If x `in` R, then the range of `(x)/(x^(2)-5x + 9)` is

Options

A. 1
B. 2
C. 3
D. 0

Answer

To find the range of the function \( f(x) = \frac{x}{x^2 - 5x + 9} \), we will follow these steps:

### Step 1: Set up the equation
Let \( y = \frac{x}{x^2 - 5x + 9} \). Rearranging gives us:
\[ yx^2 - 5yx + 9y - x = 0 \]
This can be rewritten as:
\[ yx^2 - (5y + 1)x + 9y = 0 \]

### Step 2: Apply the discriminant condition
For the quadratic equation to have real roots, the discriminant must be greater than or equal to zero. The discriminant \( D \) of the quadratic \( ax^2 + bx + c = 0 \) is given by \( D = b^2 - 4ac \). Here, \( a = y \), \( b = -(5y + 1) \), and \( c = 9y \).

Calculating the discriminant:
\[ D = (-(5y + 1))^2 - 4(y)(9y) \]
\[ D = (5y + 1)^2 - 36y^2 \]

### Step 3: Simplify the discriminant
Expanding \( (5y + 1)^2 \):
\[ D = 25y^2 + 10y + 1 - 36y^2 \]
\[ D = -11y^2 + 10y + 1 \]

### Step 4: Set the discriminant greater than or equal to zero
We need:
\[ -11y^2 + 10y + 1 \geq 0 \]

### Step 5: Solve the quadratic inequality
To find the roots of the equation \( -11y^2 + 10y + 1 = 0 \), we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-10 \pm \sqrt{(10)^2 - 4(-11)(1)}}{2(-11)} \]
\[ y = \frac{-10 \pm \sqrt{100 + 44}}{-22} \]
\[ y = \frac{-10 \pm \sqrt{144}}{-22} \]
\[ y = \frac{-10 \pm 12}{-22} \]

Calculating the two roots:
1. \( y_1 = \frac{2}{-22} = -\frac{1}{11} \)
2. \( y_2 = \frac{-22}{-22} = 1 \)

### Step 6: Determine the range
The quadratic \( -11y^2 + 10y + 1 \) opens downwards (since the coefficient of \( y^2 \) is negative). Therefore, it is greater than or equal to zero between its roots:
\[ -\frac{1}{11} \leq y \leq 1 \]

### Conclusion
Thus, the range of the function \( f(x) = \frac{x}{x^2 - 5x + 9} \) is:
\[ \boxed{\left[-\frac{1}{11}, 1\right]} \]