Let `f(x)=(sin 2n x)/(1+cos^(2)nx), n in N ` has `(pi)/(6)` as its fundamental period , then n=

Options

A. 2
B. 4
C. 6
D. none of these

Explanation

To solve the problem, we need to find the value of \( n \) such that the function

\[ f(x) = \frac{\sin(2nx)}{1 + \cos^2(nx)} \]

has a fundamental period of \( \frac{\pi}{6} \).

### Step 1: Determine the period of the numerator
The numerator of the function is \( \sin(2nx) \). The period of \( \sin(kx) \) is given by

\[ \frac{2\pi}{k}. \]

For \( \sin(2nx) \), we have \( k = 2n \), so the period \( P_1 \) is:

\[ P_1 = \frac{2\pi}{2n} = \frac{\pi}{n}. \]

### Step 2: Determine the period of the denominator
The denominator is \( 1 + \cos^2(nx) \). The function \( \cos^2(kx) \) has a period of

\[ \frac{\pi}{k}. \]

For \( \cos^2(nx) \), we have \( k = n \), so the period \( P_2 \) is:

\[ P_2 = \frac{\pi}{n}. \]

### Step 3: Find the least common multiple (LCM) of the periods
The overall period of the function \( f(x) \) will be the least common multiple of \( P_1 \) and \( P_2 \):

\[ \text{lcm}(P_1, P_2) = \text{lcm}\left(\frac{\pi}{n}, \frac{\pi}{n}\right) = \frac{\pi}{n}. \]

### Step 4: Set the LCM equal to the given fundamental period
We are given that the fundamental period is \( \frac{\pi}{6} \). Therefore, we set:

\[ \frac{\pi}{n} = \frac{\pi}{6}. \]

### Step 5: Solve for \( n \)
To find \( n \), we can multiply both sides by \( n \) and \( 6 \):

\[ 6 = n. \]

Thus, we find:

\[ n = 6. \]

### Conclusion
The value of \( n \) is \( 6 \).