<p>If \(f(x)\) is continuous and differentiable in \([-3, 9]\) and \(f'(x) \in [-2, 8]\) \(\forall x \in (-3, 9)\). Let \(N\) be the number of divisors of the greatest possible value of \(f(9) - f(-3)\), then find the sum of digits of \(N\).</p>
Step-by-Step Solution
Key Concept: By the Mean Value Theorem, f(9) - f(-3) = f'(c) · (9 - (-3)) for some c ∈ (-3, 9). Since f'(x) ∈ [-2, 8], the maximum value of f(9) - f(-3) is 8 × 12 = 96.
<p><strong>Step 1:</strong> Apply the Mean Value Theorem. Since f(x) is continuous on [-3, 9] and differentiable on (-3, 9), by MVT there exists c ∈ (-3, 9) such that:</p><p>f'(c) = [f(9) - f(-3)] / [9 - (-3)] = [f(9) - f(-3)] / 12</p><p><strong>Step 2:</strong> Rearrange to express the difference:</p><p>f(9) - f(-3) = f'(c) · 12</p><p><strong>Step 3:</strong> Determine the range of f(9) - f(-3). Since f'(x) ∈ [-2, 8] for all x ∈ (-3, 9), and f'(c) must lie in this interval:</p><p>f(9) - f(-3) = f'(c) · 12, where f'(c) ∈ [-2, 8]</p><p><strong>Step 4:</strong> Find the maximum value. The greatest possible value occurs when f'(c) = 8:</p><p>max[f(9) - f(-3)] = 8 × 12 = 96</p><p><strong>Step 5:</strong> Find the number of divisors of 96. First, find the prime factorization:</p><p>96 = 32 × 3 = 2⁵ × 3¹</p><p><strong>Step 6:</strong> Use the divisor formula. If n = p₁^a₁ × p₂^a₂ × ... × pₖ^aₖ, then the number of divisors is:</p><p>N = (a₁ + 1)(a₂ + 1)...(aₖ + 1)</p><p>N = (5 + 1)(1 + 1) = 6 × 2 = 12</p><p><strong>Step 7:</strong> Find the sum of digits of N = 12:</p><p>Sum of digits = 1 + 2 = 3</p><p><strong>∴ Answer: 3</strong></p>
Correct Answer: 3