<p>Let \(f: R \to \left(0, \dfrac{2\pi}{3}\right]\) defined as \(f(x) = \cot^{-1}(x^2 - 4x + \alpha)\). The smallest integral value of \(\alpha\) such that \(f(x)\) is an into function, is equal to:</p>
Step-by-Step Solution
Key Concept: For f(x) = cot⁻¹(u) to have range (0, 2π/3], the argument u must satisfy u ≥ cot(2π/3) = -1/√3. Since cot⁻¹ is a decreasing function, we need the minimum value of (x² - 4x + α) to be at least -1/√3 for the function to be into.
<p><strong>Step 1:</strong> For f to be an into function from ℝ to (0, 2π/3], the range of f must be a proper subset of (0, 2π/3].</p><p><strong>Step 2:</strong> Since cot⁻¹ is a decreasing function, cot⁻¹(y) ∈ (0, 2π/3] when y ≥ cot(2π/3) = cot(π - π/3) = -cot(π/3) = -1/√3.</p><p><strong>Step 3:</strong> For f to be into, we need the entire range of the quadratic u(x) = x² - 4x + α to fall within the domain that maps to (0, 2π/3].</p><p><strong>Step 4:</strong> Completing the square: u(x) = (x-2)² + α - 4. The minimum value is (α - 4) at x = 2.</p><p><strong>Step 5:</strong> For f to be into (not onto), we need: α - 4 > -1/√3, which gives α > 4 - 1/√3 ≈ 4 - 0.577 ≈ 3.42.</p><p><strong>Step 6:</strong> Alternatively, if we require strict inequality for the function to not achieve the boundary, the smallest integral value of α is <strong>4</strong>.</p><p>∴ Answer: B</p>
Correct Answer: B