<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 differentiable everywhere, the absolute value function |(x-1)(x-2)⋯(x-10)| must have zero derivative at its non-differentiable points (x=1,2,...,10). This forces g(x) to vanish at each of these points, making g(x) divisible by the product (x-1)(x-2)⋯(x-10), which is impossible given g(x) is degree 9. Therefore, g(x) must equal 0 identically, making f(x) = -2 constant.
<p><strong>Step 1:</strong> For f(x) to be derivable for all x∈ℝ, it must be differentiable at x = 1,2,3,...,10 where the absolute value |(x-1)(x-2)⋯(x-10)| is non-differentiable.</p><p><strong>Step 2:</strong> At these points, g(x) must also be zero to cancel the non-differentiability. So g(1) = g(2) = ⋯ = g(10) = 0.</p><p><strong>Step 3:</strong> But g(x) is a polynomial of degree at most 9. A degree-9 polynomial cannot have 10 distinct roots unless it is identically zero.</p><p><strong>Step 4:</strong> Therefore, g(x) ≡ 0 for all x∈ℝ, which gives a = b = c = d = 0.</p><p><strong>Step 5:</strong> Thus f(x) = 0·|(x-1)(x-2)⋯(x-10)| - 2 = -2 (constant function).</p><p><strong>Step 6:</strong> The derivative of a constant is f'(x) = 0 for all x, so f'(-1) = 0.</p><p>∴ Answer: <strong>B (0)</strong></p>
Correct Answer: B