Math Problem Statement

consider a function f(x)= [sin x] +[cos x] where [.] represents greatest integer function, range of f(x) is?

Solution

We are given the function:

f(x)=sinx+cosxf(x) = \lfloor \sin x \rfloor + \lfloor \cos x \rfloor

where .\lfloor . \rfloor represents the greatest integer function (or floor function). To determine the range of f(x)f(x), we need to analyze the behavior of sinx\lfloor \sin x \rfloor and cosx\lfloor \cos x \rfloor, since the sine and cosine functions take values between 1-1 and 11.

Step 1: Analyze the possible values of sinx\lfloor \sin x \rfloor

  • The sine function sinx\sin x takes values in the interval [1,1][-1, 1] for all xx.
  • Therefore, sinx\lfloor \sin x \rfloor will take values from the set {1,0}\{-1, 0\}, because:
    • If sinx[0,1)\sin x \in [0, 1), then sinx=0\lfloor \sin x \rfloor = 0.
    • If sinx[1,0)\sin x \in [-1, 0), then sinx=1\lfloor \sin x \rfloor = -1.
    • At sinx=1\sin x = 1, sinx=0\lfloor \sin x \rfloor = 0.

Step 2: Analyze the possible values of cosx\lfloor \cos x \rfloor

  • The cosine function cosx\cos x also takes values in the interval [1,1][-1, 1] for all xx.
  • Therefore, cosx\lfloor \cos x \rfloor will also take values from the set {1,0}\{-1, 0\}, because:
    • If cosx[0,1)\cos x \in [0, 1), then cosx=0\lfloor \cos x \rfloor = 0.
    • If cosx[1,0)\cos x \in [-1, 0), then cosx=1\lfloor \cos x \rfloor = -1.
    • At cosx=1\cos x = 1, cosx=0\lfloor \cos x \rfloor = 0.

Step 3: Possible values of f(x)f(x)

Now we combine the two possible values from sinx\lfloor \sin x \rfloor and cosx\lfloor \cos x \rfloor. The sum f(x)=sinx+cosxf(x) = \lfloor \sin x \rfloor + \lfloor \cos x \rfloor can take the following values:

  • If sinx=0\lfloor \sin x \rfloor = 0 and cosx=0\lfloor \cos x \rfloor = 0, then f(x)=0+0=0f(x) = 0 + 0 = 0.
  • If sinx=1\lfloor \sin x \rfloor = -1 and cosx=0\lfloor \cos x \rfloor = 0, then f(x)=1+0=1f(x) = -1 + 0 = -1.
  • If sinx=0\lfloor \sin x \rfloor = 0 and cosx=1\lfloor \cos x \rfloor = -1, then f(x)=0+(1)=1f(x) = 0 + (-1) = -1.
  • If sinx=1\lfloor \sin x \rfloor = -1 and cosx=1\lfloor \cos x \rfloor = -1, then f(x)=1+(1)=2f(x) = -1 + (-1) = -2.

Thus, the possible values for f(x)f(x) are {2,1,0}\{-2, -1, 0\}.

Final Answer:

The range of f(x)f(x) is {2,1,0}\{-2, -1, 0\}.

Do you need any more details or have further questions?

Here are some related questions to expand the topic:

  1. What is the range of f(x)=sinx+tanxf(x) = \lfloor \sin x \rfloor + \lfloor \tan x \rfloor?
  2. How does the range of f(x)=sinx×cosxf(x) = \lfloor \sin x \rfloor \times \lfloor \cos x \rfloor change?
  3. What is the behavior of f(x)=sinx+cosxf(x) = \lceil \sin x \rceil + \lfloor \cos x \rfloor, where .\lceil . \rceil is the ceiling function?
  4. How does the range change if f(x)=[x]+sin(x)f(x) = [x] + \sin(x)?
  5. What would be the maximum value of sinx+cosx\lfloor \sin x \rfloor + \lfloor \cos x \rfloor on the interval [0,2π][0, 2\pi]?

Tip: The greatest integer function x\lfloor x \rfloor returns the largest integer less than or equal to xx, which simplifies many trigonometric problems!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Greatest Integer Function
Floor Function

Formulas

f(x) = ⌊sin x⌋ + ⌊cos x⌋
⌊x⌋ represents the greatest integer less than or equal to x

Theorems

Greatest Integer Function Theorem

Suitable Grade Level

Grades 10-12