If x be real then find the range of the following rational expressions
(i) `y=(x^(2)+x+1)/(x^(2)+1)`
(ii) `y=(x^(2)-2x+9)/(x^(2)+2x+9)`

Answer

To find the range of the given rational expressions, we will analyze each expression step by step.

### Part (i): \( y = \frac{x^2 + x + 1}{x^2 + 1} \)

1. **Multiply both sides by the denominator**:
\[ y(x^2 + 1) = x^2 + x + 1 \]
Rearranging gives:
\[ yx^2 + y = x^2 + x + 1 \]
\[ (y - 1)x^2 - x + (y - 1) = 0 \]

2. **Identify coefficients**:
Here, \( a = y - 1 \), \( b = -1 \), and \( c = y - 1 \).

3. **Use the discriminant condition**:
For \( x \) to be real, the discriminant must be non-negative:
\[ b^2 - 4ac \geq 0 \]
Plugging in the coefficients:
\[ (-1)^2 - 4(y - 1)(y - 1) \geq 0 \]
Simplifying gives:
\[ 1 - 4(y - 1)^2 \geq 0 \]
\[ 4(y - 1)^2 \leq 1 \]
\[ (y - 1)^2 \leq \frac{1}{4} \]

4. **Taking square roots**:
\[ -\frac{1}{2} \leq y - 1 \leq \frac{1}{2} \]
Adding 1 to all parts:
\[ \frac{1}{2} \leq y \leq \frac{3}{2} \]

5. **Conclusion for part (i)**:
The range of \( y \) is:
\[ \left[\frac{1}{2}, \frac{3}{2}\right] \]

### Part (ii): \( y = \frac{x^2 - 2x + 9}{x^2 + 2x + 9} \)

1. **Multiply both sides by the denominator**:
\[ y(x^2 + 2x + 9) = x^2 - 2x + 9 \]
Rearranging gives:
\[ (y - 1)x^2 + (2y + 2)x + (9y - 9) = 0 \]

2. **Identify coefficients**:
Here, \( a = y - 1 \), \( b = 2y + 2 \), and \( c = 9y - 9 \).

3. **Use the discriminant condition**:
For \( x \) to be real, the discriminant must be non-negative:
\[ (2y + 2)^2 - 4(y - 1)(9y - 9) \geq 0 \]
Simplifying gives:
\[ 4(y + 1)^2 - 4(y - 1)(9(y - 1)) \geq 0 \]
\[ 4(y + 1)^2 - 36(y - 1)^2 \geq 0 \]
\[ 4[(y + 1)^2 - 9(y - 1)^2] \geq 0 \]

4. **Expand and simplify**:
\[ (y + 1)^2 - 9(y^2 - 2y + 1) \geq 0 \]
\[ y^2 + 2y + 1 - 9y^2 + 18y - 9 \geq 0 \]
\[ -8y^2 + 20y - 8 \geq 0 \]
Dividing by -4 (and reversing the inequality):
\[ 2y^2 - 5y + 2 \leq 0 \]

5. **Factoring the quadratic**:
\[ (2y - 1)(y - 2) \leq 0 \]

6. **Finding the intervals**:
The roots are \( y = \frac{1}{2} \) and \( y = 2 \). Testing intervals gives:
\[ \frac{1}{2} \leq y \leq 2 \]

7. **Conclusion for part (ii)**:
The range of \( y \) is:
\[ \left[\frac{1}{2}, 2\right] \]

### Final Answers:
- (i) The range of \( y = \frac{x^2 + x + 1}{x^2 + 1} \) is \( \left[\frac{1}{2}, \frac{3}{2}\right] \).
- (ii) The range of \( y = \frac{x^2 - 2x + 9}{x^2 + 2x + 9} \) is \( \left[\frac{1}{2}, 2\right] \).