Limits, Continuity & Differentiability
Absolute maximum and minimum of piecewise functions
Grade 12
Question:
<p>Let \( f(x) = \begin{cases} -x^3 + a, & 0 \leq x < 1 \\ x, & 1 \leq x \leq 3 \end{cases} \). Which of the following is/are correct?</p>
<p>if \(f(x)\) has an absolute minimum at \(x = 1\), then minimum positive integral value of \(a\) is 2.</p>
<p>if \(f(x)\) has an absolute minimum at \(x = 1\), then minimum positive integral value of \(a\) is 3.</p>
<p>if \(f(x)\) has an absolute maximum at \(x = 3\), then maximum positive integral value of \(a\) is 3.</p>
<p>if \(f(x)\) has an absolute maximum at \(x = 0\), then minimum positive integral value of \(a\) is 3.</p>
Step-by-Step Solution
Key Concept: For f to be continuous at x=1, left and right limits must equal f(1). Use this condition to find 'a', then verify differentiability by checking if left and right derivatives are equal at x=1.
<p><strong>Step 1: Apply continuity at x=1</strong></p><p>Left limit: lim(x→1⁻) f(x) = -(1)³ + a = -1 + a</p><p>Right limit: lim(x→1⁺) f(x) = |1-1| = 0</p><p>For continuity: -1 + a = 0 ⟹ <strong>a = 1</strong></p><p><strong>Step 2: Check differentiability at x=1</strong></p><p>Left derivative: f'(1⁻) = -3x²|ₓ₌₁ = -3</p><p>Right derivative: f'(1⁺) = d/dx|x-1| = sign(x-1)|ₓ₌₁⁺ = +1</p><p>Since f'(1⁻) ≠ f'(1⁺), <strong>f is NOT differentiable at x=1</strong></p><p><strong>Step 3: Analyze statements</strong></p><p><strong>A (FALSE):</strong> a = 2 contradicts continuity requirement</p><p><strong>B (TRUE):</strong> a = 1 satisfies continuity</p><p><strong>C (TRUE):</strong> f is continuous on [0,∞) when a=1</p><p><strong>D (TRUE):</strong> f is NOT differentiable at x=1 (left and right derivatives differ)</p><p>∴ Answer: BCD</p>
Correct Answer: BCD