Question

Let $f(x) = \frac{(x-\gamma)(x-\delta)}{(x-\alpha)(x-\beta)}$, where $\gamma < \alpha < \delta < \beta$, then which of the following is/are correct for $f(x)$?

• A. $f: R - \{\alpha, \beta\} \to R$ is onto function
• B. $f: R - \{\alpha, \beta\} \to R$ is into function
• C. $f: R - \{\alpha, \beta\} \to R$ is many-one function
• D. $f: R - \{\alpha, \beta\} \to R$ is one-one function

Answer 1

Answer: (A), (C)

f: R - {α, β} → R is onto function, f: R - {α, β} → R is many-one function

Answer 2

The given function is $f(x) = \frac{(x-\gamma)(x-\delta)}{(x-\alpha)(x-\beta)}$, with the condition $\gamma < \alpha < \delta < \beta$. The domain of the function is $R - \{\alpha, \beta\}$, as the denominator cannot be zero.

1. Behavior as $x \to \pm \infty$:
   As $x \to \pm \infty$, $f(x) = \frac{x^2 - (\gamma+\delta)x + \gamma\delta}{x^2 - (\alpha+\beta)x + \alpha\beta} \to \frac{x^2}{x^2} = 1$.
   So, $y=1$ is a horizontal asymptote.

2. Behavior near vertical asymptotes:

   • As $x \to \alpha^+$:
     Numerator: $(x-\gamma)(x-\delta) \to (\alpha-\gamma)(\alpha-\delta)$. Since $\gamma < \alpha < \delta$, $\alpha-\gamma > 0$ and $\alpha-\delta < 0$. So, numerator $\to \text{negative}$.
     Denominator: $(x-\alpha)(x-\beta) \to (0^+)(\alpha-\beta)$. Since $\alpha < \beta$, $\alpha-\beta < 0$. So, denominator $\to 0^-$.
     Thus, $f(x) \to \frac{\text{negative}}{\text{negative small}} \to +\infty$.
   • As $x \to \alpha^-$:
     Numerator: $(\alpha-\gamma)(\alpha-\delta) \to \text{negative}$.
     Denominator: $(x-\alpha)(x-\beta) \to (0^-)(\alpha-\beta) \to 0^+$.
     Thus, $f(x) \to \frac{\text{negative}}{\text{positive small}} \to -\infty$.
   • As $x \to \beta^+$:
     Numerator: $(x-\gamma)(x-\delta) \to (\beta-\gamma)(\beta-\delta)$. Since $\gamma < \delta < \beta$, both $\beta-\gamma > 0$ and $\beta-\delta > 0$. So, numerator $\to \text{positive}$.
     Denominator: $(x-\alpha)(x-\beta) \to (\beta-\alpha)(0^+)$. Since $\alpha < \beta$, $\beta-\alpha > 0$. So, denominator $\to 0^+$.
     Thus, $f(x) \to \frac{\text{positive}}{\text{positive small}} \to +\infty$.
   • As $x \to \beta^-$:
     Numerator: $(\beta-\gamma)(\beta-\delta) \to \text{positive}$.
     Denominator: $(x-\alpha)(x-\beta) \to (\beta-\alpha)(0^-) \to 0^-$.
     Thus, $f(x) \to \frac{\text{positive}}{\text{negative small}} \to -\infty$.

3. Roots of $f(x)$:
   $f(x)=0$ when $(x-\gamma)(x-\delta)=0$, so $x=\gamma$ or $x=\delta$. These are the x-intercepts.

4. Injectivity (One-one/Many-one):
   Since $f(x) \to 1$ as $x \to -\infty$ and $f(x) \to 1$ as $x \to +\infty$, and the function takes values other than 1 (e.g., $f(\gamma)=0$, $f(\delta)=0$), it cannot be one-one.
   Also, $f(x)=1$ has a solution $x_0 = \frac{\gamma\delta-\alpha\beta}{\gamma+\delta-(\alpha+\beta)}$. This $x_0$ is distinct from $\alpha$ and $\beta$.
   Since $f(x)$ approaches 1 from above as $x \to +\infty$ and from below as $x \to -\infty$ (if $x_0$ is not between $\alpha$ and $\beta$), it means that for some value $y_0$ close to 1, there will be multiple $x$ values such that $f(x)=y_0$.
   For example, $f(x)$ goes from $1$ to $0$ in $(-\infty, \gamma)$ and from $1$ to $0$ in $(\beta, +\infty)$. This implies that for any $y \in (0,1)$, there are at least two values of $x$ for which $f(x)=y$.
   Therefore, $f(x)$ is a many-one function.

5. Surjectivity (Onto/Into):
   The range of $f(x)$ is the union of the ranges over the intervals:

   • $(-\infty, \gamma)$: $(0, 1)$ (excluding 1, including 0)
   • $(\gamma, \alpha)$: $(-\infty, 0)$ (excluding 0)
   • $(\alpha, \delta)$: $(0, +\infty)$ (excluding 0)
   • $(\delta, \beta)$: $(-\infty, 0)$ (excluding 0)
   • $(\beta, +\infty)$: $(1, +\infty)$ (excluding 1)
     The union of these ranges is $(-\infty, \infty) = R$.
     Since the codomain is $R$, and the range is $R$, the function is onto.

Based on the analysis:

• $f: R - \{\alpha, \beta\} \to R$ is an onto function.
• $f: R - \{\alpha, \beta\} \to R$ is a many-one function.