<p>Let \(f: \mathbb{R} \to \mathbb{R}\), where \(f(x) = \frac{x^2 + ax + 1}{x^2 + x + 1}\). Then the complete set of values of \(a\) such that \(f(x)\) is onto is:</p>
<p>(a) \((-\infty, \infty)\)</p>
<p>(b) \((-\infty, 0)\)</p>
<p>(c) \((0, \infty)\)</p>
<p>(d) Empty set</p>
Step-by-Step Solution
Key Concept: For f(x) to be onto (surjective), every real number y must be in the range of f. We need to determine which values of y are achievable by solving y = f(x) for x in terms of a, and finding when this has real solutions for all y ∈ ℝ.
<p><strong>Step 1:</strong> Set up the equation for range analysis. If y is in the range of f, then:</p><p>y = (x² + ax + 1)/(x² + x + 1)</p><p>Rearranging: y(x² + x + 1) = x² + ax + 1</p><p>yx² + yx + y = x² + ax + 1</p><p>(y - 1)x² + (y - a)x + (y - 1) = 0</p><p></p><p><strong>Step 2:</strong> Analyze two cases.</p><p><strong>Case 1:</strong> If y = 1, then 0·x² + (1 - a)x + 0 = 0, which gives (1 - a)x = 0. This has a solution for all a ≠ 1. When a = 1, there's no constraint (0 = 0 always), so y = 1 is always in the range.</p><p></p><p><strong>Step 3:</strong> <strong>Case 2:</strong> If y ≠ 1, the equation is quadratic in x: (y - 1)x² + (y - a)x + (y - 1) = 0. For this to have real solutions, the discriminant must be non-negative:</p><p>Δ = (y - a)² - 4(y - 1)² ≥ 0</p><p></p><p><strong>Step 4:</strong> Expand the discriminant condition:</p><p>(y - a)² - 4(y - 1)² ≥ 0</p><p>y² - 2ay + a² - 4(y² - 2y + 1) ≥ 0</p><p>y² - 2ay + a² - 4y² + 8y - 4 ≥ 0</p><p>-3y² + (8 - 2a)y + (a² - 4) ≥ 0</p><p>3y² - (8 - 2a)y - (a² - 4) ≤ 0</p><p></p><p><strong>Step 5:</strong> For f to be onto, this inequality must hold for ALL y ∈ ℝ. However, the left side is a quadratic in y with positive leading coefficient (3 > 0). Such a quadratic can be ≤ 0 for all y only if it has no real values (which is impossible) OR if the inequality cannot hold for arbitrarily large |y|.</p><p></p><p><strong>Step 6:</strong> Since the coefficient of y² is positive (3 > 0), the parabola opens upward. For 3y² - (8 - 2a)y - (a² - 4) ≤ 0 to hold for ALL y ∈ ℝ, we would need the parabola to be non-positive everywhere, which is impossible for an upward-opening parabola.</p><p></p><p><strong>Step 7:</strong> The quadratic 3y² - (8 - 2a)y - (a² - 4) will always be positive for sufficiently large |y|. Therefore, there will always exist some value y₀ for which the discriminant condition fails, meaning y₀ is not in the range of f.</p><p></p><p><strong>Step 8:</strong> Since for any choice of a ∈ ℝ, we can find at least one real number not in the range of f, the function f is never onto for any value of a.</p><p></p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D