Limits, Continuity & Differentiability
Continuity of piecewise functions
Grade 12
Question:
<p>Let \(f(x) = \begin{cases} \sqrt{x^2 - 1}, & x \leq \sqrt{10} \\ (\sqrt{10}x - 7), & \sqrt{10} < x < 5 \\ \sin \pi x, & 5 \leq x < 6 \\ \{x\}, & 6 \leq x \leq 7 \end{cases}\), then the number of points where \(f(x)\) is discontinuous in \([1, 7]\) is: (where \(\{\cdot\}\) denotes fractional part of \(x\).)</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) 2</p>
<p>(d) 3</p>
Step-by-Step Solution
Key Concept: For differentiability at a boundary point, the function must be continuous AND the left and right derivatives must be equal. Check continuity first at x = √10, then verify if the derivative from both sides match.
<p><strong>Step 1: Check continuity at x = √10</strong></p><p>Left limit: f(√10⁻) = √((√10)² - 1) = √(10 - 1) = √9 = 3</p><p>Right limit: f(√10⁺) = √10·√10 - 7 = 10 - 7 = 3</p><p>f(√10) = 3 (using first piece)</p><p>Since all three equal 3, function is continuous at x = √10 ✓</p><p><strong>Step 2: Check left derivative at x = √10</strong></p><p>For f(x) = √(x² - 1), f'(x) = x/√(x² - 1)</p><p>Left derivative: f'(√10⁻) = √10/√(10 - 1) = √10/3</p><p><strong>Step 3: Check right derivative at x = √10</strong></p><p>For f(x) = √10·x - 7, f'(x) = √10</p><p>Right derivative: f'(√10⁺) = √10</p><p><strong>Step 4: Compare derivatives</strong></p><p>Left derivative = √10/3 ≈ 1.054</p><p>Right derivative = √10 ≈ 3.162</p><p>Since √10/3 ≠ √10, the left and right derivatives are NOT equal</p><p>∴ f is continuous but NOT differentiable at x = √10. Answer: B</p>
Correct Answer: B