If `f(x)={{:(x^(2),:,"when x is rational"),(2-x,:,"when x is irrational"):},` then

Options

A. `f(x)` is continuous for all real x
B. `f(x)` is discontinuous for all real x
C. `f(x)` is continuous only at `x=1, -2`
D. `f(x)` is discontinuous only at `x=1, -2`

Answer

To determine the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} x^2 & \text{when } x \text{ is rational} \\ 2 - x & \text{when } x \text{ is irrational} \end{cases} \]

we need to analyze the points where the function might be continuous. A function is continuous at a point \( x_0 \) if:

\[ \lim_{x \to x_0} f(x) = f(x_0) \]

This means we need to check the left-hand limit (LHS) and the right-hand limit (RHS) at various points and see if they equal the function value at that point.

### Step 1: Check for continuity at \( x = 1 \)

1. **Calculate \( f(1) \)**:
- Since \( 1 \) is rational, \( f(1) = 1^2 = 1 \).

2. **Calculate the left-hand limit as \( x \to 1^- \)**:
- As \( x \) approaches \( 1 \) from the left, it can be rational or irrational.
- If \( x \) is rational, \( \lim_{x \to 1^-} f(x) = 1^2 = 1 \).
- If \( x \) is irrational, \( \lim_{x \to 1^-} f(x) = 2 - 1 = 1 \).

3. **Calculate the right-hand limit as \( x \to 1^+ \)**:
- As \( x \) approaches \( 1 \) from the right, it can also be rational or irrational.
- If \( x \) is rational, \( \lim_{x \to 1^+} f(x) = 1^2 = 1 \).
- If \( x \) is irrational, \( \lim_{x \to 1^+} f(x) = 2 - 1 = 1 \).

4. **Conclusion at \( x = 1 \)**:
- Since \( \lim_{x \to 1} f(x) = 1 \) and \( f(1) = 1 \), the function is continuous at \( x = 1 \).

### Step 2: Check for continuity at \( x = -2 \)

1. **Calculate \( f(-2) \)**:
- Since \( -2 \) is rational, \( f(-2) = (-2)^2 = 4 \).

2. **Calculate the left-hand limit as \( x \to -2^- \)**:
- If \( x \) is rational, \( \lim_{x \to -2^-} f(x) = (-2)^2 = 4 \).
- If \( x \) is irrational, \( \lim_{x \to -2^-} f(x) = 2 - (-2) = 4 \).

3. **Calculate the right-hand limit as \( x \to -2^+ \)**:
- If \( x \) is rational, \( \lim_{x \to -2^+} f(x) = (-2)^2 = 4 \).
- If \( x \) is irrational, \( \lim_{x \to -2^+} f(x) = 2 - (-2) = 4 \).

4. **Conclusion at \( x = -2 \)**:
- Since \( \lim_{x \to -2} f(x) = 4 \) and \( f(-2) = 4 \), the function is continuous at \( x = -2 \).

### Step 3: Find points of discontinuity

To find points of discontinuity, we need to check where the two cases of the function might not equal each other:

Set \( x^2 = 2 - x \):

\[ x^2 + x - 2 = 0 \]

Factoring gives:

\[ (x - 1)(x + 2) = 0 \]

Thus, \( x = 1 \) and \( x = -2 \) are points where the two expressions are equal.

### Conclusion

The function \( f(x) \) is continuous at \( x = 1 \) and \( x = -2 \). It is discontinuous at all other points.