Sets, Relations & Functions
Range of a function
Grade 11
Question:
<p>Let \(f: R \to R\) be defined by \(f(x) = \dfrac{x}{1+x^2}\), \(x \in R\). Then the range of \(f\) is:</p>
<p>\(R - \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\)</p>
<p>\(\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\)</p>
<p>\((-1, 1) - \{0\}\)</p>
<p>\(R - [-1, 1]\)</p>
Step-by-Step Solution
Key Concept: Find the range by setting y = f(x) and solving for x in terms of y, then determine which y values yield real solutions. The discriminant condition determines the bounds.
<p><strong>Step 1:</strong> Let y = f(x) = x/(1+x²), where y ∈ ℝ and x ∈ ℝ</p><p><strong>Step 2:</strong> Rearrange to get: y(1+x²) = x → yx² - x + y = 0</p><p><strong>Step 3:</strong> For real solutions in x to exist, the discriminant must be ≥ 0:</p><p>Δ = (-1)² - 4(y)(y) ≥ 0</p><p>1 - 4y² ≥ 0</p><p>4y² ≤ 1</p><p>y² ≤ 1/4</p><p><strong>Step 4:</strong> Taking square root: |y| ≤ 1/2</p><p>Therefore: -1/2 ≤ y ≤ 1/2</p><p><strong>Step 5:</strong> Verify endpoints are achievable:</p><p>• When y = 1/2: x² - 2x + 1 = 0 → x = 1 ✓</p><p>• When y = -1/2: x² + 2x + 1 = 0 → x = -1 ✓</p><p>∴ Range of f is <strong>[-1/2, 1/2]</strong> (Answer: B)</p>
Correct Answer: B