Let `f (x)= {{:(sin ^(2) x"," , "x is rational"),(-sin ^(2)x",", " x is irrational"):},` then set of points, where f (x) is continuous, is:

Options

A. `{(2n +1) (pi)/(2)in I}`
B. a null set
C. `{n pi, n in I}`
D. set of all rational numbers

Answer

To determine the set of points where the function \( f(x) \) is continuous, we need to analyze the definition of the function:

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

### Step 1: Understand the continuity condition
A function is continuous at a point \( c \) if:
\[ \lim_{x \to c} f(x) = f(c) \]
This means that the limit of the function as \( x \) approaches \( c \) must equal the value of the function at \( c \).

### Step 2: Evaluate the limit from both sides
Consider a point \( c \). We need to evaluate the limits as \( x \) approaches \( c \) from rational and irrational values.

- If \( c \) is rational, then:
\[ f(c) = \sin^2 c \]
As \( x \) approaches \( c \) through rational numbers, \( f(x) \to \sin^2 c \).

- If \( c \) is irrational, then:
\[ f(c) = -\sin^2 c \]
As \( x \) approaches \( c \) through irrational numbers, \( f(x) \to -\sin^2 c \).

### Step 3: Set the limits equal for continuity
For \( f(x) \) to be continuous at \( c \), we need:
\[ \sin^2 c = -\sin^2 c \]
This simplifies to:
\[ 2\sin^2 c = 0 \]

### Step 4: Solve for \( c \)
The equation \( 2\sin^2 c = 0 \) implies:
\[ \sin^2 c = 0 \]
This occurs when:
\[ \sin c = 0 \]
The solutions to this equation are:
\[ c = n\pi \quad \text{where } n \text{ is any integer} \]

### Step 5: Conclusion
Thus, the set of points where the function \( f(x) \) is continuous is:
\[ \{ n\pi : n \in \mathbb{Z} \} \]