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The question asks for which values of \( a \) and \( b \) the function \( f(x) \) is continuous everywhere. The piecewise function is defined as follows:

\[
f(x) = 
\begin{cases} 
a \sin(\pi x) + b & \text{if } x \leq 2 \\
x^2 + a & \text{if } 2 < x < 5 \\
b \cos(2\pi x) + a & \text{if } x \geq 5
\end{cases}
\]

To ensure that \( f(x) \) is continuous everywhere, we need to check the continuity at the points where the definition of the function changes, specifically at \( x = 2 \) and \( x = 5 \).

### Continuity at \( x = 2 \):

For \( f(x) \) to be continuous at \( x = 2 \), the left-hand limit (as \( x \) approaches 2 from the left) must equal the right-hand limit (as \( x \) approaches 2 from the right), and both must equal \( f(2) \).

1. **Left-hand limit as \( x \to 2^- \):**
   \[
   \lim_{x \to 2^-} f(x) = a \sin(2\pi) + b = a(0) + b = b
   \]

2. **Right-hand limit as \( x \to 2^+ \):**
   \[
   \lim_{x \to 2^+} f(x) = 2^2 + a = 4 + a
   \]

3. **For continuity at \( x = 2 \), we need:**
   \[
   b = 4 + a
   \]

### Continuity at \( x = 5 \):

Similarly, for \( f(x) \) to be continuous at \( x = 5 \), the left-hand limit (as \( x \) approaches 5 from the left) must equal the right-hand limit (as \( x \) approaches 5 from the right), and both must equal \( f(5) \).

1. **Left-hand limit as \( x \to 5^- \):**
   \[
   \lim_{x \to 5^-} f(x) = 5^2 + a = 25 + a
   \]

2. **Right-hand limit as \( x \to 5^+ \):**
   \[
   \lim_{x \to 5^+} f(x) = b \cos(10\pi) + a = b(1) + a = b + a
   \]

3. **For continuity at \( x = 5 \), we need:**
   \[
   25 + a = b + a
   \]

   Simplifying, we get:
   \[
   b = 25
   \]

### Solving the System of Equations:

We now have the following system of equations:
1. \( b = 4 + a \)
2. \( b = 25 \)

Substituting \( b = 25 \) into the first equation:
\[
25 = 4 + a
\]
Solving for \( a \):
\[
a = 21
\]

Thus, the values of \( a \) and \( b \) that make \( f(x) \) continuous everywhere are:
\[
a = 21, \quad b = 25
\]

Would you like more details on this solution, or have any further questions?

Here are five related questions:
1. How do you determine if a function is continuous at a point?
2. What is the importance of limits in ensuring continuity?
3. How does the piecewise nature of the function affect its continuity?
4. Can a function be continuous everywhere but still non-differentiable? 
5. What role do trigonometric identities play in solving piecewise functions?

**Tip:** Always ensure both the left-hand and right-hand limits match the function's value at the point when checking for continuity at a specific point.

For which values of a and b is the following function continuous everywhere? f(x) = { a sin(πx) + b if x ≤ 2, x² + a if 2 < x < 5, b cos(2πx) + a if x > 5 }

Explore how to determine the values of a and b for the continuity of a piecewise function involving trigonometric and polynomial expressions. This detailed solution covers the concepts of limits, continuity, and the properties of sine and cosine functions.

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