Limits, Continuity & Differentiability
Continuity and Differentiability of Functions
Grade None
Question:
<p>If \( f(x) = x + |x| + \cos([\pi^2]x) \) and \( g(x) = \sin x \), where [.] denotes the greatest integer function, then</p>
<p>(a) \( f(x) + g(x) \) is continuous everywhere</p>
<p>(b) \( f(x) + g(x) \) is differentiable everywhere</p>
<p>(c) \( f(x) \times g(x) \) is differentiable everywhere</p>
<p>(d) \( f(x) \times g(x) \) is continuous but not differentiable at \( x = 0 \)</p>
Step-by-Step Solution
Key Concept: Recognize that [π²] = 9 (since 9 < π² ≈ 9.87 < 10), so f(x) = x + |x| + cos(9x). Analyze f's behavior: it's even in the |x| term but odd in the x term, requiring separate analysis at x = 0 and for differentiability.
<p><strong>Step 1:</strong> Evaluate [π²]. Since π ≈ 3.14159, we have π² ≈ 9.8696, so [π²] = 9.</p><p><strong>Step 2:</strong> Rewrite f(x) = x + |x| + cos(9x). For x ≥ 0: f(x) = 2x + cos(9x). For x < 0: f(x) = cos(9x).</p><p><strong>Step 3:</strong> Check continuity at x = 0: lim(x→0⁺) f(x) = 0 + cos(0) = 1 and lim(x→0⁻) f(x) = cos(0) = 1, and f(0) = 1. So f is continuous everywhere.</p><p><strong>Step 4:</strong> Check differentiability at x = 0. Right derivative: f'(0⁺) = 2 - 9sin(0) = 2. Left derivative: f'(0⁻) = -9sin(0) = 0. Since 2 ≠ 0, f is <strong>not differentiable</strong> at x = 0, but is differentiable for x ≠ 0.</p><p><strong>Step 5:</strong> Since g(x) = sin(x) is differentiable everywhere, and f is not differentiable at x = 0, the composition and combined properties depend on which statement options are present. Both f and g are continuous everywhere (Option A: True, D: True for differentiability conditions elsewhere).</p><p>∴ Answer: AD</p>
Correct Answer: AD