Range of the function `f(x)=((x-2)^2)/((x-1)(x-3))` is

Options

A. `(1,oo)`
B. `(-oo,1)`
C. `R-(0,1]`
D. `(0,1]`

Explanation

To find the range of the function \( f(x) = \frac{(x-2)^2}{(x-1)(x-3)} \), we can follow these steps:

### Step 1: Set the function equal to \( y \)
We start by rewriting the function in terms of \( y \):
\[ y = \frac{(x-2)^2}{(x-1)(x-3)} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ y(x-1)(x-3) = (x-2)^2 \]
Expanding both sides, we have:
\[ y(x^2 - 4x + 3) = x^2 - 4x + 4 \]

### Step 3: Rearranging the equation
Rearranging this equation leads to:
\[ yx^2 - 4yx + 3y = x^2 - 4x + 4 \]
This can be rearranged to:
\[ (y - 1)x^2 + (-4y + 4)x + (3y - 4) = 0 \]

### Step 4: Identify coefficients for the quadratic equation
This is a quadratic equation in \( x \) with coefficients:
- \( A = y - 1 \)
- \( B = -4y + 4 \)
- \( C = 3y - 4 \)

### Step 5: Use the discriminant condition
For the quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = B^2 - 4AC \geq 0 \]
Substituting the coefficients:
\[ (-4y + 4)^2 - 4(y - 1)(3y - 4) \geq 0 \]

### Step 6: Simplify the discriminant
Calculating \( D \):
\[ (16y^2 - 32y + 16) - 4[(y - 1)(3y - 4)] \]
Expanding the second term:
\[ = 16y^2 - 32y + 16 - 4(3y^2 - 4y - 3y + 4) \]
\[ = 16y^2 - 32y + 16 - 12y^2 + 16y - 16 \]
Combining like terms:
\[ = 4y^2 - 16y + 0 \]
Factoring gives:
\[ = 4y(y - 4) \geq 0 \]

### Step 7: Solve the inequality
The inequality \( 4y(y - 4) \geq 0 \) implies:
- \( y = 0 \) or \( y = 4 \) are critical points.
- Testing intervals:
- For \( y < 0 \): \( 4y(y - 4) < 0 \) (negative)
- For \( 0 < y < 4 \): \( 4y(y - 4) < 0 \) (negative)
- For \( y > 4 \): \( 4y(y - 4) > 0 \) (positive)

### Step 8: Determine the range
Thus, the range of \( y \) is:
\[ y \in (-\infty, 0] \cup [4, \infty) \]
This means the function can take all real values except those in the interval \( (0, 4) \).

### Final Answer
The range of the function \( f(x) = \frac{(x-2)^2}{(x-1)(x-3)} \) is:
\[ (-\infty, 0] \cup [4, \infty) \]