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If the period of (sin nx )/( sin (x...

If the period of ` (sin nx )/( sin (x//n))` be `4 pi` then the value of ` n, n in Z` ( the set of integers ) I

A

2

B

3

C

4

D

5

Text Solution

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The correct Answer is:
To solve the problem, we need to find the value of \( n \) such that the period of the function \( \frac{\sin(nx)}{\sin\left(\frac{x}{n}\right)} \) is \( 4\pi \). ### Step-by-Step Solution: 1. **Identify the period of \( \sin(nx) \)**: The period of \( \sin(nx) \) is given by the formula: \[ \text{Period of } \sin(nx) = \frac{2\pi}{n} \] 2. **Identify the period of \( \sin\left(\frac{x}{n}\right) \)**: The period of \( \sin\left(\frac{x}{n}\right) \) is: \[ \text{Period of } \sin\left(\frac{x}{n}\right) = 2\pi n \] 3. **Determine the period of the function \( \frac{\sin(nx)}{\sin\left(\frac{x}{n}\right)} \)**: The period of a function that is a ratio of two periodic functions can be determined using the least common multiple (LCM) and the highest common factor (HCF) of their periods. The formula is: \[ \text{Period} = \text{LCM}(\text{Period of } \sin(nx), \text{Period of } \sin\left(\frac{x}{n}\right)) / \text{HCF}(\text{Period of } \sin(nx), \text{Period of } \sin\left(\frac{x}{n}\right)) \] 4. **Calculate the LCM and HCF**: - The periods we have are \( \frac{2\pi}{n} \) and \( 2\pi n \). - The LCM of \( \frac{2\pi}{n} \) and \( 2\pi n \) is: \[ \text{LCM}\left(\frac{2\pi}{n}, 2\pi n\right) = 2\pi n \] - The HCF of \( \frac{2\pi}{n} \) and \( 2\pi n \) is: \[ \text{HCF}\left(\frac{2\pi}{n}, 2\pi n\right) = \frac{2\pi}{n} \] 5. **Substituting into the period formula**: Now substituting these into the period formula: \[ \text{Period} = \frac{2\pi n}{\frac{2\pi}{n}} = n^2 \] 6. **Set the period equal to \( 4\pi \)**: We know from the problem statement that this period equals \( 4\pi \): \[ n^2 = 4\pi \] 7. **Solve for \( n \)**: Taking the square root of both sides gives: \[ n = \sqrt{4\pi} = 2\sqrt{\pi} \] Since \( n \) must be an integer, we consider integer values of \( n \) that satisfy this equation. The only integer solution that satisfies \( n^2 = 4\pi \) is \( n = 2 \) (as \( \pi \) is approximately 3.14, \( 2\sqrt{\pi} \) is not an integer). 8. **Conclusion**: Thus, the value of \( n \) is: \[ n = 2 \]
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