Let `f :R to R` and `g : R rightarrow R` be defined as `f(x)={(log_e x, x gt 0), (e^(-x), x le 0):}`,
`g(x) ={(x, x ge 0), (e^x , x lt 0):}`. Then `gof : R rightarrow R` is:

Options

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

To solve the problem, we need to find the composition of the functions \( g(f(x)) \) where \( f \) and \( g \) are defined as follows:

1. **Function Definitions**:
- \( f(x) = \begin{cases} \log_e x & \text{if } x > 0 \\ e^{-x} & \text{if } x \leq 0 \end{cases} \)

- \( g(x) = \begin{cases} x & \text{if } x \geq 0 \\ e^x & \text{if } x < 0 \end{cases} \)

2. **Finding \( g(f(x)) \)**:
We will evaluate \( g(f(x)) \) for different cases based on the definition of \( f(x) \).

**Case 1**: \( x > 0 \)

Here, \( f(x) = \log_e x \). We need to determine if \( \log_e x \) is greater than or equal to 0.
- \( \log_e x \geq 0 \) when \( x \geq 1 \).
- Thus, for \( x \geq 1 \), \( g(f(x)) = g(\log_e x) = \log_e x \).
- For \( 0 < x < 1 \), \( \log_e x < 0 \), so \( g(f(x)) = g(\log_e x) = e^{\log_e x} = x \).

Therefore, for \( x > 0 \):
\[ g(f(x)) = \begin{cases} \log_e x & \text{if } x \geq 1 \\ x & \text{if } 0 < x < 1 \end{cases} \]

**Case 2**: \( x \leq 0 \)

Here, \( f(x) = e^{-x} \). Since \( e^{-x} \) is always positive for any real \( x \), we have \( g(f(x)) = g(e^{-x}) \).
- Since \( e^{-x} \geq 0 \) for all \( x \), we use the first part of \( g(x) \):
\[ g(f(x)) = e^{-x} \quad \text{for } x \leq 0. \]

3. **Combining the Results**:
Now, we can combine the results from both cases:
\[ g(f(x)) = \begin{cases} \log_e x & \text{if } x \geq 1 \\ x & \text{if } 0 < x < 1 \\ e^{-x} & \text{if } x \leq 0 \end{cases} \]

4. **Checking if \( g(f(x)) \) is One-One or Onto**:
- **One-One**: A function is one-one if different inputs give different outputs. Here, \( g(f(x)) \) is not one-one because \( g(f(x)) = x \) for \( 0 < x < 1 \) and also \( g(f(x)) = \log_e x \) for \( x \geq 1 \), which can produce the same output for different inputs.
- **Onto**: A function is onto if every possible output is covered. The range of \( g(f(x)) \) does not cover all real numbers since \( g(f(x)) \) outputs values only from \( (0, \infty) \) and does not include negative values.

Thus, we conclude that \( g(f(x)) \) is neither one-one nor onto.