If `f(x)={{:((x^(2)-9)/(x-3)", if " xne3),(m", if "x=3):}` is continuous at x = 3, then the value of 'm' is :

Options

A. 3
B. 6
C. 2
D. 1

Answer

To find the value of 'm' such that the function

\[ f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x \neq 3 \\ m & \text{if } x = 3 \end{cases} \]

is continuous at \( x = 3 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 3 is equal to \( f(3) \).

### Step-by-step solution:

1. **Find \( f(3) \)**:
Since \( f(3) = m \), we will use this in our continuity condition later.

2. **Calculate the limit as \( x \) approaches 3**:
We need to compute:
\[ \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \]

3. **Simplify the expression**:
Notice that \( x^2 - 9 \) can be factored:
\[ x^2 - 9 = (x - 3)(x + 3) \]
Therefore, we can rewrite the limit as:
\[ \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} \]
For \( x \neq 3 \), we can cancel \( (x - 3) \):
\[ \lim_{x \to 3} (x + 3) \]

4. **Evaluate the limit**:
Now substituting \( x = 3 \):
\[ \lim_{x \to 3} (x + 3) = 3 + 3 = 6 \]

5. **Set the limit equal to \( f(3) \)**:
For continuity at \( x = 3 \):
\[ \lim_{x \to 3} f(x) = f(3) \]
This gives us:
\[ 6 = m \]

6. **Conclusion**:
Thus, the value of \( m \) is:
\[ m = 6 \]

### Final Answer:
The value of \( m \) is \( 6 \).