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

The Correct Option is B

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.