Relations & Functions
One-to-One and Many-to-One Functions
Grade 12
Question:
<p><strong>Ex. 1</strong> Let $f(x) = \frac{a_{2k}x^{2k} + a_{2k-1}x^{2k-1} + \ldots + a_1x + a_0}{b_{2k}x^{2k} + b_{2k-1}x^{2k-1} + \ldots + b_1x + b_0}$ where $k$ is a positive integer, $a_i, b_i \in \mathbb{R}$ and $a_{2k} \neq 0$, $b_{2k} \neq 0$ such that $b_{2k}x^{2k} + b_{2k-1}x^{2k-1} + \ldots + b_1x + b_0 = 0$ has no real roots, then</p>
<p>(a) $f(x)$ must be one to one</p>
<p>(b) $a_{2k}x^{2k} + a_{2k-1}x^{2k-1} + \ldots + a_1x + a_0 = 0$ must have real roots</p>
<p>(c) $f(x)$ must be many to one</p>
<p>(d) Nothing can be said about the above options</p>
Step-by-Step Solution
Key Concept: A continuous function where both limits at infinity equal the same value cannot be injective, so it must be many-to-one.
<p><strong>Solution:</strong> $f(x)$ is continuous for all $x \in \mathbb{R}$.</p><p>Since the denominator $b_{2k}x^{2k} + b_{2k-1}x^{2k-1} + \ldots + b_1x + b_0 = 0$ has no real roots and $b_{2k} \neq 0$, the denominator is always positive (or always negative).</p><p>$$\lim_{x \to \infty} f(x) = \frac{a_{2k}}{b_{2k}}$$</p><p>$$\lim_{x \to -\infty} f(x) = \frac{a_{2k}}{b_{2k}}$$</p><p>Both limits approach the same finite value. Since $f(x)$ is continuous and bounded, it cannot be one-to-one. Therefore, $f(x)$ must be many to one.</p><p>∴ Answer is (c).</p>
Correct Answer: C