Sets, Relations & Functions
Domain of a function
Grade 11
Question:
<p>The domain of the real valued function \(f(x)\) for which \(4^{f(x)} + 4^{1-f(x)} = 4^x\) is</p>
<p>(a) \([-1, 1]\)</p>
<p>(b) \([1, +\infty)\)</p>
<p>(c) \((-\infty, -1]\)</p>
<p>(d) \((-\infty, -1]\)</p>
Step-by-Step Solution
Key Concept: Rewrite the equation using substitution t = 4^(f(x)), then recognize it as a quadratic in t. For real f(x), the discriminant of the resulting quadratic must be non-negative, which constrains x.
<p><strong>Step 1:</strong> Let t = 4^(f(x)). Since f(x) is real-valued, t must be positive (t > 0).</p><p><strong>Step 2:</strong> Rewrite the equation: 4^(f(x)) + 4^(1-f(x)) = 4^x becomes t + 4/t = 4^x</p><p><strong>Step 3:</strong> Multiply by t: t² - 4^x · t + 4 = 0</p><p><strong>Step 4:</strong> For t to be real and positive, the discriminant must be non-negative:<br/>Δ = (4^x)² - 16 ≥ 0<br/>4^(2x) - 16 ≥ 0<br/>4^(2x) ≥ 16 = 4²<br/>2x ≥ 2<br/>x ≥ 1</p><p><strong>Step 5:</strong> Also verify both roots are positive using Vieta's formulas: product = 4 > 0 ✓, and sum = 4^x > 0 ✓</p><p><strong>Step 6:</strong> Check that t ≠ 4^(1/2) = 2 at boundary (which would make the discriminant = 0 and f(x) = 1/2 uniquely). At x = 1: t² - 4t + 4 = (t-2)² = 0, so t = 2 is valid.</p><p>∴ Domain: <strong>x ≥ 1</strong> or <strong>[1, ∞)</strong></p>
Correct Answer: B