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The problem asks to determine the value of \( a \) such that the given piecewise function is continuous everywhere. The function is defined as:

\[
f(t) =
\begin{cases}
2t + k & \text{if } t < 0 \\
2 \cos(t) - 3 & \text{if } 0 \leq t \leq \frac{\pi}{2} \\
a \sin(t) + 5k & \text{if } t > \frac{\pi}{2}
\end{cases}
\]

To ensure the function is continuous everywhere, we need to check for continuity at the points where the pieces of the function change, specifically at \( t = 0 \) and \( t = \frac{\pi}{2} \).

### Step 1: Continuity at \( t = 0 \)
The function must satisfy:

\[
\lim_{t \to 0^-} f(t) = \lim_{t \to 0^+} f(t) = f(0)
\]

- For \( t < 0 \), \( f(t) = 2t + k \). As \( t \to 0^- \), we get:
  \[
  \lim_{t \to 0^-} f(t) = 2(0) + k = k
  \]
  
- For \( 0 \leq t \leq \frac{\pi}{2} \), \( f(0) = 2 \cos(0) - 3 = 2(1) - 3 = -1 \).

For continuity at \( t = 0 \):
\[
k = -1
\]

### Step 2: Continuity at \( t = \frac{\pi}{2} \)
Next, the function must be continuous at \( t = \frac{\pi}{2} \), so:

\[
\lim_{t \to \left( \frac{\pi}{2} \right)^-} f(t) = \lim_{t \to \left( \frac{\pi}{2} \right)^+} f(t)
\]

- For \( 0 \leq t \leq \frac{\pi}{2} \), \( f(t) = 2 \cos(t) - 3 \). As \( t \to \frac{\pi}{2}^- \):
  \[
  f\left(\frac{\pi}{2}\right) = 2 \cos\left(\frac{\pi}{2}\right) - 3 = 2(0) - 3 = -3
  \]
  
- For \( t > \frac{\pi}{2} \), \( f(t) = a \sin(t) + 5k \). As \( t \to \frac{\pi}{2}^+ \), \( \sin\left(\frac{\pi}{2}\right) = 1 \), so:
  \[
  \lim_{t \to \left(\frac{\pi}{2}\right)^+} f(t) = a(1) + 5k = a + 5k
  \]

For continuity at \( t = \frac{\pi}{2} \):
\[
a + 5k = -3
\]

Substituting \( k = -1 \) into the equation:
\[
a + 5(-1) = -3 \implies a - 5 = -3 \implies a = 2
\]

Thus, the value of \( a \) is \( \boxed{2} \).

Would you like further clarification or details?

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Here are 5 questions related to this solution:

1. Why is checking the continuity at \( t = 0 \) and \( t = \frac{\pi}{2} \) important?
2. How would the solution change if \( k \) had a different value?
3. What does it mean for a function to be continuous at a point?
4. How would we check differentiability at the transition points?
5. Can the same approach be applied to piecewise functions with more complex definitions?

**Tip:** Always remember that for continuity at a point, both the left-hand limit and right-hand limit must equal the function value at that point.

If the following function is continuous everywhere and a and k are constants, then what is the value of a?
f(t) =
{
2t + k, if t < 0
2 cos(t) - 3, if 0 ≤ t ≤ π/2
a sin(t) + 5k, if t > π/2
}

Solve a piecewise function for continuity at key transition points. Learn how to find the values of constants that ensure the function is continuous, using limits and continuity principles. Suitable for advanced high school math (Pre-Calculus/Calculus).

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