Correct Option: C

Step-by-step Solution:

1. Solve the quadratic inequality for $x$: We are given the inequality $2x^2 - 3x + 1 Factor the quadratic expression: $(2x - 1)(x - 1) The roots of the quadratic equation $2x^2 - 3x + 1 = 0$ are $x = \frac{1}{2}$ and $x = 1$. Since the parabola opens upwards (coefficient of $x^2$ is positive), the inequality $(2x - 1)(x - 1) Thus, the range of $x$ is $\frac{1}{2} 2. Express $x$ in terms of $y$: We are given $y = \frac{x+1}{x-1}$. To find $x$ in terms of $y$, multiply both sides by $(x-1)$: $y(x-1) = x+1$. $yx - y = x + 1$. Rearrange to isolate $x$: $yx - x = y + 1$. Factor out $x$: $x(y - 1) = y + 1$. So, $x = \frac{y+1}{y-1}$. Note that $y \neq 1$, otherwise $1=0$ which is impossible.3. Substitute $x$ back into the inequality for $x$: We have $\frac{1}{2} $\frac{1}{2} This can be into two separate inequalities: a) $\frac{y+1}{y-1} > \frac{1}{2}$ b) $\frac{y+1}{y-1} 4. Solve inequality (a): $\frac{y+1}{y-1} - \frac{1}{2} > 0$ $\frac{2(y+1) - (y-1)}{2(y-1)} > 0$ $\frac{2y + 2 - y + 1}{2(y-1)} > 0$ $\frac{y+3}{2(y-1)} > 0$ For this inequality to hold, the numerator and denominator must have the same sign. This means $(y+3)(y-1) > 0$. The critical points are $y = -3$ and $y = 1$. Therefore, $y 1$.5. Solve inequality (b): $\frac{y+1}{y-1} $\frac{y+1}{y-1} - 1 $\frac{(y+1) - (y-1)}{y-1} $\frac{2}{y-1} For this inequality to hold, the denominator must be negative (since the numerator is positive). So, $y-1 6. Combine the solutions: We need to find the values of $y$ that satisfy both conditions: ($y 1$) AND ($y Let's analyze the intersection: If $y If $y > 1$, then $y$ is not less than $1$. So $y > 1$ does not satisfy both. Therefore, the combined solution is $y In interval notation, this is $(-\infty, -3)$.The final answer is $\boxed{\text{C}}$