<p>If the range of function \(f(x) = \dfrac{x+1}{k+x^2}\) contains the interval \([0, 1]\), then the value of \(k\) can be equal to</p>
Step-by-Step Solution
Key Concept: For the range of f(x) to contain [0,1], the equation f(x)=y must have real solutions for every y∈[0,1]. This requires analyzing when y = (x+1)/(k+x²) is solvable, which transforms to yx² - x + (yk-1) = 0.
<p><strong>Step 1:</strong> Set y = (x+1)/(k+x²). For y to be in the range, the equation yx² - x + (yk-1) = 0 must have real solutions in x.</p><p><strong>Step 2:</strong> For y = 0: Setting 0 = (x+1)/(k+x²) gives x = -1 (valid for any k > 0).</p><p><strong>Step 3:</strong> For y ≠ 0: Rearrange to yx² - x + (yk-1) = 0. Discriminant Δ = 1 - 4y(yk-1) = 1 - 4y²k + 4y ≥ 0.</p><p><strong>Step 4:</strong> This must hold for all y ∈ [0,1]. The critical constraint is at y = 1: Δ = 1 - 4k + 4 = 5 - 4k ≥ 0, giving k ≤ 5/4.</p><p><strong>Step 5:</strong> Also need k > 0 to ensure the denominator is positive. Additionally verify that at y = 1, we get x² - x + (k-1) = 0, which requires k ≤ 5/4 for real solutions.</p><p><strong>Step 6:</strong> For the range to contain [0,1], we need 0 < k ≤ 5/4. Therefore k can equal values like 1/4, 1/2, 1, or 5/4.</p><p>∴ Answer: A, B, C (typical options within (0, 5/4])</p>
Correct Answer: A,B,C