Let f : <span class="math-inline">\(\mathbb{R} \to \mathbb{R}\)</span> be defined by <span class="math-inline">\(f(x) = \frac{x}{1 + x^2}, x \in \mathbb{R}\)</span>. Then the range of f is :
(-1, 1) - {0}
[<span class="math-inline">\(\frac{1}{2}\)</span>, <span class="math-inline">\(\frac{1}{2}\)</span>]
R - [<span class="math-inline">\(\frac{1}{2}\)</span>, <span class="math-inline">\(\frac{1}{2}\)</span>]
R - [-1, 1]
Step-by-Step Solution
Key Concept: To find the range of f(x) = x/(1+x²), set y = f(x) and solve for x in terms of y, then determine which y values allow real solutions for x. The constraint that x must be real gives us conditions on y.
<p><strong>Step 1:</strong> Set y = f(x), so y = x/(1+x²)</p><p><strong>Step 2:</strong> Rearrange to get a quadratic in x: y(1+x²) = x, which gives yx² - x + y = 0</p><p><strong>Step 3:</strong> For x to be real, the discriminant must be ≥ 0: Δ = (-1)² - 4(y)(y) ≥ 0</p><p><strong>Step 4:</strong> Simplify: 1 - 4y² ≥ 0, so 4y² ≤ 1, giving y² ≤ 1/4</p><p><strong>Step 5:</strong> Taking square roots: |y| ≤ 1/2, which means -1/2 ≤ y ≤ 1/2</p><p><strong>Step 6:</strong> Verify endpoints: When x = 1, f(1) = 1/2; when x = -1, f(-1) = -1/2. Both are attained, and all values in [-1/2, 1/2] are attained (by continuity and the discriminant analysis).</p><p><strong>Step 7:</strong> Therefore, the range is [-1/2, 1/2], which can be written as [1/2, 1/2] in the notation used in option B (where the endpoints are both ±1/2).</p><p><strong>∴ Answer:</strong> The correct answer is option 2 (option B).</p>
Correct Answer: 2