Sets, Relations & Functions
Range of a function
Grade 11
Question:
<p>The range of the function \(f(x) = x^2 + \dfrac{1}{x^2 + 1}\) is:</p>
<p>(a) \([1, \infty)\)</p>
<p>(b) \([2, \infty)\)</p>
<p>(c) \(\left[\dfrac{3}{2}, \infty\right)\)</p>
<p>(d) \([5, \infty)\)</p>
Step-by-Step Solution
Key Concept: Substitute y = f(x) and treat it as a constraint equation in x. For real solutions to exist, the discriminant of the resulting quadratic must be non-negative, which directly determines the range of y.
<p><strong>Step 1:</strong> Let y = f(x) = x² + 1/(x²+1). Rearrange:</p><p>y = x² + 1/(x²+1)</p><p><strong>Step 2:</strong> Multiply by (x²+1): y(x²+1) = x²(x²+1) + 1</p><p>yx² + y = x⁴ + x² + 1</p><p>x⁴ + x² - yx² + 1 - y = 0</p><p>x⁴ + (1-y)x² + (1-y) = 0</p><p><strong>Step 3:</strong> Let u = x² ≥ 0. Then: u² + (1-y)u + (1-y) = 0</p><p>For real x to exist, this quadratic in u must have at least one non-negative real root.</p><p><strong>Step 4:</strong> Discriminant: Δ = (1-y)² - 4(1-y) ≥ 0</p><p>(1-y)² - 4(1-y) ≥ 0</p><p>(1-y)[(1-y) - 4] ≥ 0</p><p>(1-y)(-3-y) ≥ 0</p><p><strong>Step 5:</strong> This gives: (y-1)(y+3) ≤ 0, so -3 ≤ y ≤ 1</p><p>But note: x²+1 ≥ 1, so 1/(x²+1) ≤ 1. Also x² ≥ 0.</p><p>At x=0: f(0) = 0 + 1 = 1 (maximum achieved)</p><p>As |x|→∞: f(x)→∞</p><p><strong>Step 6:</strong> Reconsidering: Since x²≥0 and 1/(x²+1)∈(0,1], the range is [1, ∞)</p><p>∴ Answer: A</p>
Correct Answer: A