Question

The period of \(f(x) = \sin\left(\frac{\pi x}{n-1}\right) + \cos\left(\frac{\pi x}{n}\right)\), \(n \in \mathbb{Z}\), \(n>2\) is

• A. \(2\pi n(n-1)\)
• B. \(4n(n-1)\)
• C. \(2n(n-1)\)
• D. None of the above

Hint:

Period of \(\sin(ax)\) = \(\frac{2\pi}{|a|}\).

Answer 1

The Correct Option is C

To determine the period of the function \(f(x) = \sin\left(\frac{\pi x}{n-1}\right) + \cos\left(\frac{\pi x}{n}\right)\), let's analyze each component separately.

1. The period of the sine function \(\sin\left(\frac{\pi x}{n-1}\right)\) is calculated as follows:
   • The standard period of \(\sin(kx)\) is \(\frac{2\pi}{k}\).
   • Here, \(k = \frac{\pi}{n-1}\), so the period becomes \(\frac{2\pi}{\frac{\pi}{n-1}} = 2(n-1)\).
2. The period of the cosine function \(\cos\left(\frac{\pi x}{n}\right)\) is calculated similarly:
   • The period of \(\cos(kx)\) is \(\frac{2\pi}{k}\).
   • Here, \(k = \frac{\pi}{n}\), so the period becomes \(\frac{2\pi}{\frac{\pi}{n}} = 2n\).
3. To find the period of \(f(x)\), we need the Least Common Multiple (LCM) of the two periods, \(2(n-1)\) and \(2n\).
   • Factor out the common \(2\): We are left with periods \(n-1\) and \(n\).
   • The LCM of \(n\) and \(n-1\) is \(n(n-1)\).
   • Thus, the LCM of the original periods, \(2(n-1)\) and \(2n\), is \(2 \times n(n-1) = 2n(n-1)\).

Therefore, the period of the function \(f(x) = \sin\left(\frac{\pi x}{n-1}\right) + \cos\left(\frac{\pi x}{n}\right)\) is \(2n(n-1)\), which matches option 3.