Let `f(x)=x^2` if `x` is rational and `f(x)=1-x^2` if `x` is irrational then the number of points of continuity of f is

Answer

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

- \( f(x) = x^2 \) if \( x \) is rational
- \( f(x) = 1 - x^2 \) if \( x \) is irrational

we need to analyze the continuity of the function at various points.

### Step 1: Definition of Continuity
A function \( f \) is continuous at a point \( a \) if:
\[ \lim_{x \to a} f(x) = f(a) \]
This means that the left-hand limit and right-hand limit must both exist and be equal to \( f(a) \).

### Step 2: Finding Limits
Let’s consider a point \( a \). We will evaluate the left-hand limit and right-hand limit as \( x \) approaches \( a \).

- **Left-hand limit**:
\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^-} x^2 \quad \text{(if } x \text{ is rational)} \]
- **Right-hand limit**:
\[ \lim_{x \to a^+} f(x) = \lim_{x \to a^+} (1 - x^2) \quad \text{(if } x \text{ is irrational)} \]

### Step 3: Setting the Limits Equal
For \( f \) to be continuous at \( a \), we need:
\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \]
This gives us:
\[ a^2 = 1 - a^2 \]

### Step 4: Solving the Equation
Rearranging the equation:
\[ a^2 + a^2 = 1 \implies 2a^2 = 1 \implies a^2 = \frac{1}{2} \implies a = \pm \frac{1}{\sqrt{2}} \]

### Step 5: Points of Continuity
The points of continuity are:
- \( a = \frac{1}{\sqrt{2}} \)
- \( a = -\frac{1}{\sqrt{2}} \)

### Step 6: Conclusion
Thus, the function \( f(x) \) is continuous at exactly **two points**: \( \frac{1}{\sqrt{2}} \) and \( -\frac{1}{\sqrt{2}} \).

### Final Answer
The number of points of continuity of \( f \) is **2**.

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