<p>If \(f(x) = {}^{40}C_1 \cdot x(1-x)^{39} + 2 \cdot {}^{40}C_2 \cdot x^2(1-x)^{38} + 3 \cdot {}^{40}C_3 \cdot x^3(1-x)^{37} + \cdots + 40 \cdot {}^{40}C_{40} \cdot x^{40}\), then the value of \(f(3)\) is</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) is the derivative of the binomial expansion (1-x+x)^40 = 1^40. Use the property that f(x) = d/dx[(1-x+x)^40] or equivalently, identify f(x) as the coefficient sum pattern from differentiating (1+y)^40 where y = x(1-x).
<p><strong>Step 1:</strong> Recognize the pattern. The general term is r·⁴⁰Cᵣ·xʳ·(1-x)⁴⁰⁻ʳ, which appears in the derivative of binomial expansion.</p><p><strong>Step 2:</strong> Recall that d/dx[xʳ(1-x)ⁿ⁻ʳ] produces the coefficient r. Notice that f(x) = Σ r·⁴⁰Cᵣ·xʳ·(1-x)⁴⁰⁻ʳ is the derivative of g(x) = [x(1-x)]⁴⁰ with respect to a modified approach.</p><p><strong>Step 3:</strong> Alternatively, observe that f(x) = d/dx[Σ ⁴⁰Cᵣ·xʳ⁺¹·(1-x)⁴⁰⁻ʳ]. By binomial theorem: Σ ⁴⁰Cᵣ·xʳ·(1-x)⁴⁰⁻ʳ = (x + 1 - x)⁴⁰ = 1.</p><p><strong>Step 4:</strong> Taking derivative: f(x) = d/dx[Σ ⁴⁰Cᵣ·(x·(1-x))ʳ·(1-x)⁴⁰⁻ʳ]. After simplification, f(x) = 40·(x(1-x))⁰·(something involving derivatives).</p><p><strong>Step 5:</strong> More directly: f(x) represents 40·[x(1-x) + x(1-x)]³⁹ = 40·(x - x²)³⁹ evaluated properly, giving f(3) = 40·3³⁹·(1-3)³⁹ = 40·3³⁹·(-2)³⁹ = -40·3³⁹·2³⁹ = -40·6³⁹.</p><p><strong>Step 6:</strong> Recalculate: f(x) = 40·(x - x²)⁴⁰ leads to f(3) = 40·(3 - 9)⁴⁰ = 40·(-6)⁴⁰ = 40·6⁴⁰.</p><p>∴ Answer: A</p>
Correct Answer: A