Two functions `f:R to R and g:Rto R` are defined as follows:
`f(x)={{:(,0,x in Q),(,1,x ne Q):},g(x)={{:(,-1,x in Q),(,0,x in Q):}`
Then, fof (e)+fog`(pi)`

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem, we need to evaluate the expression \( g(f(e)) + f(g(\pi)) \) using the definitions of the functions \( f \) and \( g \).

### Step-by-Step Solution:

1. **Identify the nature of \( e \) and \( \pi \)**:
- Both \( e \) and \( \pi \) are irrational numbers.

2. **Evaluate \( f(e) \)**:
- The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \notin \mathbb{Q} \end{cases} \]
- Since \( e \) is irrational, we have:
\[ f(e) = 1 \]

3. **Evaluate \( g(f(e)) \)**:
- Now we need to find \( g(f(e)) = g(1) \).
- The function \( g(x) \) is defined as:
\[ g(x) = \begin{cases} -1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases} \]
- Since \( 1 \) is a rational number, we have:
\[ g(1) = -1 \]

4. **Evaluate \( g(\pi) \)**:
- We now need to find \( g(\pi) \).
- Since \( \pi \) is also irrational, we have:
\[ g(\pi) = 0 \]

5. **Evaluate \( f(g(\pi)) \)**:
- Now we need to find \( f(g(\pi)) = f(0) \).
- Since \( 0 \) is a rational number, we have:
\[ f(0) = 0 \]

6. **Combine the results**:
- Now we can combine the results to find the final expression:
\[ g(f(e)) + f(g(\pi)) = -1 + 0 = -1 \]

### Final Answer:
The value of \( g(f(e)) + f(g(\pi)) \) is \( -1 \).