<p>Let \(f(x) = 1 - \dfrac{2x+c}{x^2+x+2c}\) where \(P(x) = x^2 + x + 2c\). If the range of \(f(x)\) is all real numbers \(\mathbb{R}\), then which of the following is/are correct regarding the value of \(c\)?</p>
<p>(a) \(c \in (-6, 0)\)</p>
<p>(b) \(c \in (-6, 0)\)</p>
<p>(c) \(P\left(\dfrac{-c}{2}\right) < 0\)</p>
<p>(d) \(c \geq 0\)</p>
Step-by-Step Solution
Key Concept: For the range of f(x) to be all real numbers, the equation y = 1 - (2x+c)/(x²+x+2c) must be solvable for all real y. Rearranging to (2x+c) = (1-y)(x²+x+2c), this quadratic in x must have real solutions for every y ∈ ℝ, requiring the discriminant condition to hold for all y.
<p><strong>Step 1: Set up the inverse relationship.</strong> Let y = 1 - (2x+c)/(x²+x+2c). Rearrange: (2x+c) = (1-y)(x²+x+2c), giving (1-y)x² + (1-y-2)x + (2c(1-y)-c) = 0, or (1-y)x² + (-1-y)x + c(2-2y-1) = 0.</p><p><strong>Step 2: Apply discriminant condition.</strong> Simplify to: (1-y)x² - (1+y)x + c(1-2y) = 0. For range = ℝ, this must have real solutions for all y. When y = 1, we get -2x + (-c) = 0, so x = -c/2 works. For y ≠ 1, discriminant Δ = (1+y)² - 4(1-y)·c(1-2y) ≥ 0 for all y.</p><p><strong>Step 3: Expand discriminant inequality.</strong> Δ = (1+y)² - 4c(1-y)(1-2y) = 1 + 2y + y² - 4c(1-3y+2y²) = y²(1-8c) + y(2+12c) + (1-4c) ≥ 0 for all y ∈ ℝ.</p><p><strong>Step 4: Determine constraint on c.</strong> For a quadratic in y to be non-negative for all y: either the coefficient of y² is positive and its discriminant ≤ 0, or the quadratic is a perfect square or degenerate. This gives: 1-8c > 0 (so c < 1/8) AND (2+12c)² - 4(1-8c)(1-4c) ≤ 0. Computing: 4 + 48c + 144c² - 4(1-4c-8c+32c²) = 4 + 48c + 144c² - 4 + 48c - 128c² = 16c² + 96c = 16c(c+6) ≤ 0. This gives -6 ≤ c ≤ 0.</p><p><strong>Step 5: Combine conditions.</strong> The binding constraint is -6 ≤ c ≤ 0 (which automatically satisfies c < 1/8). Therefore c ∈ [-6, 0].</p><p>∴ Answer: B,C</p>
Correct Answer: B,C