`f : R -> R`, where `f(x)=(x^2+ax+1)/(x^2+x+1)` Complete set of values of 'a' such that `f (x)` is onto, is

Options

A. `(-oo, oo)`
B. `(-oo,0)`
C. `(0,oo)`
D. Empty set

Answer

To determine the complete set of values of 'a' such that the function \( f(x) = \frac{x^2 + ax + 1}{x^2 + x + 1} \) is onto (surjective), we will analyze the function step by step.

### Step 1: Analyze the Denominator
The denominator of the function is \( g(x) = x^2 + x + 1 \). We need to check if this expression is always positive.

**Hint:** Check the discriminant of the quadratic to determine if it has real roots.

The discriminant \( D \) of \( g(x) \) is given by:
\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, \( g(x) \) has no real roots and is always positive for all \( x \in \mathbb{R} \).

### Step 2: Analyze the Numerator
Next, we analyze the numerator \( h(x) = x^2 + ax + 1 \). The coefficient of \( x^2 \) is 1, which is greater than 0, indicating that this is also a parabola opening upwards.

**Hint:** Check the vertex of the parabola to determine its minimum value.

The vertex of the quadratic \( h(x) \) occurs at:
\[ x = -\frac{b}{2a} = -\frac{a}{2} \]
Substituting this back into \( h(x) \) gives:
\[ h\left(-\frac{a}{2}\right) = \left(-\frac{a}{2}\right)^2 + a\left(-\frac{a}{2}\right) + 1 = \frac{a^2}{4} - \frac{a^2}{2} + 1 = \frac{a^2 - 2a^2 + 4}{4} = \frac{-a^2 + 4}{4} \]
This shows that the minimum value of \( h(x) \) is \( \frac{-a^2 + 4}{4} \).

### Step 3: Determine Conditions for Onto
For \( f(x) \) to be onto, the range of \( f(x) \) must cover all real numbers. Since the denominator is always positive, the range of \( f(x) \) will depend on the minimum value of the numerator \( h(x) \).

**Hint:** Set the minimum value of \( h(x) \) to be less than or equal to 0.

To ensure that \( f(x) \) can take all real values, we need:
\[ \frac{-a^2 + 4}{4} \leq 0 \]
This simplifies to:
\[ -a^2 + 4 \leq 0 \implies a^2 \geq 4 \]
Thus, we have:
\[ |a| \geq 2 \implies a \leq -2 \text{ or } a \geq 2 \]

### Conclusion
The complete set of values of \( a \) such that \( f(x) \) is onto is:
\[ (-\infty, -2] \cup [2, \infty) \]