Question

Let $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ be defined as: $f(x) = \begin{cases} \log_e x, & \text{if } x > 0, \\\ e^{-x}, & \text{if } x \leq 0, \end{cases}$ and
g(x) = \begin{cases} x, & \text{if } x \geq 0, \\\ e^x, & \text{if } x < 0\. \end{cases} Then $g \circ f: \mathbb{R} \to \mathbb{R}$ is:

• A. one-one but not onto
• B. neither one-one nor onto
• C. onto but not one-one
• D. both one-one and onto

Answer 1

Answer: (B)

neither one-one nor onto

Answer 2

Consider:

$g(f(x)) = \begin{cases} g(\log_e x), & x > 0 \\\ g(e^{-x}), & x \leq 0 \end{cases}$

For $x > 0$, we have:

$f(x) = \log_e x \implies g(f(x)) = g(\log_e x) = \log_e x \quad (\text{since } \log_e x \geq 0)$

For $x \leq 0$, we have:

$f(x) = e^{-x} \implies g(f(x)) = g(e^{-x}) = e^{-x} \quad (\text{since } e^{-x} > 0 \text{ for all } x \leq 0)$

Thus, the function $g(f(x))$ is given by:

$g(f(x)) = \begin{cases} \log_e x, & x > 0 \\\ e^{-x}, & x \leq 0 \end{cases}$

Analyzing this function, we observe:

For $x > 0$, $g(f(x)) = \log_e x$ is an increasing function but not onto as it maps to $(0, \infty)$.

For $x \leq 0$, $g(f(x)) = e^{-x}$ is a decreasing function and does not cover the entire range of real numbers.
Therefore, $g \circ f$ is neither one-one nor onto.