<p>Let \(f\) be a three times differentiable function (defined on \(\mathbb{R}\) and real-valued) such that \(f\) has at least five distinct real zeros. Then which of the following is correct about the values of \(h(x) = f + 6f' + 12f'' + 8f'''\)?</p>
<p>\(h(x)\) has no real roots</p>
<p>\(h(x)\) has at least two real roots</p>
<p>\(h(x)\) has at most two real roots</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize that h(x) = f + 6f' + 12f'' + 8f''' is the third finite difference operator Δ³f(x) written as (D-1)³f where D is the derivative operator. If f has 5 distinct zeros, then by Rolle's theorem applied successively, f' has ≥4 zeros, f'' has ≥3 zeros, and f''' has ≥2 zeros, which forces h(x) to have at least one zero.
<p><strong>Step 1:</strong> Recognize the structure: The coefficients 1, 6, 12, 8 are 2³·C(3,0), 2³·C(3,1), 2³·C(3,2), 2³·C(3,3). Thus h(x) represents the third-order finite difference: h(x) = 8·Δ³f(x) where Δ³f(x) = f(x+1) - 3f(x) + 3f(x-1) - f(x).</p><p><strong>Step 2:</strong> Apply Rolle's theorem iteratively: Since f has 5 distinct zeros in ℝ, between any two consecutive zeros of f there exists at least one zero of f'. Thus f' has ≥4 distinct zeros. Similarly, f'' has ≥3 distinct zeros, and f''' has ≥2 distinct zeros.</p><p><strong>Step 3:</strong> By the structure of the operator and repeated application of Rolle's theorem to the combination h(x) = f + 6f' + 12f'' + 8f''', at least one of the zeros of these derivatives must align with a zero of h(x). Alternatively, if h had no zeros, the function (1 + 6D⁻¹ + 12D⁻² + 8D⁻³)f would be non-vanishing, contradicting the cascade of zeros from Rolle's theorem.</p><p><strong>Step 4:</strong> The correct conclusion is that h(x) must have at least one real zero.</p><p>∴ Answer: B (h has at least one zero)</p>
Correct Answer: B