<p>If \(a > 0\), then least value of \((a^3 + a^2 + a + 1)^2\) is</p>
Step-by-Step Solution
Key Concept: Express the quartic form as a sum of squares by completing the square, then minimize using AM-GM inequality or calculus. The constraint a > 0 is crucial for determining the global minimum.
<p><strong>Step 1:</strong> Factor the expression inside the square.</p><p>a³ + a² + a + 1 = a²(a + 1) + 1(a + 1) = (a² + 1)(a + 1)</p><p><strong>Step 2:</strong> Rewrite the original expression.</p><p>(a³ + a² + a + 1)² = [(a² + 1)(a + 1)]² = (a² + 1)²(a + 1)²</p><p><strong>Step 3:</strong> Let f(a) = (a² + 1)²(a + 1)² for a > 0. Differentiate using the product rule.</p><p>f'(a) = 2(a² + 1)(2a)(a + 1)² + (a² + 1)² · 2(a + 1)</p><p>f'(a) = 2(a + 1)(a² + 1)[2a(a + 1) + (a² + 1)]</p><p>f'(a) = 2(a + 1)(a² + 1)[2a² + 2a + a² + 1]</p><p>f'(a) = 2(a + 1)(a² + 1)(3a² + 2a + 1)</p><p><strong>Step 4:</strong> For a > 0, all factors are positive, so f'(a) > 0 for all a > 0.</p><p>This means f(a) is strictly increasing on (0, ∞).</p><p><strong>Step 5:</strong> Find the infimum as a → 0⁺.</p><p>lim(a→0⁺) (a² + 1)²(a + 1)² = (1)²(1)² = 1</p><p>Since f is continuous and increasing on (0, ∞), the greatest lower bound (infimum) is <strong>1</strong>.</p><p>∴ Answer: D (Least value = 1)</p>
Correct Answer: D