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To solve this problem, we need to use the fact that the function \( f \) is continuous at \( x = 6 \). Since \( f \) is continuous, the left-hand limit and the right-hand limit at \( x = 6 \) must be equal.

Given:
1. For \( x < 6 \): \( f(x) = \cos(x) \).
2. For \( x \geq 6 \): \( f(x) = x + a \).

To ensure continuity at \( x = 6 \), we set up the following equation:

\[
\lim_{x \to 6^-} f(x) = \lim_{x \to 6^+} f(x)
\]

### Step 1: Calculate the left-hand limit at \( x = 6 \)
For \( x < 6 \), \( f(x) = \cos(x) \). So, the left-hand limit is:
\[
\lim_{x \to 6^-} f(x) = \cos(6)
\]

### Step 2: Calculate the right-hand limit at \( x = 6 \)
For \( x \geq 6 \), \( f(x) = x + a \). So, the right-hand limit is:
\[
\lim_{x \to 6^+} f(x) = 6 + a
\]

### Step 3: Set the limits equal to ensure continuity
Since \( f \) is continuous at \( x = 6 \), we equate the left-hand and right-hand limits:
\[
\cos(6) = 6 + a
\]

### Step 4: Solve for \( a \)
Rearrange the equation to solve for \( a \):
\[
a = \cos(6) - 6
\]

Now, calculating the numerical value of \( \cos(6) \) (in radians):
\[
\cos(6) \approx 0.9602
\]
\[
a \approx 0.9602 - 6 = -5.0398
\]

### Final Answer:
\[
a \approx -5.04
\]

Would you like additional details, or do you have any questions?

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### Related Questions
1. What does continuity mean for a function at a specific point?
2. How can we calculate limits from the left-hand side and right-hand side?
3. How does one determine if a piecewise function is continuous?
4. What are the general properties of the cosine function?
5. How can we approximate trigonometric values like \( \cos(6) \) without a calculator?

### Tip
For piecewise functions, always ensure continuity by matching the values from the left and right at the boundaries.

The function f is continuous in 6. For x < 6, f(x) = cos(x) applies. For x > 6, f(x) = x + a applies. Determine a.

Learn how to determine the constant 'a' to ensure continuity of a piecewise function at x = 6. This problem involves understanding limits, continuity, and trigonometric functions with a step-by-step solution. Suitable for advanced high school students in Grades 11-12.

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