The sum of an infinite G.P. whose first term is the greatest value of the function f(x) = sinx + cos2x in [0, n] and whose common ratio is the gradient of the curve y = = (ex/2 + e-x/2) at the point 2 with abscissa x = In 4 is:

Let's break down the problem step by step.

### 1. Greatest Value of the Function \( f(x) = \sin(x) + \cos(2x) \) in the interval \([0, n]\)
First, we need to determine the greatest value of the function \( f(x) = \sin(x) + \cos(2x) \).

To find the critical points, we'll differentiate \( f(x) \) with respect to \( x \) and set the derivative to zero.

\[
f'(x) = \cos(x) - 2\sin(2x)
\]
Using the chain rule for \( \cos(2x) \):
\[
f'(x) = \cos(x) - 4\sin(x)\cos(x)
\]

Set \( f'(x) = 0 \) to find the critical points:
\[
\cos(x)(1 - 4\sin(x)) = 0
\]
This gives us:
\[
\cos(x) = 0 \quad \text{or} \quad \sin(x) = \frac{1}{4}
\]

The function values corresponding to these points are evaluated to find the maximum within the interval \([0, n]\). However, let's denote the maximum value as \( a \) (since the specific interval is not explicitly given).

### 2. Common Ratio of the G.P.
Next, we need to find the gradient of the curve \( y = \frac{e^{x/2} + e^{-x/2}}{2} \) at the point where \( x = \ln 4 \).

The curve \( y \) can be rewritten as:
\[
y = \cosh\left(\frac{x}{2}\right)
\]
The derivative (gradient) is:
\[
\frac{dy}{dx} = \frac{1}{2}\sinh\left(\frac{x}{2}\right)
\]
At \( x = \ln 4 \):
\[
\frac{dy}{dx}\bigg|_{x = \ln 4} = \frac{1}{2}\sinh\left(\frac{\ln 4}{2}\right)
\]

Since \( \sinh(\frac{\ln 4}{2}) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4} \), the gradient is \( \frac{3}{8} \).

### 3. Sum of the Infinite G.P.
The sum \( S \) of an infinite geometric progression is given by:
\[
S = \frac{a}{1 - r}
\]
Where \( a \) is the first term and \( r \) is the common ratio.

Here:
- First term \( a \) is the greatest value of \( f(x) = \sin(x) + \cos(2x) \) in \([0, n]\).
- Common ratio \( r = \frac{3}{8} \).

So, the sum is:
\[
S = \frac{a}{1 - \frac{3}{8}} = \frac{a}{\frac{5}{8}} = \frac{8a}{5}
\]

Thus, the sum of the infinite G.P. is \( \frac{8a}{5} \), where \( a \) is the maximum value of \( \sin(x) + \cos(2x) \) in the interval.

### Would you like details or have any questions?
If you need specific details or calculations of \( a \), please provide the interval \( n \) or any other specific conditions. 

### Relative Questions:
1. What is the maximum value of \( f(x) = \sin(x) + \cos(2x) \) over a specific interval?
2. How do you differentiate trigonometric functions like \( \sin(x) \) and \( \cos(2x) \)?
3. What is the significance of finding the gradient of a curve at a point?
4. How is the derivative of \( \cosh(x) \) calculated?
5. What is the general form of a hyperbolic function, and how is it related to exponential functions?
6. Can you explain the concept of an infinite geometric progression?
7. How do you compute the sum of an infinite geometric series?
8. What are common applications of geometric progressions in real-world problems?

### Tip:
For finding the maximum or minimum values of a function, always start by taking the derivative and setting it to zero to locate the critical points.

Explore how to find the sum of an infinite geometric progression with the greatest value of the function f(x) = sin(x) + cos(2x) within a specified interval. Understand the differentiation of trigonometric functions and the application of geometric series in calculus. Suitable for advanced high school students or college-level learners.

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