If `theta={(3^(2n))/(8)}`, where {x} = Fractional part of x then the value of `sec^(-1)(8theta)` is

Options

A. `pi`
B. `2pi`
C. `3pi`
D. 0

Answer

To solve the problem, we need to find the value of \( \sec^{-1}(8\theta) \) where \( \theta = \{ \frac{3^{2n}}{8} \} \).

### Step-by-Step Solution:

1. **Understanding the Fractional Part**:
\[ \theta = \left\{ \frac{3^{2n}}{8} \right\} \]
The fractional part function \( \{x\} \) gives us the part of \( x \) that remains after subtracting the greatest integer less than or equal to \( x \).

2. **Expressing \( 3^{2n} \)**:
We can express \( 3^{2n} \) as:
\[ 3^{2n} = 9^n \]
Therefore, we have:
\[ \theta = \left\{ \frac{9^n}{8} \right\} \]

3. **Finding the Integer and Fractional Parts**:
The expression \( \frac{9^n}{8} \) can be split into its integer part and fractional part. The integer part \( \lfloor \frac{9^n}{8} \rfloor \) is the largest integer less than or equal to \( \frac{9^n}{8} \).

4. **Calculating \( 8\theta \)**:
Now, we need to calculate \( 8\theta \):
\[ 8\theta = 8 \left\{ \frac{9^n}{8} \right\} = 9^n - 8 \lfloor \frac{9^n}{8} \rfloor \]
Since \( \{x\} = x - \lfloor x \rfloor \), we can see that \( 8\theta \) will be a value between 0 and 8.

5. **Finding \( \sec^{-1}(8\theta) \)**:
We need to find \( \sec^{-1}(8\theta) \). Since \( 8\theta \) is a fractional part, it will be less than 8. The maximum value of \( 8\theta \) will be just under 8.

6. **Determining the Value of \( \sec^{-1}(x) \)**:
The function \( \sec^{-1}(x) \) is defined for \( |x| \geq 1 \). Since \( 8\theta \) is less than 8 but greater than 0, we need to find the value of \( \sec^{-1}(1) \) since \( 8\theta \) can approach 1 when \( n \) is large.

\[ \sec^{-1}(1) = 0 \]

7. **Final Answer**:
Therefore, the value of \( \sec^{-1}(8\theta) \) is:
\[ \boxed{0} \]