Limits, Continuity & Differentiability
Continuity and greatest integer function
Grade 12
Question:
<p>Let \(f(x) = \left[\dfrac{4^x + 2^x + 1}{2^x - 2^{x/2} + 1}\right]\) and \(g(x) = \left[\dfrac{9}{x^2 + 5}\right]\). Identify which of the following statement(s) is(are) <strong>correct</strong>?</p><p>[<strong>Note:</strong> where \([y]\) denotes greatest integer function less than or equal to \(y\).]</p>
<p>Number of points of discontinuities of \(f(x)\) in \((-\infty, 0]\) is 2.</p>
<p>Number of points of discontinuities of \(g(x)\) in \((-\infty, \infty)\) is 2.</p>
<p>Number of points of discontinuities of \(f(x) \cdot g(x)\) in \((-\infty, \infty)\) is 7.</p>
<p>Number of points of discontinuities of \(f(x) \cdot g(x)\) in \((-\infty, \infty)\) is 6.</p>
Step-by-Step Solution
Key Concept: Simplify f(x) by substituting t = 2^(x/2) to get f(x) = [t² + 1], then analyze the range of g(x) using calculus to find where both greatest integer functions have specific values.
<p><strong>Step 1: Simplify f(x)</strong></p><p>Let u = 2^(x/2), so 2^x = u². Then:</p><p>f(x) = [(u² + u + 1)/(u - 1 + 1)] = [(u² + u + 1)/u] = [u + 1 + 1/u]</p><p>Since u > 0, by AM-GM: u + 1/u ≥ 2, with equality when u = 1 (x = 0).</p><p>∴ f(x) ∈ {3, 4, 5, ...} for all real x, and f(0) = [1 + 1 + 1] = [3] = 3</p><p><strong>Step 2: Analyze g(x)</strong></p><p>g(x) = [9/(x² + 5)]. Since x² ≥ 0, we have x² + 5 ≥ 5.</p><p>Therefore: 9/(x² + 5) ≤ 9/5 = 1.8</p><p>Also, 9/(x² + 5) > 0 for all x.</p><p>∴ g(x) ∈ {0, 1} for all real x, with g(x) = 1 when x² + 5 ≤ 9, i.e., |x| ≤ 2</p><p><strong>Step 3: Check statements</strong></p><p><strong>A:</strong> f(0) = 3 ✓ (Since f(0) = [3] = 3)</p><p><strong>B:</strong> f is continuous everywhere ✓ (f(x) = [u + 1 + 1/u] where u = 2^(x/2) is continuous and u + 1/u is strictly monotonic on (0,∞), so greatest integer jumps at discrete points only, but composition maintains GIF continuity properties)</p><p><strong>C:</strong> g(x) is continuous everywhere ✗ (g has jump discontinuities at x = ±2 where 9/(x²+5) = 1.8)</p><p><strong>D:</strong> f(x) ≥ 3 for all x ∈ ℝ ✓ (Minimum value of u + 1/u is 2 at u = 1, so f(x) ≥ [2 + 1] = [3] = 3)</p><p>∴ Answer: <strong>A, B, D</strong></p>
Correct Answer: A,B,D