<p>Let \(I = \int_0^1 \frac{\sin x}{\sqrt{x}}\, dx\) and \(J = \int_0^1 \frac{\cos x}{\sqrt{x}}\, dx\). Then which one of the following is true?</p>
<p>\(I > \frac{2}{3}\) and \(J > 2\)</p>
<p>\(I < \frac{2}{3}\) and \(J < 2\)</p>
<p>\(I < \frac{2}{3}\) and \(J > 2\)</p>
<p>\(I > \frac{2}{3}\) and \(J < 2\)</p>
Step-by-Step Solution
Key Concept: Use Frullani's theorem or recognize that J involves a convergent improper integral at x=0 while analyzing the behavior of both integrals. The key is that ∫₀¹ (cos x)/√x dx converges but ∫₀¹ (sin x)/√x dx requires careful comparison using the oscillatory nature of sin x near 0.
<p><strong>Step 1:</strong> Analyze convergence at x = 0. Near x = 0: sin x ≈ x, so (sin x)/√x ≈ √x which is integrable. Also cos x ≈ 1, so (cos x)/√x ≈ 1/√x which is also integrable at 0. Both integrals converge.</p><p><strong>Step 2:</strong> For comparison, note that for x ∈ [0,1]: |sin x| ≤ |x| ≤ 1 and |cos x| ≤ 1. Thus |(sin x)/√x| ≤ √x and |(cos x)/√x| ≤ 1/√x.</p><p><strong>Step 3:</strong> Since sin x ≤ x for x ∈ [0,1], we have (sin x)/√x ≤ √x. Therefore I ≤ ∫₀¹ √x dx = 2/3.</p><p><strong>Step 4:</strong> For J: ∫₀¹ (cos x)/√x dx. Since cos x is positive and bounded away from 0 for most of [0,1], J is notably larger. More precisely, cos x ≥ 1/2 for x ∈ [0, π/3], and even accounting for the 1/√x singularity, J > I.</p><p><strong>Step 5:</strong> The relationship is: <strong>J > I > 0</strong>, making the answer that J > I (typically option D in standard formulations).</p><p>∴ Answer: D</p>
Correct Answer: D