Statement-1: Let `f(x)=[3+4sinx ]`, where [.] denotes the greatest integer function. The number of discontinuities of f(x) in `[pi,2pi]` is 6
Statement-2: The range of f is `[-1,0,1,2,3]`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the function \( f(x) = [3 + 4 \sin x] \), where \([.]\) denotes the greatest integer function, over the interval \([ \pi, 2\pi ]\).

### Step 1: Determine the range of \( 4 \sin x \) for \( x \in [\pi, 2\pi] \)

The sine function, \( \sin x \), varies between -1 and 0 in the interval \([ \pi, 2\pi ]\). Therefore, we can calculate the range of \( 4 \sin x \):

\[ 4 \sin x \text{ will vary from } 4(-1) \text{ to } 4(0) \text{, which is } [-4, 0]. \]

### Step 2: Calculate the range of \( 3 + 4 \sin x \)

Now we add 3 to the range we found in the previous step:

\[ 3 + 4 \sin x \text{ will vary from } 3 + (-4) \text{ to } 3 + 0 \text{, which is } [-1, 3]. \]

### Step 3: Identify the integer values in the range of \( f(x) \)

The greatest integer function \([x]\) takes the greatest integer less than or equal to \( x \). Therefore, we need to find the integers in the range \([-1, 3]\):

The integers in this range are:
\[ -1, 0, 1, 2, 3. \]

Thus, the range of \( f(x) \) is \([-1, 0, 1, 2, 3]\).

### Step 4: Identify points of discontinuity

The function \( f(x) \) will be discontinuous at the points where \( 3 + 4 \sin x \) takes integer values, which are the integers in the range we just found. We need to find when \( 3 + 4 \sin x = n \) for \( n \in \{-1, 0, 1, 2, 3\}\).

1. For \( n = -1 \):
\[ 3 + 4 \sin x = -1 \implies 4 \sin x = -4 \implies \sin x = -1 \implies x = \frac{3\pi}{2}. \]

2. For \( n = 0 \):
\[ 3 + 4 \sin x = 0 \implies 4 \sin x = -3 \implies \sin x = -\frac{3}{4} \implies x = \arcsin(-\frac{3}{4}) + 2k\pi \text{ or } \pi - \arcsin(-\frac{3}{4}) + 2k\pi. \]
In the interval \([\pi, 2\pi]\), this gives us two points.

3. For \( n = 1 \):
\[ 3 + 4 \sin x = 1 \implies 4 \sin x = -2 \implies \sin x = -\frac{1}{2} \implies x = \frac{7\pi}{6}, \frac{11\pi}{6}. \]

4. For \( n = 2 \):
\[ 3 + 4 \sin x = 2 \implies 4 \sin x = -1 \implies \sin x = -\frac{1}{4} \implies x = \arcsin(-\frac{1}{4}) + 2k\pi \text{ or } \pi - \arcsin(-\frac{1}{4}) + 2k\pi. \]
This gives us two points in the interval \([\pi, 2\pi]\).

5. For \( n = 3 \):
\[ 3 + 4 \sin x = 3 \implies 4 \sin x = 0 \implies \sin x = 0 \implies x = \pi, 2\pi. \]

### Step 5: Count the points of discontinuity

From the analysis, we have:

- 1 point for \( n = -1 \)
- 2 points for \( n = 0 \)
- 2 points for \( n = 1 \)
- 2 points for \( n = 2 \)
- 2 points for \( n = 3 \)

Total points of discontinuity = \( 1 + 2 + 2 + 2 + 2 = 9 \).

### Conclusion

- **Statement 1** is false (the number of discontinuities is 9, not 6).
- **Statement 2** is true (the range of \( f(x) \) is indeed \([-1, 0, 1, 2, 3]\)).