<p>If \(I = \int_0^{1/2} \dfrac{1}{\sqrt{1-x^{2n}}}\,dx\), for \(n \ge 1\), then</p>
Step-by-Step Solution
Key Concept: As n increases, the integrand 1/√(1-x^(2n)) decreases for 0 < x < 1/2 (since x^(2n) decreases), making the integral decrease monotonically with n. The key is recognizing that 1/√(1-x^(2n)) ≤ 1/√(1-x²) for all n ≥ 1 when x ∈ [0,1/2].
<p><strong>Step 1:</strong> Analyze the integrand behavior. For x ∈ (0, 1/2), we have 0 < x < 1, so x^(2n) decreases as n increases.</p><p><strong>Step 2:</strong> Since x^(2n) decreases with n, the quantity (1 - x^(2n)) increases with n, making √(1 - x^(2n)) increase with n.</p><p><strong>Step 3:</strong> Therefore, 1/√(1 - x^(2n)) decreases as n increases for each fixed x ∈ (0, 1/2).</p><p><strong>Step 4:</strong> By the monotone convergence principle for integrals, I(n) is a decreasing function of n. That is: I₁ > I₂ > I₃ > ... as n = 1, 2, 3, ...</p><p><strong>Step 5:</strong> We can also verify: as n → ∞, x^(2n) → 0 for x ∈ [0, 1/2), so I(n) → π/6 (a finite limit). This confirms I decreases monotonically.</p><p>∴ Answer: <strong>B</strong> (The sequence is strictly decreasing)</p>
Correct Answer: B