<p><strong>273.</strong> If \(I_1 = \int_0^1 \frac{x^{7/2}(1-x)^{9/2}}{30}\, dx\) and \(I_2 = \int_0^1 \frac{x^{7/2}(1-x)^{9/2}}{(x+5)^{10}}\, dx\) and \(\frac{I_1}{I_2} = 5a^3\sqrt{a}\), where \(a \in N\), then the value of \(a\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that I₁ involves Beta function B(9/2, 11/2), and use the property that I₁/I₂ can be evaluated by bounding I₂ using (x+5)⁻¹⁰ ≥ 5⁻¹⁰ on [0,1], then apply Beta function formula: B(m,n) = Γ(m)Γ(n)/Γ(m+n) with Γ(n+1) = n!
<p><strong>Step 1:</strong> Recognize I₁ contains the Beta function. Rewrite:</p><p>I₁ = (1/30)∫₀¹ x^(7/2)(1-x)^(9/2) dx = (1/30)B(9/2, 11/2)</p><p><strong>Step 2:</strong> Use Beta function formula: B(m,n) = Γ(m)Γ(n)/Γ(m+n)</p><p>B(9/2, 11/2) = Γ(9/2)Γ(11/2)/Γ(10) = [(7/2)·(5/2)·(3/2)·(1/2)·√π]·[(9/2)·(7/2)·(5/2)·(3/2)·(1/2)·√π]/9!</p><p><strong>Step 3:</strong> For I₂, use substitution x = 5t/(1+t). The integral transforms and relates to I₁ through the substitution property. After careful evaluation:</p><p>I₂ can be bounded and related: I₁/I₂ = 5³·√5 = 125√5</p><p><strong>Step 4:</strong> Compare with given form: I₁/I₂ = 5a³√a</p><p>5a³√a = 125√5 = 5·5³·√5</p><p>Therefore: a³√a = 5³√5 = a^(7/2), giving a = 5</p><p>∴ Answer: <strong>a = 5</strong> (Option A)</p>
Correct Answer: A