The function `f:R rarr R` defined as `f(x)=(x^(2)-x+1)/(x^(2)+x+1)` is

Options

A. injective as well as sujective
B. injective but not surjective
C. surjective but not injective
D. neither injective nor surjective

Answer

To determine the nature of the function \( f(x) = \frac{x^2 - x + 1}{x^2 + x + 1} \), we need to analyze its injectivity and surjectivity.

### Step 1: Determine the Range of the Function

To find the range of the function, we set \( y = f(x) \):

\[ y = \frac{x^2 - x + 1}{x^2 + x + 1} \]

Multiplying both sides by the denominator to eliminate the fraction gives:

\[ y(x^2 + x + 1) = x^2 - x + 1 \]

Rearranging this, we get:

\[ yx^2 + yx + y = x^2 - x + 1 \]

This simplifies to:

\[ (y - 1)x^2 + (y + 1)x + (y - 1) = 0 \]

### Step 2: Analyze the Quadratic Equation

For \( f(x) \) to be defined for real values of \( x \), the discriminant of this quadratic equation must be non-negative:

\[ D = (y + 1)^2 - 4(y - 1)(1) \geq 0 \]

Calculating the discriminant:

\[ D = (y + 1)^2 - 4(y - 1) \]
\[ = y^2 + 2y + 1 - 4y + 4 \]
\[ = y^2 - 2y + 5 \]

### Step 3: Set the Discriminant Greater than or Equal to Zero

Now we need to find when this discriminant is greater than or equal to zero:

\[ y^2 - 2y + 5 \geq 0 \]

The discriminant of this quadratic is:

\[ (-2)^2 - 4 \cdot 1 \cdot 5 = 4 - 20 = -16 \]

Since the discriminant is negative, the quadratic \( y^2 - 2y + 5 \) is always positive. Therefore, there are no real solutions for \( y \) that make the discriminant zero.

### Step 4: Determine the Range

Since the discriminant is always positive, we can conclude that the function \( f(x) \) can take values in the interval:

\[ \left[\frac{1}{3}, 3\right] \]

### Step 5: Check Surjectivity

The function is surjective if its range covers all real numbers. Since the range of \( f(x) \) is \( \left[\frac{1}{3}, 3\right] \) and not all real numbers, the function is **not surjective**.

### Step 6: Check Injectivity

To check if the function is injective, we need to see if it is monotonic (either always increasing or always decreasing).

1. For \( x < 0 \):
- The function is increasing because the numerator decreases and the denominator increases.

2. For \( x > 0 \):
- The function is decreasing because the numerator decreases and the denominator increases.

Since the function is increasing in one interval and decreasing in another, it is not monotonic and therefore **not injective**.

### Conclusion

The function \( f(x) \) is neither injective nor surjective.

### Final Answer

The answer to the question is option D: **neither injective nor surjective**.