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.

Answer 2

Determine the properties of the composite function \( g \circ f: \mathbb{R} \to \mathbb{R} \), where \( f \) and \( g \) are piecewise defined.

Concept Used:

To analyze \( g \circ f \), we compute \( (g \circ f)(x) = g(f(x)) \) for different ranges of \( x \), using the definitions of \( f \) and \( g \). We then check for continuity and differentiability at the transition points.

Step-by-Step Solution:

Step 1: Write definitions of \( f \) and \( g \).

\[ f(x) = \begin{cases} \ln x, & x > 0, \\ e^{-x}, & x \leq 0, \end{cases} \quad g(x) = \begin{cases} x, & x \geq 0, \\ e^x, & x < 0. \end{cases} \]

Step 2: Compute \( g(f(x)) \) for \( x > 0 \).

For \( x > 0 \), \( f(x) = \ln x \).

We need \( g(\ln x) \), which depends on whether \( \ln x \geq 0 \) or \( \ln x < 0 \).

• If \( \ln x \geq 0 \) i.e. \( x \geq 1 \): \( g(\ln x) = \ln x \) (since \( \ln x \geq 0 \)).
• If \( \ln x < 0 \) i.e. \( 0 < x < 1 \): \( g(\ln x) = e^{\ln x} = x \) (since \( \ln x < 0 \)).

So for \( x > 0 \): \[ (g \circ f)(x) = \begin{cases} x, & 0 < x < 1, \\ \ln x, & x \geq 1. \end{cases} \]

Step 3: Compute \( g(f(x)) \) for \( x \leq 0 \).

For \( x \leq 0 \), \( f(x) = e^{-x} \).

Since \( x \leq 0 \), \( -x \geq 0 \), so \( e^{-x} \geq 1 > 0 \). Thus \( f(x) > 0 \) for \( x \leq 0 \).

Since \( f(x) > 0 \), we use \( g(y) = y \) for \( y \geq 0 \).

So \( g(f(x)) = f(x) = e^{-x} \) for \( x \leq 0 \).

Step 4: Write complete definition of \( g \circ f \).

\[ (g \circ f)(x) = \begin{cases} e^{-x}, & x \leq 0, \\ x, & 0 < x < 1, \\ \ln x, & x \geq 1. \end{cases} \]

Step 5: Check continuity at \( x = 0 \).

Left-hand limit as \( x \to 0^- \): \( \lim_{x \to 0^-} e^{-x} = e^0 = 1 \).

Right-hand limit as \( x \to 0^+ \): \( \lim_{x \to 0^+} x = 0 \).

Since \( 1 \neq 0 \), \( g \circ f \) is discontinuous at \( x = 0 \).

Step 6: Check continuity at \( x = 1 \).

Left-hand limit as \( x \to 1^- \): \( \lim_{x \to 1^-} x = 1 \).

Right-hand limit as \( x \to 1^+ \): \( \lim_{x \to 1^+} \ln x = \ln 1 = 0 \).

Since \( 1 \neq 0 \), \( g \circ f \) is discontinuous at \( x = 1 \).

Step 7: Check differentiability.

Since the function is discontinuous at \( x = 0 \) and \( x = 1 \), it is not differentiable at these points.

On intervals \( (-\infty, 0) \), \( (0, 1) \), \( (1, \infty) \), the function is differentiable (as composition of differentiable functions).

Step 8: Conclusion.

The function \( g \circ f \) is discontinuous at \( x = 0 \) and \( x = 1 \), hence not one-one, not onto, and not differentiable at these points.

From the definition: For \( x \leq 0 \), \( (g \circ f)(x) = e^{-x} \in [1, \infty) \); for \( 0 < x < 1 \), \( (g \circ f)(x) = x \in (0, 1) \); for \( x \geq 1 \), \( (g \circ f)(x) = \ln x \in [0, \infty) \). So range is \( (0, 1) \cup [1, \infty) = (0, \infty) \), not all \( \mathbb{R} \), so not onto.

Also, not one-one since multiple \( x \) give same value (e.g., \( x = -1 \) gives \( e^1 \), \( x = e^e \) gives \( \ln(e^e) = e \), etc.).

Therefore, \( g \circ f \) is neither one-one nor onto.