<p>The positive value of \(\lambda\) for which the co-efficient of \(x^2\) in the expression \(x^2\left(\sqrt{x}+\dfrac{\lambda}{x^2}\right)^{10}\) is 720, is:</p>
Step-by-Step Solution
Key Concept: Simplify the expression to x^2(x^(1/2) + λx^(-2))^10, then use the binomial theorem to find the general term. The power of x in the final expansion is 2 + (1/2)r - 2(10-r), which must equal 2 to get the coefficient of x^2.
<p><strong>Step 1:</strong> Rewrite the expression as x^2·(√x + λ/x²)^10 = x^2·(x^(1/2) + λx^(-2))^10</p><p><strong>Step 2:</strong> The general term in the binomial expansion of (x^(1/2) + λx^(-2))^10 is: T_(r+1) = C(10,r)·(x^(1/2))^(10-r)·(λx^(-2))^r = C(10,r)·λ^r·x^((10-r)/2 - 2r)</p><p><strong>Step 3:</strong> When multiplied by x^2, the power of x becomes: 2 + (10-r)/2 - 2r = 2 + 5 - r/2 - 2r = 7 - 5r/2</p><p><strong>Step 4:</strong> For coefficient of x^2: 7 - 5r/2 = 2 ⟹ 5r/2 = 5 ⟹ r = 2</p><p><strong>Step 5:</strong> The coefficient of x^2 is: C(10,2)·λ^2 = 45λ^2 = 720</p><p><strong>Step 6:</strong> λ^2 = 16 ⟹ λ = 4 (taking positive value)</p><p>∴ Answer: A</p>
Correct Answer: A