<p><strong>154.</strong> 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 into function, is equal to:</p>
Step-by-Step Solution
Key Concept: For f to be an 'into' function (not surjective), the range of cot⁻¹(x² - 4x + α) must be strictly smaller than (0, 2π/3]. This requires that x² - 4x + α never achieves values in (cot(2π/3), ∞) = (-1/√3, ∞), meaning x² - 4x + α must stay bounded below cot(2π/3) = -1/√3.
<p><strong>Step 1:</strong> For f: ℝ → (0, 2π/3] to be into, the range of cot⁻¹(x² - 4x + α) must be a proper subset of (0, 2π/3].</p><p><strong>Step 2:</strong> The function cot⁻¹ is strictly decreasing. At the boundary: cot(2π/3) = -1/√3. For the range to not include values near 0 (which correspond to large arguments), we need x² - 4x + α to never go below -1/√3.</p><p><strong>Step 3:</strong> The minimum of g(x) = x² - 4x + α occurs at x = 2: g(2) = 4 - 8 + α = α - 4.</p><p><strong>Step 4:</strong> We require: α - 4 ≥ -1/√3, which gives α ≥ 4 - 1/√3 = 4 - √3/3 ≈ 4 - 0.577 ≈ 3.423.</p><p><strong>Step 5:</strong> The smallest integral value of α is <strong>4</strong>.</p><p>∴ Answer: B</p>
Correct Answer: B