Let f(x)=[cosx+ sin x], `0 lt x lt 2pi`, where [x] denotes the greatest integer less than or equal to x. The number of points of discontinuity of f(x) is

Options

A. 6
B. 5
C. 4
D. 3

Explanation

To solve the problem, we need to analyze the function \( f(x) = [\cos x + \sin x] \) where \( [x] \) denotes the greatest integer less than or equal to \( x \), and \( 0 < x < 2\pi \).

### Step 1: Determine the range of \( \cos x + \sin x \)

The expression \( \cos x + \sin x \) can be rewritten using the sine addition formula. We have:

\[ \cos x + \sin x = \sqrt{2} \left( \sin\left(x + \frac{\pi}{4}\right) \right) \]

The maximum value of \( \sin \) function is 1 and the minimum value is -1. Therefore, the range of \( \cos x + \sin x \) is:

\[ -\sqrt{2} \leq \cos x + \sin x \leq \sqrt{2} \]

### Step 2: Calculate the approximate values of the range

Calculating the numerical values:

- The maximum value \( \sqrt{2} \approx 1.414 \)
- The minimum value \( -\sqrt{2} \approx -1.414 \)

Thus, the range of \( \cos x + \sin x \) is approximately:

\[ [-1.414, 1.414] \]

### Step 3: Determine the integer values in the range

The greatest integer function \( [x] \) will take integer values within this range. The integers that fall within the range \( [-1.414, 1.414] \) are:

\[ -1, 0, 1 \]

### Step 4: Find the points where \( f(x) \) changes value

Next, we need to find the points where \( f(x) \) changes its value, which corresponds to the points where \( \cos x + \sin x \) crosses the integers -1, 0, and 1.

1. **For \( f(x) = -1 \)**:
- Solve \( \cos x + \sin x = -1 \):
- This occurs when \( \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = -1 \), which gives no solutions in \( (0, 2\pi) \).

2. **For \( f(x) = 0 \)**:
- Solve \( \cos x + \sin x = 0 \):
- This occurs when \( \tan x = 1 \) or \( x = \frac{\pi}{4} + n\pi \). In \( (0, 2\pi) \), the solutions are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

3. **For \( f(x) = 1 \)**:
- Solve \( \cos x + \sin x = 1 \):
- This occurs when \( \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), which gives \( x = \frac{\pi}{4} - \frac{\pi}{4} + 2n\pi \) and \( x = \frac{3\pi}{4} - \frac{\pi}{4} + 2n\pi \). In \( (0, 2\pi) \), the solutions are \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \).

### Step 5: Count the points of discontinuity

The points of discontinuity occur at the transitions between the integer values of \( f(x) \). We found the following points where \( f(x) \) changes value:

- \( x = \frac{\pi}{4} \) (from 0 to 1)
- \( x = \frac{5\pi}{4} \) (from 1 to 0)

Thus, the total number of points of discontinuity of \( f(x) \) is **2**.

### Final Answer

The number of points of discontinuity of \( f(x) \) is **2**.

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