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 whether the function \( f: \mathbb{R} \to \mathbb{R} \) defined by

\[ f(x) = \frac{x^2 - x + 1}{x^2 + x + 1} \]

is injective or surjective, we will analyze its properties step by step.

### Step 1: Check for Injectivity

A function is injective (one-to-one) if \( f(a) = f(b) \) implies \( a = b \). To check this, we can analyze the derivative of the function.

1. **Find the derivative \( f'(x) \)** using the quotient rule:

\[ f'(x) = \frac{(x^2 + x + 1)(2x - 1) - (x^2 - x + 1)(2x + 1)}{(x^2 + x + 1)^2} \]

Here, let \( u = x^2 - x + 1 \) and \( v = x^2 + x + 1 \). Then, using the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), we calculate \( \frac{du}{dx} = 2x - 1 \) and \( \frac{dv}{dx} = 2x + 1 \).

2. **Simplify the derivative**:

After substituting and simplifying, we find:

\[ f'(x) = \frac{(x^2 + x + 1)(2x - 1) - (x^2 - x + 1)(2x + 1)}{(x^2 + x + 1)^2} \]

This leads to:

\[ f'(x) = \frac{-2x^2 + 4x - 2}{(x^2 + x + 1)^2} \]

3. **Determine the sign of \( f'(x) \)**:

The numerator \( -2x^2 + 4x - 2 \) can be factored or analyzed using the quadratic formula. The roots of the equation \( -2x^2 + 4x - 2 = 0 \) are:

\[ x = 1 \quad \text{and} \quad x = 3 \]

The sign of \( f'(x) \) changes at these points, indicating that the function is increasing on \( (-\infty, 1) \) and \( (3, \infty) \), and decreasing on \( (1, 3) \). Thus, it is not monotonic over its entire domain, which means it is not injective.

### Step 2: Check for Surjectivity

A function is surjective (onto) if for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \).

1. **Set the equation**:

We need to solve for \( y \):

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

Rearranging gives:

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

This simplifies to:

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

2. **Analyze the discriminant**:

For this quadratic equation in \( x \) to have real solutions, the discriminant must be non-negative:

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

Simplifying gives:

\[ D = (y + 1)^2 - 4(y^2 - 2y + 1) = -3y^2 + 10y - 3 \]

Setting \( D \geq 0 \) leads to finding the roots of the quadratic \( -3y^2 + 10y - 3 = 0 \). The roots are:

\[ y = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6} \]

This results in \( y = 3 \) and \( y = \frac{1}{3} \).

3. **Determine the range**:

The function is surjective if the range equals the codomain \( \mathbb{R} \). Since the discriminant is non-negative only for \( y \in \left[\frac{1}{3}, 3\right] \), the function does not cover all real numbers, hence it is not surjective.

### Conclusion

The function \( f(x) = \frac{x^2 - x + 1}{x^2 + x + 1} \) is neither injective nor surjective.