Relations & Functions
Injective and Surjective Functions
Grade 12
Question:
<p>Let a function <mi>f</mi> <mo>:</mo> <mo>(</mo> <mn>0</mn> <mo>,</mo> <mi>∞</mi> <mo>)</mo> <mo>→</mo> <mo>(</mo> <mn>0</mn> <mo>,</mo> <mi>∞</mi> <mo>)</mo> be defined by \(f(x) = 1 - \frac{1}{x}\). Then, <mi>f</mi> is</p>
<p>(a) injective only</p>
<p>(b) both injective as well as surjective</p>
<p>(c) not injective but it is surjective</p>
<p>(d) neither injective nor surjective</p>
Step-by-Step Solution
Key Concept: Analyze the piecewise behavior of the function by examining its graph to determine if it is injective (one-to-one) and surjective (onto).
<p><strong>Solution:</strong> We have $f(x) = \frac{|x-1|}{x} = \begin{cases} \frac{1}{x} - 1, & \text{if } 0 < x \leq 1 \\ 1 - \frac{1}{x}, & \text{if } x > 1 \end{cases}$</p><p>From the graph: when $x \to 0$, then $f(x) \to \infty$; when $x = 1$, then $f(x) = 0$; and when $x \to \infty$, then $f(x) \to 1$.</p><p>Clearly, $f(x)$ is not injective because if $f(x) < 1$, then $f$ is many-one, as shown in the figure.</p><p>Also, $f(x)$ is not surjective because the range of $f(x)$ is $[0, \infty)$, not $(0, \infty)$.</p><p>∴ Answer is (d).</p>
Correct Answer: D