Let `f:R rarr R ` be given by `f(x)={"5x ,if x in Q ", "x^ 2 + 6",if x in R-Q` then

Options

A. f is continuous at x=2 and x=3
B. f is not continuous at x=2 and x=3
C. f is continuous at x=2 but not at x=3
D. f is continuous at x=3 but not at x=2

Answer

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

\[ f(x) = \begin{cases} 5x & \text{if } x \in \mathbb{Q} \\ x^2 + 6 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} \]

we will check the continuity at \( x = 2 \) and \( x = 3 \).

### Step 1: Check continuity at \( x = 2 \)

1. **Calculate \( f(2) \)**:
Since \( 2 \) is a rational number, we use the first case of the function:
\[ f(2) = 5 \times 2 = 10 \]

2. **Calculate \( \lim_{x \to 2} f(x) \)**:
We need to find the limit as \( x \) approaches \( 2 \). We will consider both rational and irrational approaches.

- For rational \( x \) approaching \( 2 \):
\[ \lim_{x \to 2, x \in \mathbb{Q}} f(x) = \lim_{x \to 2} 5x = 5 \times 2 = 10 \]

- For irrational \( x \) approaching \( 2 \):
\[ \lim_{x \to 2, x \in \mathbb{R} \setminus \mathbb{Q}} f(x) = \lim_{x \to 2} (x^2 + 6) = 2^2 + 6 = 4 + 6 = 10 \]

3. **Conclusion for \( x = 2 \)**:
Since both limits equal \( 10 \) and \( f(2) = 10 \):
\[ \lim_{x \to 2} f(x) = f(2) \]
Therefore, \( f(x) \) is continuous at \( x = 2 \).

### Step 2: Check continuity at \( x = 3 \)

1. **Calculate \( f(3) \)**:
Since \( 3 \) is a rational number, we use the first case of the function:
\[ f(3) = 5 \times 3 = 15 \]

2. **Calculate \( \lim_{x \to 3} f(x) \)**:
We will again consider both rational and irrational approaches.

- For rational \( x \) approaching \( 3 \):
\[ \lim_{x \to 3, x \in \mathbb{Q}} f(x) = \lim_{x \to 3} 5x = 5 \times 3 = 15 \]

- For irrational \( x \) approaching \( 3 \):
\[ \lim_{x \to 3, x \in \mathbb{R} \setminus \mathbb{Q}} f(x) = \lim_{x \to 3} (x^2 + 6) = 3^2 + 6 = 9 + 6 = 15 \]

3. **Conclusion for \( x = 3 \)**:
Since both limits equal \( 15 \) and \( f(3) = 15 \):
\[ \lim_{x \to 3} f(x) = f(3) \]
Therefore, \( f(x) \) is continuous at \( x = 3 \).

### Final Conclusion
The function \( f(x) \) is continuous at both \( x = 2 \) and \( x = 3 \).