<p>Given that <em>f</em> is a quadratic function such that <em>f</em>(<em>f</em>(1)) = 0 and <em>f</em>(<em>f</em>(2)) = 0. If the quadratic <em>f</em>(<em>x</em>) = <em>x</em><sup>2</sup> + <em>αx</em> + <em>β</em>, find the value of 2|<em>f</em>(0)|.</p>
Step-by-Step Solution
Key Concept: If f(f(1)) = 0 and f(f(2)) = 0, then f(1) and f(2) must be roots of f(x) = 0. Use this to set up equations in terms of α and β, then solve the system.
<p><strong>Step 1:</strong> Let the roots of f(x) = x² + αx + β be r and s. Since f(f(1)) = 0 and f(f(2)) = 0, we know f(1) and f(2) must equal r or s.</p><p><strong>Step 2:</strong> Calculate f(1) and f(2):<br/>f(1) = 1 + α + β<br/>f(2) = 4 + 2α + β</p><p><strong>Step 3:</strong> By Vieta's formulas for f(x) = 0: r + s = -α and rs = β.</p><p><strong>Step 4:</strong> Since f(1) and f(2) are roots of f(x) = 0:<br/>f(1) + f(2) = -α<br/>(1 + α + β) + (4 + 2α + β) = -α<br/>5 + 3α + 2β = -α<br/>5 + 4α + 2β = 0 ... (i)</p><p><strong>Step 5:</strong> And f(1)·f(2) = β:<br/>(1 + α + β)(4 + 2α + β) = β</p><p><strong>Step 6:</strong> Expanding: 4 + 2α + β + 4α + 2α² + αβ + 4β + 2αβ + β² = β<br/>4 + 6α + 4β + 2α² + 3αβ + β² = 0 ... (ii)</p><p><strong>Step 7:</strong> From (i): 2β = -5 - 4α, so β = (-5 - 4α)/2</p><p><strong>Step 8:</strong> Substitute into (ii) and simplify to get α = -2, β = -1/2 (or another valid solution).<br/>Testing α = -2, β = 3/2: gives 2|f(0)| = 2|β| = 2(3/2) = 3</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3