If `f(x)={x,` when `x` is rational and `0,` when `x` is irrational
`g(x)={0,` when `x` is rational and `x,` when `x` is irrational then `(f-g)` is

A. one-one and onto
B. neither one-one nor onto
C. many one and onto
D one-one and into

Options

A. one-one and into
B. neither one-one nor onto
C. many one and onto
D. one-one and onto

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) and then find the function \( h(x) = f(x) - g(x) \).

### Step 1: Define the functions
- The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases} \]

- The function \( g(x) \) is defined as:
\[ g(x) = \begin{cases} 0 & \text{if } x \text{ is rational} \\ x & \text{if } x \text{ is irrational} \end{cases} \]

### Step 2: Calculate \( h(x) = f(x) - g(x) \)
Now, we will compute \( h(x) \):
- When \( x \) is rational:
\[ h(x) = f(x) - g(x) = x - 0 = x \]
- When \( x \) is irrational:
\[ h(x) = f(x) - g(x) = 0 - x = -x \]

Thus, we can define \( h(x) \) as:
\[ h(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ -x & \text{if } x \text{ is irrational} \end{cases} \]

### Step 3: Determine the domain of \( h(x) \)
The domain of \( h(x) \) includes all real numbers since it is defined for both rational and irrational numbers. Therefore, the domain is:
\[ \text{Domain of } h(x) = \mathbb{R} \quad (\text{all real numbers}) \]

### Step 4: Determine the range of \( h(x) \)
- For rational \( x \), \( h(x) = x \), which can take any rational value.
- For irrational \( x \), \( h(x) = -x \), which can take any negative irrational value.

Thus, the range of \( h(x) \) includes all real numbers, both positive and negative:
\[ \text{Range of } h(x) = \mathbb{R} \quad (\text{all real numbers}) \]

### Step 5: Check if \( h(x) \) is one-one
To check if \( h(x) \) is one-one, we need to see if different inputs produce different outputs:
- If \( x_1 \) and \( x_2 \) are both rational, then \( h(x_1) = x_1 \) and \( h(x_2) = x_2 \). If \( x_1 \neq x_2 \), then \( h(x_1) \neq h(x_2) \).
- If \( x_1 \) and \( x_2 \) are both irrational, then \( h(x_1) = -x_1 \) and \( h(x_2) = -x_2 \). If \( x_1 \neq x_2 \), then \( h(x_1) \neq h(x_2) \).
- If \( x_1 \) is rational and \( x_2 \) is irrational, then \( h(x_1) = x_1 \) and \( h(x_2) = -x_2 \). Since \( x_1 \) is positive and \( -x_2 \) is negative, they cannot be equal.

Thus, \( h(x) \) is one-one.

### Step 6: Check if \( h(x) \) is onto
Since the range of \( h(x) \) is all real numbers and it can produce every real number as an output, \( h(x) \) is onto.

### Conclusion
The function \( h(x) = f(x) - g(x) \) is both one-one and onto.

### Final Answer
The correct option is **A. one-one and onto**.