<p>Let \(f(x) = \begin{cases} x^2 - 2|x| + a, & x \leq 1 \\ 6 + x, & x > 1 \end{cases}\), then number of positive integral value(s) of \(a\) for which \(f(x)\) has local minima at \(x = 1\), is/are:</p>
Step-by-Step Solution
Key Concept: For f(x) to have a local minimum at x=1, the left derivative at x=1 must be non-negative and the right derivative must be positive, combined with continuity/comparison of values. The critical constraint is that the absolute value function creates different behaviors for x<0 and 0≤x≤1.
<p><strong>Step 1:</strong> For x≤1, split by absolute value sign:</p><p>• If x<0: f(x) = x² + 2x + a, so f'(x) = 2x + 2</p><p>• If 0≤x≤1: f(x) = x² - 2x + a, so f'(x) = 2x - 2</p><p><strong>Step 2:</strong> Check left derivative at x=1: f'(1⁻) = 2(1) - 2 = 0</p><p><strong>Step 3:</strong> Check right derivative at x=1: For x>1, f(x) = 6+x, so f'(1⁺) = 1 > 0</p><p><strong>Step 4:</strong> For local minimum at x=1, we need f(1) ≤ f(x) for x near 1.</p><p>• f(1) = 1 - 2 + a = a - 1</p><p>• For x slightly less than 1 (in the region [0,1]): f(x) = x² - 2x + a has minimum at x=1 in this region since f'(x) = 2x-2 ≤ 0 for x≤1</p><p>• For x slightly greater than 1: f(x) = 6+x, so f(1⁺) = 7</p><p><strong>Step 5:</strong> Need a - 1 ≤ 7, giving a ≤ 8. Also need a - 1 ≤ f(0) = a (automatically satisfied). Since a must be positive integer: a ∈ {1, 2, 3, 4, 5, 6, 7, 8}</p><p><strong>Step 6:</strong> Verify continuity isn't required but monotonicity from left (non-positive derivative) and to right (positive derivative) confirms local minimum.</p><p>∴ Answer: A (8 positive integral values)
Correct Answer: A