Let `F(n)=(sin1)xx(sin2)xx....sin(n),AA"n"in"N"` then number of elements in the set `A={f(1),f(2),..........,f(6)}` that are positive are

Options

A. 3
B. 4
C. 5
D. 2

Answer

To solve the problem, we need to analyze the function \( F(n) = \sin(1) \cdot \sin(2) \cdots \sin(n) \) for \( n = 1, 2, \ldots, 6 \) and determine how many of these values are positive.

### Step-by-step Solution:

1. **Calculate \( F(1) \)**:
\[ F(1) = \sin(1) \]
Since \( \sin(1) > 0 \), \( F(1) \) is positive.

2. **Calculate \( F(2) \)**:
\[ F(2) = \sin(1) \cdot \sin(2) \]
Both \( \sin(1) > 0 \) and \( \sin(2) > 0 \), hence \( F(2) > 0 \).

3. **Calculate \( F(3) \)**:
\[ F(3) = \sin(1) \cdot \sin(2) \cdot \sin(3) \]
Since \( \sin(1) > 0 \), \( \sin(2) > 0 \), and \( \sin(3) > 0 \), we conclude that \( F(3) > 0 \).

4. **Calculate \( F(4) \)**:
\[ F(4) = \sin(1) \cdot \sin(2) \cdot \sin(3) \cdot \sin(4) \]
Here, \( \sin(4) < 0 \) (as \( 4 \) radians is in the range where sine is negative). Therefore, \( F(4) < 0 \).

5. **Calculate \( F(5) \)**:
\[ F(5) = \sin(1) \cdot \sin(2) \cdot \sin(3) \cdot \sin(4) \cdot \sin(5) \]
Since \( F(4) < 0 \) and \( \sin(5) > 0 \), we have \( F(5) < 0 \) (negative times positive is negative).

6. **Calculate \( F(6) \)**:
\[ F(6) = \sin(1) \cdot \sin(2) \cdot \sin(3) \cdot \sin(4) \cdot \sin(5) \cdot \sin(6) \]
Again, \( F(5) < 0 \) and \( \sin(6) < 0 \), so \( F(6) > 0 \) (negative times negative is positive).

### Summary of Results:
- \( F(1) > 0 \)
- \( F(2) > 0 \)
- \( F(3) > 0 \)
- \( F(4) < 0 \)
- \( F(5) < 0 \)
- \( F(6) > 0 \)

### Count of Positive Values:
The positive values are \( F(1), F(2), F(3), \) and \( F(6) \). Thus, there are **4 positive elements** in the set \( A = \{ F(1), F(2), F(3), F(4), F(5), F(6) \} \).

### Final Answer:
The number of elements in the set \( A \) that are positive is **4**.

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