Which of the following functions has period `2pi` ?

Options

A. `f(x)=sin(2pix+(pi)/(3))+2 sin (3pix+(pi)/(4))+3 sin5pix`
B. `f(x)=sin""(pix)/(3)+sin""(pix)/(4)`
C. `f(x)=sinx+cos2x`
D. none of these

Answer

To determine which of the given functions has a period of \(2\pi\), we need to check each function to see if it satisfies the condition for periodicity. A function \(f(x)\) is periodic with period \(T\) if:

\[ f(x) = f(x + T) \]

for all \(x\). In this case, we are looking for \(T = 2\pi\).

### Step 1: Check Option 1

The first function is:

\[ f(x) = \sin(2\pi x + \frac{\pi}{3}) + 2\sin(3\pi x + \frac{\pi}{4}) + 3\sin(5\pi x) \]

We need to compute \(f(x + 2\pi)\):

\[ f(x + 2\pi) = \sin(2\pi(x + 2\pi) + \frac{\pi}{3}) + 2\sin(3\pi(x + 2\pi) + \frac{\pi}{4}) + 3\sin(5\pi(x + 2\pi)) \]

Calculating each term:

1. \( \sin(2\pi(x + 2\pi) + \frac{\pi}{3}) = \sin(2\pi x + 4\pi + \frac{\pi}{3}) = \sin(2\pi x + \frac{\pi}{3})\) (since sine is periodic with period \(2\pi\))
2. \( 2\sin(3\pi(x + 2\pi) + \frac{\pi}{4}) = 2\sin(3\pi x + 6\pi + \frac{\pi}{4}) = 2\sin(3\pi x + \frac{\pi}{4})\)
3. \( 3\sin(5\pi(x + 2\pi)) = 3\sin(5\pi x + 10\pi) = 3\sin(5\pi x)\)

Thus,

\[ f(x + 2\pi) = \sin(2\pi x + \frac{\pi}{3}) + 2\sin(3\pi x + \frac{\pi}{4}) + 3\sin(5\pi x) \neq f(x) \]

So, **Option 1 does not have a period of \(2\pi\)**.

### Step 2: Check Option 2

The second function is:

\[ f(x) = \sin\left(\frac{\pi x}{3}\right) + \sin\left(\frac{\pi x}{4}\right) \]

Now we compute \(f(x + 2\pi)\):

\[ f(x + 2\pi) = \sin\left(\frac{\pi (x + 2\pi)}{3}\right) + \sin\left(\frac{\pi (x + 2\pi)}{4}\right) \]

Calculating each term:

1. \( \sin\left(\frac{\pi (x + 2\pi)}{3}\right) = \sin\left(\frac{\pi x}{3} + \frac{2\pi^2}{3}\right) \neq \sin\left(\frac{\pi x}{3}\right)\)
2. \( \sin\left(\frac{\pi (x + 2\pi)}{4}\right) = \sin\left(\frac{\pi x}{4} + \frac{\pi^2}{2}\right) \neq \sin\left(\frac{\pi x}{4}\right)\)

Thus,

\[ f(x + 2\pi) \neq f(x) \]

So, **Option 2 does not have a period of \(2\pi\)**.

### Step 3: Check Option 3

The third function is:

\[ f(x) = \sin(x) + \cos(2x) \]

Now we compute \(f(x + 2\pi)\):

\[ f(x + 2\pi) = \sin(x + 2\pi) + \cos(2(x + 2\pi)) \]

Calculating each term:

1. \( \sin(x + 2\pi) = \sin(x)\) (since sine is periodic with period \(2\pi\))
2. \( \cos(2(x + 2\pi)) = \cos(2x + 4\pi) = \cos(2x)\) (since cosine is periodic with period \(2\pi\))

Thus,

\[ f(x + 2\pi) = \sin(x) + \cos(2x) = f(x) \]

So, **Option 3 has a period of \(2\pi\)**.

### Conclusion

The function that has a period of \(2\pi\) is:

\[ \boxed{f(x) = \sin(x) + \cos(2x)} \]