Two mappings `f:R to R` and `g:R to R` are defined as follows:
`f(x)={(0,"when x is rational"),(1,"when x is irrational"):}` and
`g(x)={(-1,"wnen x is rational"),(0,"when x is irrational"):}` then the value of `[(gof)(e)+(fog)(pi)]` is -

Options

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

Answer

To solve the problem, we need to evaluate the expression \((g \circ f)(e) + (f \circ g)(\pi)\) using the given mappings \(f\) and \(g\).

### Step 1: Understand the mappings
The mappings are defined as follows:
- \(f(x) = 0\) when \(x\) is rational, and \(f(x) = 1\) when \(x\) is irrational.
- \(g(x) = -1\) when \(x\) is rational, and \(g(x) = 0\) when \(x\) is irrational.

### Step 2: Evaluate \(g \circ f\) at \(e\)
We need to find \((g \circ f)(e)\):
1. First, evaluate \(f(e)\):
- Since \(e\) is an irrational number, we have \(f(e) = 1\).

2. Now, substitute \(f(e)\) into \(g\):
- We need to find \(g(f(e)) = g(1)\).
- Since \(1\) is a rational number, we have \(g(1) = -1\).

Thus, \((g \circ f)(e) = -1\).

### Step 3: Evaluate \(f \circ g\) at \(\pi\)
Next, we need to find \((f \circ g)(\pi)\):
1. First, evaluate \(g(\pi)\):
- Since \(\pi\) is an irrational number, we have \(g(\pi) = 0\).

2. Now, substitute \(g(\pi)\) into \(f\):
- We need to find \(f(g(\pi)) = f(0)\).
- Since \(0\) is a rational number, we have \(f(0) = 0\).

Thus, \((f \circ g)(\pi) = 0\).

### Step 4: Combine the results
Now we can combine the results:
\[ (g \circ f)(e) + (f \circ g)(\pi) = -1 + 0 = -1. \]

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

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