Question:

<p><strong>Step 1: Set up the onto condition.</strong></p><p>For f to be onto ℝ, for every y ∈ ℝ, the equation y = f(x) must have at least one real solution x.</p><p>y = (x² + ax + 1)/(x² + x + 1)</p><p><strong>Step 2: Rearrange to find when real solutions exist.</strong></p><p>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><strong>Step 3: Analyze the equation for all y ∈ ℝ.</strong></p><p>Case 1: If y = 1, the equation becomes (1 - a)x = 0.</p><p>For this to have a solution, we need 1 - a ≠ 0, i.e., a ≠ 1.</p><p><strong>Step 4: For y ≠ 1, use discriminant condition.</strong></p><p>For real solutions: Δ = (y - a)² - 4(y - 1)² ≥ 0</p><p>Expanding: (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><strong>Step 5: Check if this inequality holds for ALL y ≠ 1.</strong></p><p>For the quadratic -3y² + (8 - 2a)y + (a² - 4) to be non-negative for all y (except possibly one point), the coefficient of y² is negative (-3 < 0).</p><p>A downward-opening parabola cannot be ≥ 0 for all real y. It can be ≥ 0 only on a bounded interval.</p><p>For f to be onto, we need the inequality to hold for all y, which is impossible when the leading coefficient is negative.</p><p><strong>Step 6: Conclusion.</strong></p><p>Since the discriminant analysis yields a condition that cannot be satisfied for all y ∈ ℝ, no value of a makes f onto.</p><p><strong>∴ Answer: d (not possible)</strong></p>