To determine which binary operation is associative on the set of integers, we need to analyze each given operation. A binary operation \(*\)is associative if for all integers \(a\), \(b\), and \(c\), the following condition holds:

\((a * b) * c = a * (b * c)\)

Let's analyze each operation:

1. \(a * b = \max\{a, b\}\):

This operation is associative because:

\(((a * b) * c = \max(\max(a, b), c) = \max(a, b, c)\)

\(a * (b * c) = \max(a, \max(b, c)) = \max(a, b, c)\)

Since both expressions resolve to \(\max(a, b, c)\), this operation is associative.

1. \(a * b = a^b\):

This operation is not associative because exponentiation doesn't satisfy the associative property:

Example: Let \(a = 2\), \(b = 3\), \(c = 2\).

\((a^b)^c = (2^3)^2 = 8^2 = 64\)

\(a^(b^c) = 2^{(3^2)} = 2^9 = 512\)

Clearly, \(64 \neq 512\), so this operation is not associative.

1. \(a * b = 2a + b\):

This operation is not associative because:

Example: Let \(a = 1\), \(b = 2\), \(c = 3\).

\(((2a + b) * c = (2 \times 1 + 2) + 3 = 7\)

\(2a + (b + c) = 2 \times 1 + (2 + 3) = 7\)

\(a * (b * c) = 2 \times 1 + (2 \times 2 + 3) = 9\)

These expressions do not match, thus this operation is not associative.

1. \(a * b = a - b\):

This operation is not associative because subtraction is inherently non-associative.

Example: Let \(a = 3\), \(b = 2\), \(c = 1\).

\(((a - b) - c = (3 - 2) - 1 = 0\)

\(a - (b - c) = 3 - (2 - 1) = 2\)

Since \(0 \neq 2\), the operation is not associative.

Thus, the operation \(a * b = \max\{a, b\}\) is the only associative binary operation on integers among the given options.