If `x` is real, then `x//(x^2-5x+9)` lies between `-1a n d-1//11` b. `1a n d-1//11` c. `1a n d1//11` d. none of these

Options

A. `-1 and -1//11`
B. `1 and -1//11`
C. `1 and 1//11`
D. none of these

Explanation

To solve the problem, we need to determine the range of the expression \( y = \frac{x}{x^2 - 5x + 9} \) for real values of \( x \).

### 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: Identify the coefficients
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = y \)
- \( b = -(5y + 1) \)
- \( c = 9y \)

### Step 3: Apply the discriminant condition
For \( x \) to be real, the discriminant \( D \) must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Calculating the discriminant:
\[ D = (-(5y + 1))^2 - 4(y)(9y) = (5y + 1)^2 - 36y^2 \]
Expanding this gives:
\[ D = 25y^2 + 10y + 1 - 36y^2 = -11y^2 + 10y + 1 \]

### Step 4: Set up the inequality
We need to solve the inequality:
\[ -11y^2 + 10y + 1 \geq 0 \]
Rearranging gives:
\[ 11y^2 - 10y - 1 \leq 0 \]

### Step 5: Find the roots of the quadratic
To find the roots of \( 11y^2 - 10y - 1 = 0 \), we use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 11 \cdot (-1)}}{2 \cdot 11} \]
Calculating the discriminant:
\[ D = 100 + 44 = 144 \]
Thus, the roots are:
\[ y = \frac{10 \pm 12}{22} \]
Calculating the two roots:
1. \( y_1 = \frac{22}{22} = 1 \)
2. \( y_2 = \frac{-2}{22} = -\frac{1}{11} \)

### Step 6: Determine the intervals
The quadratic \( 11y^2 - 10y - 1 \) opens upwards (since the coefficient of \( y^2 \) is positive). Therefore, the expression is less than or equal to zero between the roots:
\[ -\frac{1}{11} \leq y \leq 1 \]

### Conclusion
Thus, the range of \( y \) is:
\[ y \in \left[-\frac{1}{11}, 1\right] \]
This corresponds to option (b): \( 1 \) and \( -\frac{1}{11} \).