If `f(x)={{:(mx+1,xle5),(3x-5,xgt5):}` is continuous, then the value of m is :

Options

A. `9/5`
B. `5/9`
C. `5/3`
D. `3/5`

Answer

To determine the value of \( m \) for which the function

\[ f(x) = \begin{cases} mx + 1 & \text{if } x \leq 5 \\ 3x - 5 & \text{if } x > 5 \end{cases} \]

is continuous, we need to ensure that the left-hand limit and the right-hand limit at \( x = 5 \) are equal, and that they are equal to the value of the function at that point.

### Step-by-step Solution:

1. **Identify the function values at \( x = 5 \)**:
- For \( x \leq 5 \), the function is given by \( f(x) = mx + 1 \).
- Therefore, at \( x = 5 \):
\[ f(5) = m(5) + 1 = 5m + 1 \]

2. **Calculate the left-hand limit as \( x \) approaches 5**:
- The left-hand limit is calculated using the expression for \( f(x) \) when \( x \leq 5 \):
\[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (mx + 1) = m(5) + 1 = 5m + 1 \]

3. **Calculate the right-hand limit as \( x \) approaches 5**:
- The right-hand limit is calculated using the expression for \( f(x) \) when \( x > 5 \):
\[ \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x - 5) = 3(5) - 5 = 15 - 5 = 10 \]

4. **Set the left-hand limit equal to the right-hand limit**:
- For the function to be continuous at \( x = 5 \), we need:
\[ 5m + 1 = 10 \]

5. **Solve for \( m \)**:
- Rearranging the equation:
\[ 5m = 10 - 1 \]
\[ 5m = 9 \]
\[ m = \frac{9}{5} \]

### Final Answer:
The value of \( m \) is \( \frac{9}{5} \).