<p>The domain of definition of \(f(x) = \log_2\left(\frac{2x^2 - 7x + 9}{x^2 - x + 1}\right)\) is:</p>
<p>(a) \(\mathbb{R}\)</p>
<p>(b) \(\mathbb{R} - \{0\}\)</p>
<p>(c) \(\mathbb{R} - \{0, 1\}\)</p>
<p>(d) \(\mathbb{R} - \{1\}\)</p>
Step-by-Step Solution
Key Concept: For f(x) = log₂(g(x)) to be defined, we need g(x) > 0. We must check that the argument of the logarithm is always positive by analyzing both the numerator and denominator.
<p><strong>Step 1: Identify the condition for the domain.</strong></p><p>For f(x) = log₂(h(x)) to be defined, we need h(x) > 0, where h(x) = (2x² - 7x + 9)/(x² - x + 1).</p><p><strong>Step 2: Check if the denominator can be zero.</strong></p><p>Examine x² - x + 1. Its discriminant is Δ = (-1)² - 4(1)(1) = 1 - 4 = -3 < 0.</p><p>Since the discriminant is negative and the leading coefficient is positive, x² - x + 1 > 0 for all real x.</p><p>Therefore, the denominator is never zero and always positive.</p><p><strong>Step 3: Check if the numerator is always positive.</strong></p><p>Examine 2x² - 7x + 9. Its discriminant is Δ = (-7)² - 4(2)(9) = 49 - 72 = -23 < 0.</p><p>Since the discriminant is negative and the leading coefficient is positive (2 > 0), we have 2x² - 7x + 9 > 0 for all real x.</p><p><strong>Step 4: Conclude about the fraction.</strong></p><p>Since both numerator > 0 and denominator > 0 for all x ∈ ℝ, the fraction (2x² - 7x + 9)/(x² - x + 1) > 0 for all x ∈ ℝ.</p><p>Therefore, the argument of the logarithm is always positive, and f(x) is defined for all real numbers.</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A