<p>The term independent of \(x\) in the binomial expansion of \(\left(1-\dfrac{1}{x}+3x^5\right)\left(2x^2-\dfrac{1}{x}\right)^8\) is</p>
Step-by-Step Solution
Key Concept: Expand (2x² - 1/x)⁸ using binomial theorem to find the general term, then multiply by (1 - 1/x + 3x⁵) and identify which combinations yield x⁰ (constant terms).
<p><strong>Step 1:</strong> Find general term in (2x² - 1/x)⁸</p><p>T_{r+1} = C(8,r)(2x²)^{8-r}(-1/x)^r = C(8,r)·2^{8-r}·(-1)^r·x^{16-2r-r} = C(8,r)·2^{8-r}·(-1)^r·x^{16-3r}</p><p><strong>Step 2:</strong> For constant term from x^{16-3r}, we need 16-3r = 0 ⟹ r = 16/3 (not integer, so no constant from this alone)</p><p><strong>Step 3:</strong> Multiply by (1 - 1/x + 3x⁵). The product yields constants when:</p><p>• 1 × [term with x⁰] from expansion</p><p>• (-1/x) × [term with x¹] from expansion</p><p>• 3x⁵ × [term with x⁻⁵] from expansion</p><p><strong>Step 4:</strong> Find required powers in (2x² - 1/x)⁸:</p><p>• x⁰: 16-3r=0 ⟹ no integer r</p><p>• x¹: 16-3r=1 ⟹ r=5, term = C(8,5)·2³·(-1)⁵ = 56·8·(-1) = -448</p><p>• x⁻⁵: 16-3r=-5 ⟹ r=7, term = C(8,7)·2¹·(-1)⁷ = 8·2·(-1) = -16</p><p><strong>Step 5:</strong> Constant term = 1·0 + (-1/x)·(-448)x + 3x⁵·(-16)x⁻⁵ = 0 + 448 - 48 = 400</p><p>∴ Answer: B</p>
Correct Answer: B