<p>Find the area of the region containing the points satisfying \(|y| + \dfrac{1}{2} \leq e^{-|x|}\); \(\max(|x|, |y|) \leq 2\).</p>
Step-by-Step Solution
Key Concept: The region is bounded by the exponential curve e^(-|x|) shifted down by 1/2 and the square |x| ≤ 2, |y| ≤ 2. Due to symmetry about both axes, calculate area in the first quadrant and multiply by 4.
<p><strong>Step 1:</strong> Analyze the constraints. The region is defined by: |y| + 1/2 ≤ e^(-|x|) and max(|x|, |y|) ≤ 2. This means |y| ≤ e^(-|x|) - 1/2 within the square [-2,2] × [-2,2].</p><p><strong>Step 2:</strong> Note that e^(-|x|) - 1/2 must be positive. When x = 0: e^0 - 1/2 = 1/2 > 0. When |x| = 2: e^(-2) - 1/2 ≈ 0.135 - 0.5 < 0, so the constraint is inactive for |x| ≥ ln(2) ≈ 0.693.</p><p><strong>Step 3:</strong> By symmetry about both axes, calculate for x ≥ 0, y ≥ 0 and multiply by 4. For 0 ≤ x ≤ ln(2), the region extends from y = 0 to y = e^(-x) - 1/2. For ln(2) < x ≤ 2, the constraint e^(-x) - 1/2 < 0 means no region exists above the x-axis.</p><p><strong>Step 4:</strong> Area in first quadrant = ∫₀^(ln2) (e^(-x) - 1/2) dx = [-e^(-x)]₀^(ln2) - (1/2)·ln(2) = (-e^(-ln2) + 1) - (ln2)/2 = (1 - 1/2) - (ln2)/2 = 1/2 - (ln2)/2.</p><p><strong>Step 5:</strong> Total area = 4[1/2 - (ln2)/2] = 2 - 2ln(2) = 2(1 - ln2).</p><p>∴ Answer: B</p>
Correct Answer: B