$) to be negative, its denominator must be negative. Therefore: $$n-1 < 0$$ $$n < 1$$ Step 3: Determine the range of $n$ that satisfies the conditions. From Step 2, the condition on the product of roots, $\frac{n}{n(n-1)} < 0$, simplifies to $n < 1$ (with the exclusion of $n=0$). More rigorously, the inequality $\frac{n}{n(n-1)} < 0$ implies that $n(n-1)$ must be negative. The product $n(n-1)$ is negative if and only if $n$ and $n-1$ have opposite signs. There are two cases: 1. $n > 0$ and $n-1 < 0$: This implies $n > 0$ and $n < 1$, which combines to $0 < n < 1$. 2. $n < 0$ and $n-1 > 0$: This implies $n < 0$ and $n > 1$, which is an impossible condition. Therefore, the only range for $n$ that satisfies the condition $n(n-1) < 0$ is $0 < n < 1$. The given question also states "If $0 < n < 1$", which is consistent with the derived condition. Hence, the conditions imply that $n$ must be in the interval $(0,1)$. The final answer is $n \in (0,1)$. The correct option is 1.