The function <span class="math-inline">\( f : \mathbb{R} \to \left[-\frac{1}{2}, \frac{1}{2}\right] \)</span> defined as <span class="math-inline">\( f(x) = \frac{x}{1+x^2} \)</span> is :
(1) neither injective nor surjective.
(2) invertible.
(3) injective but not surjective.
(4) surjective but not injective.
Step-by-Step Solution
Key Concept: To classify a function, we must verify injectivity (one-to-one) and surjectivity (onto). Injectivity requires f(a)=f(b)⟹a=b, while surjectivity requires every element in the codomain [-1/2, 1/2] has a preimage in ℝ.
<p><strong>Step 1: Check Injectivity</strong></p><p>For f to be injective, f(a) = f(b) must imply a = b.</p><p>Assume f(a) = f(b): $\frac{a}{1+a^2} = \frac{b}{1+b^2}$</p><p>Cross-multiplying: $a(1+b^2) = b(1+a^2)$</p><p>Expanding: $a + ab^2 = b + ba^2$</p><p>Rearranging: $a - b + ab^2 - ba^2 = 0$</p><p>$(a-b) + ab(b-a) = 0$</p><p>$(a-b)(1 - ab) = 0$</p><p>This gives either a = b OR ab = 1 (i.e., b = 1/a).</p><p>Counter-example: Let a = 2, then b = 1/2. Check: $f(2) = \frac{2}{1+4} = \frac{2}{5}$ and $f(1/2) = \frac{1/2}{1+1/4} = \frac{1/2}{5/4} = \frac{2}{5}$.</p><p>So f(2) = f(1/2) but 2 ≠ 1/2. Therefore, <strong>f is NOT injective</strong>.</p><p><strong>Step 2: Check Surjectivity</strong></p><p>We must verify that for every y ∈ [-1/2, 1/2], there exists x ∈ ℝ such that f(x) = y.</p><p>Let y = $\frac{x}{1+x^2}$. Rearranging: $y(1+x^2) = x$</p><p>$yx^2 - x + y = 0$</p><p>Using the quadratic formula (for y ≠ 0): $x = \frac{1 ± \sqrt{1-4y^2}}{2y}$</p><p>For real solutions to exist, we need: $1 - 4y^2 ≥ 0$</p><p>This gives: $4y^2 ≤ 1$, so $|y| ≤ \frac{1}{2}$, which means $y ∈ [-\frac{1}{2}, \frac{1}{2}]$.</p><p>Since this is exactly our codomain, every y in the codomain has at least one preimage in ℝ. Therefore, <strong>f is surjective</strong>.</p><p><strong>Step 3: Conclusion</strong></p><p>f is surjective but NOT injective. The function fails to be invertible because it is not bijective.</p><p><strong>∴ Answer: 4</strong></p>
Correct Answer: 4