Binomial Theorem
Approximations using Binomial expansion
Grade 11
Question:
<p>If \(x\) is so small that \(x^3\) and higher powers of \(x\) may be neglected, then \(\dfrac{(1+x)^{3/2} - \left(1 + \dfrac{1}{2}x\right)^3}{(1-x)^{1/2}}\) may be approximated as</p>
<p>(1) \(3x + \dfrac{3}{8}x^2\)</p>
<p>(2) \(1 - \dfrac{3}{8}x^2\)</p>
<p>(3) \(\dfrac{x}{2} - \dfrac{3}{8}x^2\)</p>
<p>(4) \(-\dfrac{3}{8}x^2\)</p>
Step-by-Step Solution
Key Concept: Use binomial expansion for fractional powers up to x² terms only, then divide by the expansion of (1-x)^(-1/2) to get the final approximation.
<p><strong>Step 1:</strong> Expand (1+x)^(3/2) using binomial theorem: (1+x)^(3/2) = 1 + (3/2)x + (3/2)(1/2)/2! · x² + ... = 1 + (3/2)x + (3/8)x² + ...</p><p><strong>Step 2:</strong> Expand (1 + x/2)³ = 1 + 3(x/2) + 3(x/2)² + ... = 1 + (3/2)x + (3/4)x² + ...</p><p><strong>Step 3:</strong> Find the numerator: (1+x)^(3/2) - (1 + x/2)³ = [1 + (3/2)x + (3/8)x²] - [1 + (3/2)x + (3/4)x²] = (3/8 - 3/4)x² = -(3/8)x²</p><p><strong>Step 4:</strong> Expand denominator (1-x)^(1/2) = 1 - (1/2)x - (1/8)x² + ..., so (1-x)^(-1/2) = 1 + (1/2)x + (3/8)x² + ...</p><p><strong>Step 5:</strong> Divide numerator by denominator: [-(3/8)x²] · [1 + (1/2)x + ...] ≈ -(3/8)x²</p><p>∴ Answer: <strong>D</strong> (which should be <strong>-3x²/8</strong>)</p>
Correct Answer: D