<p>Let \(f(x) = (e^x - a)(3ax + 1)\). Number of possible values of \(a\) satisfying \(f(x) \geq 0\) for all \(x \in R\).</p>
Step-by-Step Solution
Key Concept: For f(x) = (e^x - a)(3ax + 1) ≥ 0 for all x ∈ ℝ, we need either both factors non-negative everywhere or both non-positive everywhere. Since e^x ranges over (0, ∞), we must analyze when the linear factor's sign complements e^x - a's behavior.
<p><strong>Step 1:</strong> Analyze the factors. e^x - a = 0 when x = ln(a) (for a > 0). The line 3ax + 1 = 0 when x = -1/(3a) (for a ≠ 0).</p><p><strong>Step 2:</strong> For a linear function 3ax + 1 to maintain a consistent sign relationship with e^x - a everywhere:</p><p><strong>Case 1 (a = 0):</strong> f(x) = e^x · 1 = e^x > 0 for all x. ✓ This works.</p><p><strong>Case 2 (a > 0):</strong> At x → -∞: e^x → 0⁺ (so e^x - a < 0) and 3ax + 1 → -∞ (negative). Product → 0⁺. At x → ∞: both factors → ∞ (positive). Product → ∞. The product could be negative in between unless the zero of the line occurs exactly at the zero of e^x - a.</p><p>Setting ln(a) = -1/(3a), we get ln(a) + 1/(3a) = 0. Let g(a) = ln(a) + 1/(3a) for a > 0. g'(a) = 1/a - 1/(3a²) = (3a-1)/(3a²). For a = 1/3: g(1/3) = ln(1/3) + 1 = -ln(3) + 1 ≈ -0.099 < 0. For a = 1: g(1) = 0 + 1/3 > 0. By continuity, exactly one solution exists in (1/3, 1), say a = a₀.</p><p>At a = a₀, both factors vanish at the same point x₀ = ln(a₀), making f(x₀) = 0 and f(x) ≥ 0 for all x. ✓</p><p><strong>Case 3 (a < 0):</strong> Then 3ax + 1 is a decreasing line, and e^x - a > 0 for all x. The product can be negative where 3ax + 1 < 0. ✗ This doesn't work.</p><p><strong>∴ Answer: 2</strong> (a = 0 and a = a₀ where ln(a₀) + 1/(3a₀) = 0)</p>
Correct Answer: 2