<p>If \(f(x) = g(x)|(x-1)(x-2)\cdots(x-10)| - 2\) is derivable for all \(x \in R\), where \(g(x) = ax^9 + bx^6 + cx^3 + d,\; a, b, c, d \in R\), then \(f'(-1)\) is equal to:</p>
Step-by-Step Solution
Key Concept: For f(x) to be derivable at points where |(x-1)(x-2)···(x-10)| has corners (at x=1,2,...,10), the function g(x)|(x-1)(x-2)···(x-10)| must have zero value AND zero derivative at these points. This constrains the coefficients of g(x).
<p><strong>Step 1: Condition for Differentiability</strong></p><p>The product |(x-1)(x-2)···(x-10)| has corners at x=1,2,...,10. For f(x) = g(x)|(x-1)(x-2)···(x-10)| - 2 to be derivable everywhere, g(x)|(x-1)(x-2)···(x-10)| must be differentiable at these points.</p><p><strong>Step 2: Apply Continuity and Differentiability Conditions</strong></p><p>At x=k (k=1,2,...,10), the absolute value has a corner. For differentiability, we need:</p><ul><li>g(k)·0 = 0 (automatically satisfied)</li><li>g(k) must equal 0 for the derivative to exist at the corner</li></ul><p>So g(1) = g(2) = g(3) = ··· = g(10) = 0.</p><p><strong>Step 3: Determine g(x)</strong></p><p>Since g(x) = ax⁹ + bx⁶ + cx³ + d has degree 9 and has 10 roots (1,2,...,10), this is impossible unless we reconsider. Actually, g(x) must have (x-1)(x-2)···(x-10) as a factor, but g has degree only 9. This forces g(x) ≡ 0.</p><p><strong>Step 4: Reconsider Using Symmetry</strong></p><p>Note that g(x) = ax⁹ + bx⁶ + cx³ + d contains only odd and constant powers. For g(x) to vanish at 10 distinct points with degree 9 is impossible, so actually the constraint is that g(x) = 0.</p><p>Therefore: f(x) = 0·|(x-1)(x-2)···(x-10)| - 2 = -2</p><p><strong>Step 5: Calculate f'(-1)</strong></p><p>Since f(x) = -2 (constant), f'(-1) = 0</p><p>∴ Answer: <strong>C (which is 0)</strong></p>
Correct Answer: C