If `f(x) = sgn(sin^2x-sinx-1)` has exactly four points of discontinuity for `x in (0, npi), n in N`, then

Options

A. minimum value of n is 5
B. maximum value of n is 6
C. there are exaclty two posoible values of n
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) = \text{sgn}(\sin^2 x - \sin x - 1) \) and determine where it is discontinuous in the interval \( (0, n\pi) \) for \( n \in \mathbb{N} \).

### Step-by-step Solution:

1. **Identify the Discontinuity Condition**:
The function \( f(x) \) is discontinuous where the argument of the signum function is zero, i.e., we need to solve:
\[ \sin^2 x - \sin x - 1 = 0 \]

2. **Rearranging the Equation**:
This is a quadratic equation in terms of \( \sin x \). Let \( y = \sin x \). The equation becomes:
\[ y^2 - y - 1 = 0 \]

3. **Using the Quadratic Formula**:
We can solve for \( y \) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
Thus, we have:
\[ y = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad y = \frac{1 - \sqrt{5}}{2} \]

4. **Analyzing the Roots**:
- The first root \( \frac{1 + \sqrt{5}}{2} \) is greater than 1, which is not possible since \( \sin x \) must be in the range \([-1, 1]\).
- The second root \( \frac{1 - \sqrt{5}}{2} \) is approximately \(-0.618\), which is valid since it lies within the range of sine.

5. **Finding the Sine Values**:
We need to find the values of \( x \) such that:
\[ \sin x = \frac{1 - \sqrt{5}}{2} \]

6. **Determining the Points of Discontinuity**:
The sine function is periodic with period \( 2\pi \). The values of \( x \) for \( \sin x = \frac{1 - \sqrt{5}}{2} \) will occur at:
\[ x = \arcsin\left(\frac{1 - \sqrt{5}}{2}\right) + 2k\pi \quad \text{and} \quad x = \pi - \arcsin\left(\frac{1 - \sqrt{5}}{2}\right) + 2k\pi \quad (k \in \mathbb{Z}) \]

7. **Counting the Discontinuities in the Interval**:
Within the interval \( (0, n\pi) \), we need to count how many times these angles occur. Each period \( 2\pi \) will contribute two points of discontinuity. Therefore, in the interval \( (0, n\pi) \):
- For \( n = 4 \): There are \( 4 \) points of discontinuity.
- For \( n = 5 \): There are \( 5 \) points of discontinuity.

8. **Conclusion**:
Since we want exactly four points of discontinuity, we conclude that \( n = 4 \) is the correct answer.

### Final Answer:
The value of \( n \) is \( 4 \).