Sets, Relations & Functions
Domain and Range of Functions
Grade 11
<p>Let \(f(x) = \ln(x^2 + ax + 1)\). If \(f(x)\) is defined \(\forall\, x \in R\), then the number of integers in the range of \(a\) is:</p>
Step-by-Step Solution
Key Concept: For f(x) = ln(x² + ax + 1) to be defined for all x ∈ ℝ, the argument x² + ax + 1 must be positive for all x. This requires the quadratic to have no real roots, meaning its discriminant must be negative.
<p><strong>Step 1:</strong> For f(x) = ln(x² + ax + 1) to be defined ∀x ∈ ℝ, we need x² + ax + 1 > 0 for all real x.</p><p><strong>Step 2:</strong> For a quadratic Ax² + Bx + C with A > 0 to be always positive, its discriminant must be negative: Δ < 0.</p><p><strong>Step 3:</strong> Here, Δ = a² - 4(1)(1) = a² - 4. We need: a² - 4 < 0</p><p><strong>Step 4:</strong> Solving a² < 4 gives |a| < 2, which means -2 < a < 2.</p><p><strong>Step 5:</strong> The integers in the interval (-2, 2) are: -1, 0, 1.</p><p><strong>Step 6:</strong> Number of integers = 3.</p><p>∴ Answer: B</p>
Correct Answer: B