Math Problem Statement

If ( f(x)=cos left[e^{2}right] x+cos left[-e^{2}right] x ) where ( [x] ) stands for greatest integer function, then 1) ( f(pi)=1 ) 2) ( f(2 pi)=1 ) 3) ( f(pi / 2)=1 ) 4) ( f(pi / 4)=1 ) If

Solution

Given the function f(x)=cos(e2)[x]+cos(e2)[x]f(x) = \cos\left(e^2\right) \cdot [x] + \cos\left(-e^2\right) \cdot [x], where [x][x] represents the greatest integer function (also known as the floor function), we need to analyze the value of f(x)f(x) for different values of xx.

Step 1: Simplify the Function

We start by recognizing that cos(e2)=cos(e2)\cos\left(e^2\right) = \cos\left(-e^2\right) because the cosine function is an even function, meaning cos(θ)=cos(θ)\cos(-\theta) = \cos(\theta). So the function simplifies to:

[ f(x) = \cos\left(e^2\right) \cdot [x] + \cos\left(e^2\right) \cdot [x] ] f(x)=2cos(e2)[x]f(x) = 2 \cdot \cos\left(e^2\right) \cdot [x]

Step 2: Evaluate f(x)f(x) for Specific Values

Let's evaluate the function for the given specific values of xx.

  1. For x=πx = \pi: f(π)=2cos(e2)[π]f(\pi) = 2 \cdot \cos\left(e^2\right) \cdot [\pi] Since π\pi is approximately 3.14159, [π]=3[\pi] = 3: f(π)=2cos(e2)3f(\pi) = 2 \cdot \cos\left(e^2\right) \cdot 3 This is not necessarily equal to 1 unless cos(e2)\cos\left(e^2\right) is a specific value, but it's not generally 1.

  2. For x=2πx = 2\pi: f(2π)=2cos(e2)[2π]f(2\pi) = 2 \cdot \cos\left(e^2\right) \cdot [2\pi] Since 2π2\pi is approximately 6.28318, [2π]=6[2\pi] = 6: f(2π)=2cos(e2)6f(2\pi) = 2 \cdot \cos\left(e^2\right) \cdot 6 Again, this is not necessarily equal to 1.

  3. For x=π2x = \frac{\pi}{2}: f(π2)=2cos(e2)[π2]f\left(\frac{\pi}{2}\right) = 2 \cdot \cos\left(e^2\right) \cdot \left[\frac{\pi}{2}\right] Since π2\frac{\pi}{2} is approximately 1.5708, [π2]=1\left[\frac{\pi}{2}\right] = 1: f(π2)=2cos(e2)1=2cos(e2)f\left(\frac{\pi}{2}\right) = 2 \cdot \cos\left(e^2\right) \cdot 1 = 2 \cdot \cos\left(e^2\right) This is not necessarily equal to 1.

  4. For x=π4x = \frac{\pi}{4}: f(π4)=2cos(e2)[π4]f\left(\frac{\pi}{4}\right) = 2 \cdot \cos\left(e^2\right) \cdot \left[\frac{\pi}{4}\right] Since π4\frac{\pi}{4} is approximately 0.7854, [π4]=0\left[\frac{\pi}{4}\right] = 0: f(π4)=2cos(e2)0=0f\left(\frac{\pi}{4}\right) = 2 \cdot \cos\left(e^2\right) \cdot 0 = 0 This is certainly not equal to 1.

Conclusion

From this analysis:

  • f(π)=6cos(e2)f(\pi) = 6 \cdot \cos(e^2)
  • f(2π)=12cos(e2)f(2\pi) = 12 \cdot \cos(e^2)
  • f(π2)=2cos(e2)f(\frac{\pi}{2}) = 2 \cdot \cos(e^2)
  • f(π4)=0f(\frac{\pi}{4}) = 0

None of these values are guaranteed to be 1 under general circumstances, so none of the options given are correct.

Would you like any further details or explanations?

Related Questions

  1. What is the greatest integer function, and how does it work?
  2. How does the even nature of the cosine function affect the simplification of functions?
  3. What is the value of cos(e2)\cos(e^2) approximately?
  4. Can f(x)f(x) be 1 for some specific value of xx?
  5. What other properties of the floor function are useful in simplifying expressions?
  6. How do you calculate the floor function for various values?
  7. How does the choice of xx affect the value of f(x)f(x)?
  8. Can you find a value of xx for which f(x)=1f(x) = 1?

Tip: Always check if the function in a trigonometric expression simplifies due to the properties of the trigonometric functions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Greatest Integer Function (Floor Function)
Cosine Function

Formulas

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Theorems

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Suitable Grade Level

Grades 11-12