Question:

<p><strong>Step 1: Verify Continuity at x = 0</strong></p><p>Left piece at x = 0: f(0) = 1 + sin(0) = 1 + 0 = 1</p><p>Right piece limit: lim(x→0⁺) (x² - x + 1) = 0 - 0 + 1 = 1</p><p>Since f(0) = 1 and the right limit equals 1, f is continuous at x = 0. ✓</p><p><strong>Step 2: Find Left Derivative at x = 0</strong></p><p>For x ≤ 0: f(x) = 1 + sin x</p><p>f'(x) = cos x</p><p>Left derivative: f'(0⁻) = cos(0) = 1 > 0</p><p>This means f is increasing as we approach x = 0 from the left.</p><p><strong>Step 3: Find Right Derivative at x = 0</strong></p><p>For x > 0: f(x) = x² - x + 1</p><p>f'(x) = 2x - 1</p><p>Right derivative: f'(0⁺) = 2(0) - 1 = -1 < 0</p><p>This means f is decreasing immediately after x = 0.</p><p><strong>Step 4: Analyze the Transition</strong></p><p>At x = 0, the function transitions from:</p><p>• Increasing (f'(0⁻) = 1 > 0) on the left</p><p>• To decreasing (f'(0⁺) = -1 < 0) on the right</p><p>This is the definition of a local maximum: the function reaches a peak at x = 0.</p><p><strong>Step 5: Verify Other Options are Incorrect</strong></p><p>Option (b): f has a local minimum — FALSE (we showed it's a local maximum)</p><p>Option (c): f is increasing everywhere — FALSE (it decreases for x > 0 when 2x - 1 < 0, i.e., when 0 < x < 1/2)</p><p>Option (d): f is decreasing everywhere — FALSE (it increases for x ≤ 0 since cos x ≥ -1 for x in [-π/2, 0])</p><p><strong>∴ Answer: A</strong></p>