Limits, Continuity & Differentiability
Continuity and Differentiability of Functions
Grade 12
<p>\( f(x) = [\log_e x] + \sqrt{\{\log_e x\}}, x > 1 \), where [.] and {.} denote the greatest integer function and the fractional part function respectively, then</p>
<p>(a) \( f(x) \) is continuous but non-differentiable at \( x = e \)</p>
<p>(b) \( f(x) \) is differentiable at \( x = e \)</p>
<p>(c) \( f(x) \) is discontinuous at \( x = e \)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: The function f(x) is defined using the greatest integer [ln x] and fractional part {ln x} = ln x - [ln x]. Since {ln x} ∈ [0,1), the square root is always real, and f is discontinuous wherever ln x crosses an integer value.
<p><strong>Step 1:</strong> Identify the domain and structure. For x > 1, we have ln x > 0. Let ln x = n + α where n = [ln x] ≥ 0 and α = {ln x} ∈ [0,1).</p><p><strong>Step 2:</strong> Express f(x) = n + √α where n is a non-negative integer and α ∈ [0,1).</p><p><strong>Step 3:</strong> Check continuity at x = e^k (where k is a positive integer). As x → e^k⁻, ln x → k⁻, so [ln x] = k-1 and {ln x} → 1⁻, giving f(x) → (k-1) + 1 = k. As x → e^k⁺, ln x → k⁺, so [ln x] = k and {ln x} → 0⁺, giving f(x) → k + 0 = k. Thus f is continuous from left and right at jump points.</p><p><strong>Step 4:</strong> Check differentiability. Within each interval [e^n, e^(n+1)), f(x) = n + √(ln x - n) is differentiable with f'(x) = 1/[2x√(ln x - n)]. At x = e^n, the left and right derivatives are different (left derivative involves √(1⁻) while right derivative involves √(0⁺)), so f is non-differentiable at x = e^n for n ≥ 1.</p><p><strong>Step 5:</strong> f is continuous for all x > 1 but not differentiable at x = e, e², e³, ... </p><p>∴ f is continuous everywhere on (1, ∞) but has non-differentiable points at x = eⁿ (n ∈ ℕ).</p>
Correct Answer: A