Limits, Continuity & Differentiability
Continuity and Differentiability
Grade 12
Question:
<p>If \(g(x) = \begin{cases} [f(x)], & x \in \left(0,\dfrac{\pi}{2}\right) \cup \left(\dfrac{\pi}{2},\pi\right) \\ 3, & x = \dfrac{\pi}{2} \end{cases}\) where \([x]\) denotes the greatest integer function<br><br>and \(f(x) = \dfrac{2(\sin x - \sin^n x) + |\sin x - \sin^n x|}{2(\sin x - \sin^n x) - |\sin x - \sin^n x|},\ n \in R\), then</p>
<p>\(g(x)\) is continuous and differentiable at \(x = \pi/2\), when \(0 < n < 1\)</p>
<p>\(g(x)\) is continuous and differentiable at \(x = \pi/2\), when \(n > 1\)</p>
<p>\(g(x)\) is continuous but not differentiable at \(x = \pi/2\), when \(0 < n < 1\)</p>
<p>\(g(x)\) is continuous but not differentiable, at \(x = \pi/2\), when \(n > 1\)</p>
Step-by-Step Solution
Key Concept: Analyze the sign of (sin x - sin^n x) in different regions of (0,π) to simplify the absolute value expression, then use the floor function to find [f(x)], and finally check continuity at x = π/2.
<p><strong>Step 1: Analyze sin x - sin^n x</strong></p><p>For x ∈ (0,π), we have sin x ∈ (0,1]. Since 0 < sin x ≤ 1, we compare sin x with sin^n x.</p><p>• If 0 < sin x < 1 and n > 1: sin^n x < sin x, so sin x - sin^n x > 0</p><p>• If sin x = 1 (at x = π/2): sin^n x = 1, so sin x - sin^n x = 0</p><p><strong>Step 2: Simplify f(x) using |A| = A when A > 0</strong></p><p>For x ∈ (0, π/2) ∪ (π/2, π), let A = sin x - sin^n x > 0:</p><p>$$f(x) = \frac{2A + A}{2A - A} = \frac{3A}{A} = 3$$</p><p><strong>Step 3: Find [f(x)]</strong></p><p>Since f(x) = 3 for all x ∈ (0, π/2) ∪ (π/2, π):</p><p>$$[f(x)] = [3] = 3$$</p><p><strong>Step 4: Verify continuity at x = π/2</strong></p><p>• Left limit: lim(x→π/2⁻) g(x) = [f(x)] = 3</p><p>• Right limit: lim(x→π/2⁺) g(x) = [f(x)] = 3</p><p>• Value at x = π/2: g(π/2) = 3</p><p>All three are equal, so g(x) is continuous at x = π/2.</p><p>∴ Answer: B</p>
Correct Answer: B