Range of the function `f(x)=(x^2-3x+2)/(x^2+x-6)` is

Options

A. `R-[1//5,1] `
B. `R`
C. `R-{1}`
D. none of these

Explanation

To find the range of the function \( f(x) = \frac{x^2 - 3x + 2}{x^2 + x - 6} \), we will follow these steps:

### Step 1: Factor the numerator and denominator
First, we need to factor both the numerator and the denominator.

**Numerator:**
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

**Denominator:**
\[ x^2 + x - 6 = (x - 2)(x + 3) \]

Thus, we can rewrite the function as:
\[ f(x) = \frac{(x - 1)(x - 2)}{(x - 2)(x + 3)} \]

### Step 2: Simplify the function
We can cancel the common factor \( (x - 2) \) from the numerator and denominator (as long as \( x \neq 2 \)):
\[ f(x) = \frac{x - 1}{x + 3}, \quad x \neq 2 \]

### Step 3: Set \( f(x) = y \)
To find the range, we set \( f(x) = y \):
\[ y = \frac{x - 1}{x + 3} \]

### Step 4: Cross-multiply to solve for \( x \)
Cross-multiplying gives:
\[ y(x + 3) = x - 1 \]
Expanding this:
\[ yx + 3y = x - 1 \]

### Step 5: Rearrange to isolate \( x \)
Rearranging the equation:
\[ yx - x = -1 - 3y \]
Factoring out \( x \):
\[ x(y - 1) = -1 - 3y \]
Thus, we can express \( x \) as:
\[ x = \frac{-1 - 3y}{y - 1} \]

### Step 6: Identify restrictions for \( y \)
For \( x \) to be defined, the denominator \( y - 1 \) must not be zero:
\[ y - 1 \neq 0 \implies y \neq 1 \]

### Step 7: Determine the range
The function \( f(x) = \frac{x - 1}{x + 3} \) can take all real values except \( y = 1 \). Therefore, the range of the function is:
\[ \text{Range of } f(x) = \mathbb{R} \setminus \{1\} \]

### Final Answer
The range of the function \( f(x) = \frac{x^2 - 3x + 2}{x^2 + x - 6} \) is all real numbers except \( 1 \).

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