Math Problem Statement

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:

Solution

Let's break down the problem step by step.

1. Greatest Value of the Function f(x)=sin(x)+cos(2x)f(x) = \sin(x) + \cos(2x) in the interval [0,n][0, n]

First, we need to determine the greatest value of the function f(x)=sin(x)+cos(2x)f(x) = \sin(x) + \cos(2x).

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

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

Set f(x)=0f'(x) = 0 to find the critical points: cos(x)(14sin(x))=0\cos(x)(1 - 4\sin(x)) = 0 This gives us: cos(x)=0orsin(x)=14\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][0, n]. However, let's denote the maximum value as aa (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=ex/2+ex/22y = \frac{e^{x/2} + e^{-x/2}}{2} at the point where x=ln4x = \ln 4.

The curve yy can be rewritten as: y=cosh(x2)y = \cosh\left(\frac{x}{2}\right) The derivative (gradient) is: dydx=12sinh(x2)\frac{dy}{dx} = \frac{1}{2}\sinh\left(\frac{x}{2}\right) At x=ln4x = \ln 4: dydxx=ln4=12sinh(ln42)\frac{dy}{dx}\bigg|_{x = \ln 4} = \frac{1}{2}\sinh\left(\frac{\ln 4}{2}\right)

Since sinh(ln42)=eln2eln22=2122=34\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 38\frac{3}{8}.

3. Sum of the Infinite G.P.

The sum SS of an infinite geometric progression is given by: S=a1rS = \frac{a}{1 - r} Where aa is the first term and rr is the common ratio.

Here:

  • First term aa is the greatest value of f(x)=sin(x)+cos(2x)f(x) = \sin(x) + \cos(2x) in [0,n][0, n].
  • Common ratio r=38r = \frac{3}{8}.

So, the sum is: S=a138=a58=8a5S = \frac{a}{1 - \frac{3}{8}} = \frac{a}{\frac{5}{8}} = \frac{8a}{5}

Thus, the sum of the infinite G.P. is 8a5\frac{8a}{5}, where aa is the maximum value of sin(x)+cos(2x)\sin(x) + \cos(2x) in the interval.

Would you like details or have any questions?

If you need specific details or calculations of aa, please provide the interval nn or any other specific conditions.

Relative Questions:

  1. What is the maximum value of f(x)=sin(x)+cos(2x)f(x) = \sin(x) + \cos(2x) over a specific interval?
  2. How do you differentiate trigonometric functions like sin(x)\sin(x) and cos(2x)\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)\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Geometric Progressions
Exponential Functions

Formulas

Derivative of trigonometric functions
Sum of an infinite geometric progression

Theorems

-

Suitable Grade Level

Advanced High School or College