https://file.aiquickdraw.com/mathbot/file/1728652049609qj7oyhck.jpg

The function provided is:

\[
F(x) =
\begin{cases} 
\frac{\sin(3x)}{2x} + (k + 1), & x \neq 0 \\
k, & x = 0
\end{cases}
\]
The function is continuous at \(x = 0\), and we need to determine the value of \(k\).

### Steps to solve:
1. **Continuity Condition**: For \(F(x)\) to be continuous at \(x = 0\), the left-hand limit, right-hand limit, and the value of the function at \(x = 0\) must be equal.
   
   This means:
   \[
   \lim_{x \to 0} F(x) = F(0)
   \]

2. **Evaluate the limit**:
   For \(x \neq 0\), we have the function:
   \[
   F(x) = \frac{\sin(3x)}{2x} + (k + 1)
   \]
   We need to find:
   \[
   \lim_{x \to 0} \left( \frac{\sin(3x)}{2x} + (k + 1) \right)
   \]

   Using the standard limit result \(\lim_{x \to 0} \frac{\sin(ax)}{x} = a\), we can calculate:
   \[
   \lim_{x \to 0} \frac{\sin(3x)}{2x} = \frac{3}{2}
   \]
   So:
   \[
   \lim_{x \to 0} F(x) = \frac{3}{2} + (k + 1)
   \]

3. **Continuity condition at \(x = 0\)**:
   \[
   \frac{3}{2} + (k + 1) = k
   \]

4. **Solve for \(k\)**:
   \[
   \frac{3}{2} + k + 1 = k
   \]
   Simplifying:
   \[
   \frac{3}{2} + 1 = 0
   \]
   \[
   \frac{5}{2} = 0
   \]
   This leads to an inconsistency since the equation cannot be true, indicating a different approach

If function defined by F(x) = { (sin3x) / (2x) + (k + 1), for x ≠ 0; k, for x = 0 } is continuous at x = 0, then k equals:

Explore how to determine the continuity of a piecewise function at x = 0. This problem involves calculus concepts such as limits and continuity, suitable for Grades 11-12.

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