Question

Using the discriminant method, find the range of $\frac{x^2-2x+1}{x^2-3x+2}$.

• A. $\mathbb{R} \setminus \{0, 1\}$
• B. $\mathbb{R} \setminus \{1\}$
• C. $\mathbb{R} \setminus \{0\}$
• D. $\mathbb{R}$

Answer 1

Answer: (A)

$\mathbb{R} \setminus \{0, 1\}$

Answer 2

Let $y = \frac{x^2-2x+1}{x^2-3x+2}$. Rearranging yields $(y-1)x^2 + (2-3y)x + (2y-1) = 0$.

Case 1: $y=1$. The equation becomes $-x+1=0$, so $x=1$. However, $x=1$ is not in the domain of the original function, so $y \neq 1$.

Case 2: $y \neq 1$. For real solutions of $x$, the discriminant $\Delta = (2-3y)^2 - 4(y-1)(2y-1) = y^2$ must be non-negative. This is always true ($y^2 \ge 0$).

The roots for $x$ are $x = \frac{-(2-3y) \pm \sqrt{y^2}}{2(y-1)} = \frac{3y-2 \pm y}{2(y-1)}$. The roots are $x_1 = \frac{2y-1}{y-1}$ and $x_2 = 1$. Since $x \neq 1$, $x_2$ is an extraneous solution.

The valid root $x_1 = \frac{2y-1}{y-1}$ must not be equal to the excluded values of the original function's domain, which are $x=1$ and $x=2$.

• $x_1 \neq 1 \implies \frac{2y-1}{y-1} \neq 1 \implies 2y-1 \neq y-1 \implies y \neq 0$.
• $x_1 \neq 2 \implies \frac{2y-1}{y-1} \neq 2 \implies 2y-1 \neq 2y-2 \implies -1 \neq -2$, which is always true.

Therefore, $y \neq 1$ and $y \neq 0$. The range is $\mathbb{R} \setminus \{0, 1\}$.