The number of vlaues of x where the function `f(x)= cos x + cos (sqrt(2) x)` attains its maximum is

Options

A. 0
B. 1
C. 2
D. Infinite

Explanation

To solve the problem of finding the number of values of \( x \) where the function \( f(x) = \cos x + \cos(\sqrt{2} x) \) attains its maximum, we can follow these steps:

### Step 1: Understand the Function
The function given is:
\[ f(x) = \cos x + \cos(\sqrt{2} x) \]
We know that the maximum value of the cosine function is 1. Therefore, the maximum value of \( f(x) \) can be:
\[ f(x)_{\text{max}} = 1 + 1 = 2 \]

### Step 2: Set Up Conditions for Maximum
For \( f(x) \) to attain its maximum value of 2, both cosine terms must equal 1:
\[ \cos x = 1 \quad \text{and} \quad \cos(\sqrt{2} x) = 1 \]

### Step 3: Solve for \( x \)
The cosine function equals 1 at integer multiples of \( 2\pi \):
1. From \( \cos x = 1 \):
\[ x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]
2. From \( \cos(\sqrt{2} x) = 1 \):
\[ \sqrt{2} x = 2m\pi \quad \text{for } m \in \mathbb{Z} \implies x = \frac{2m\pi}{\sqrt{2}} = \frac{m\sqrt{2}\pi}{1} \quad \text{for } m \in \mathbb{Z} \]

### Step 4: Find Common Values of \( x \)
We need to find the common values of \( x \) from both equations:
1. From \( x = 2n\pi \)
2. From \( x = m\sqrt{2}\pi \)

Setting them equal gives:
\[ 2n\pi = m\sqrt{2}\pi \]
Dividing both sides by \( \pi \) (assuming \( \pi \neq 0 \)):
\[ 2n = m\sqrt{2} \implies m = \frac{2n}{\sqrt{2}} = n\sqrt{2} \]

### Step 5: Determine Integer Solutions
For \( m \) to be an integer, \( n\sqrt{2} \) must also be an integer. Since \( \sqrt{2} \) is irrational, the only integer \( n \) that satisfies this is \( n = 0 \). Thus:
\[ n = 0 \implies m = 0 \]

### Conclusion
The only solution is when \( n = 0 \) and \( m = 0 \), which gives:
\[ x = 0 \]
Thus, the function \( f(x) \) attains its maximum value at only one point.

### Final Answer
The number of values of \( x \) where the function \( f(x) \) attains its maximum is:
\[ \boxed{1} \]