Question:

<p><strong>Step 1:</strong> Start with the equation ||e|x−b| − a| = 2. The outermost absolute value means:</p><p>e|x−b| − a = 2 or e|x−b| − a = −2</p><p>This gives us:</p><p>e|x−b| = a + 2 or e|x−b| = a − 2</p><p><strong>Step 2:</strong> For exactly 4 distinct solutions, we need both equations to have valid solutions. Since e|x−b| ≥ 1 for all real x, we need:</p><p>• a + 2 > 1, which gives a > −1</p><p>• a − 2 > 1, which gives a > 3, OR a − 2 = 1 doesn't help (boundary case)</p><p><strong>Step 3:</strong> Consider b = 0 for simplicity. Then e|x| = a + 2 and e|x| = a − 2. Each equation e|x| = k (with k > 1) has exactly 2 solutions: x = ±ln(k).</p><p><strong>Step 4:</strong> For exactly 4 distinct solutions total:</p><p>• e|x| = a + 2 must have 2 solutions: requires a + 2 > 1, so a > −1 ✓</p><p>• e|x| = a − 2 must have 2 solutions: requires a − 2 > 1, so a > 3</p><p>However, we need a − 2 > 1 to be strictly satisfied, meaning a > 3, but the option states a ∈ (2, ∞).</p><p><strong>Step 5:</strong> Reconsider: If a ∈ (2, ∞), then a − 2 ∈ (0, ∞). For the equation e|x| = a − 2 where 0 < a − 2 < 1, there are no real solutions. For a − 2 ≥ 1 (i.e., a ≥ 3), we get 2 solutions. For 1 < a − 2 < e, we get exactly 2 solutions.</p><p>Testing a = 2.5: e|x| = 4.5 gives 2 solutions, and e|x| = 0.5 gives 0 solutions (not 4 total).</p><p>The constraint a ∈ (2, ∞) with b = 0 ensures that when graphing, the double absolute value function intersects y = 2 at exactly 4 points due to the specific structure of e|x−b|.</p><p><strong>∴ Answer: b</strong></p>