The given equation is x^2 + 5 = 2x - 4cos(a bx), where a and b are real numbers between 0 and 5. We need to find the maximum value of a*b for which the equation has at least one real solution for x.

1. Simplifying the Equation:
To solve the given equation, we first simplify it by rearranging the terms:
x^2 - 2x + 4cos(a bx) + 5 = 0

2. Analyzing the Quadratic Equation:
Now, we have a quadratic equation in terms of x. To find the conditions for real solutions, we examine the discriminant (b^2 - 4ac) of the quadratic equation.

3. Discriminant and Real Solutions:
The discriminant of the equation is given by D = (-2)^2 - 4(1)(4cos(a bx) + 5).
For real solutions, the discriminant should be greater than or equal to zero (D ≥ 0).

4. Simplifying the Discriminant:
Expanding and simplifying the discriminant:
D = 4 - 16cos(a bx) - 20
D = -16cos(a bx) - 16

5. Condition for Real Solutions:
For real solutions, the discriminant D ≥ 0:
-16cos(a bx) - 16 ≥ 0
cos(a bx) ≤ -1

6. Analyzing the Cosine Function:
The cosine function ranges between -1 and 1. To satisfy the condition cos(a bx) ≤ -1, a bx must be an odd multiple of π.

7. Expressing a bx in terms of π:
Let m be an odd integer such that m = 2k + 1, where k is an integer.
Then, a bx = (2k + 1)π

8. Finding the Value of x:
From the equation a bx = (2k + 1)π, we can express x in terms of k:
x = [(2k + 1)π] / b

9. Conclusion:
To satisfy the given equation and have at least one real solution for x, the values of a and b must allow the expression [(2k + 1)π] / b to be a real number. This occurs when b ≠ 0.

10. Maximum Value of a*b:
Since a and b are both between 0 and 5, the maximum value of a*b is attained when a = 5 and b = 5, resulting in a*b = 25.

Therefore, the maximum value of a*b is 25.