Maxima And Minima

We are tasked with finding the maximum value of the function: where .

Step 1: Analyzing the function
The function is the sum of two exponential functions, one in terms of and the other in terms of . To find the maximum value of , we need to analyze the behavior of these terms. Both and are bounded by and , which means and will be bounded by and , respectively.

Step 2: Maximize the sum
To maximize , we look for the values of where and are maximized simultaneously. The values of and both reach their maximum at , where: Thus, at , we have:

Step 3: Conclusion
Thus, the maximum value of is:

Answer:

Let , where . We need to analyze the continuity of this function.

Understanding the Components:

* is the greatest integer function of . It is discontinuous at points where is an integer, which means , where n is a non-negative integer.

* is a continuous function for all real numbers.

Analyzing Discontinuities:

The function can only be discontinuous at points where is discontinuous, which are , where n is a non-negative integer.

Now, we need to check the continuity at these points.

Let , where n is an integer.

For f(x) to be continuous at , we need to have:

Case 1: n is a perfect square

If n is a perfect square, then is an integer, so .

Therefore, . Now, we need to check the limits. Since is continuous and equal to zero when x is an integer, then when x equals the square root of an integer we expect continuity at the points which are perfect squares. We already have 0 for these points: .

Case 2: n is not a perfect square

Now let's analyze the limit at a point where , where n is an integer: .

Since is continuous,

If we choose n as a non perfect square integer, the main factor defining continuity is whether is continuous. But by our earlier reasoning this function does have discrete values. However, is equal to , and .

Conclusion:

The greatest integer function is not continuous, thus, at points which are not perfect integers, there exists a discrete step. Since the non integer points go to that means they are equal to 0. This also can never happen due to the domain condition ( ). Hence, this only holds true for perfect squares, thus the function can only be continuous at perfect square points. f is continuous at only those points which are perfect squares.

Step 1: Let

Step 2: Since both and are differentiable functions, their difference is also differentiable.

Step 3: To find maximum or minimum values of , we find critical points by setting:


Step 4: Differentiate :
Suppose and are such that:


Step 5: Solve the equation:


Step 6: Depending on the functions and , this equation might have infinitely many solutions. For example, if and are trigonometric functions that intersect periodically, at infinitely many values of .

Step 7: Therefore, can attain a maximum value at infinitely many points.

Final Answer: (C) exists at infinitely many points