Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Let \(f: [-1,\,3] \to R\) be defined as \[f(x) = \begin{cases} |x| + [x], & -1 \leq x < 1 \\ x + |x|, & 1 \leq x < 2 \\ x + [x], & 2 \leq x \leq 3 \end{cases}\] where \([t]\) denotes the greatest integer less than or equal to \(t\). Then, \(f\) is discontinuous at</p>
<p>only one point.</p>
<p>only two points.</p>
<p>only three points.</p>
<p>four or more points.</p>
Step-by-Step Solution
Key Concept: To find differentiability, check left and right derivatives at boundary points of piecewise functions. The function's differentiability fails where the left and right derivatives are unequal, even if the function is continuous.
<p><strong>Step 1: Identify the function pieces</strong></p><p>f(x) = |x| + [x] for -1 ≤ x < 0 gives f(x) = -x - 1</p><p>f(x) = |x| + [x] for 0 ≤ x < 1 gives f(x) = x + 0 = x</p><p>f(x) = |x| + [x] for 1 ≤ x ≤ 3 gives f(x) = x + 1</p><p><strong>Step 2: Check continuity at x = 0</strong></p><p>lim(x→0⁻) f(x) = 0, lim(x→0⁺) f(x) = 0, f(0) = 0 ✓ Continuous</p><p><strong>Step 3: Check left and right derivatives at x = 0</strong></p><p>Left derivative: f'(0⁻) = d/dx(-x - 1) = -1</p><p>Right derivative: f'(0⁺) = d/dx(x) = 1</p><p>Since -1 ≠ 1, f is NOT differentiable at x = 0</p><p><strong>Step 4: Check continuity at x = 1</strong></p><p>lim(x→1⁻) f(x) = 1, lim(x→1⁺) f(x) = 2 ✗ NOT continuous</p><p>Therefore, f is not differentiable at x = 1 (not even continuous)</p><p><strong>Step 5: Conclusion</strong></p><p>f is differentiable on [-1, 3] except at x = 0 and x = 1.</p><p>∴ Answer: C</p>
Correct Answer: C