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(B) Given the function $f(x) = \cos x + \cos (\sqrt{2} x)$.To find the maxima,we calculate the first derivative: $f'(x) = -\sin x - \sqrt{2} \sin (\sq

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$1$

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(B) Given the function $f(x) = \cos x + \cos (\sqrt{2} x)$.To find the maxima,we calculate the first derivative: $f'(x) = -\sin x - \sqrt{2} \sin (\sqrt{2} x)$.Setting $f'(x) = 0$,we get $\sin x + \sqrt{2} \sin (\sqrt{2} x) = 0$.At $x = 0$,$f'(0) = -\sin(0) - \sqrt{2} \sin(0) = 0$.Now,check the second derivative: $f''(x) = -\cos x - 2 \cos (\sqrt{2} x)$.At $x = 0$,$f''(0) = -\cos(0) - 2 \cos(0) = -1 - 2 = -3$.Since $f''(0) Because $\sqrt{2}$ is an irrational number,the function $f(x)$ is not periodic. The values of $\cos x$ and $\cos (\sqrt{2} x)$ can only both be $1$ simultaneously at $x = 0$. For any other $x 
eq 0$,the sum $\cos x + \cos (\sqrt{2} x)$ will be strictly less than $2$. Thus,$x = 0$ is the unique point where the maximum value of $2$ is attained.

The number of values of $x$ where the function $f(x) = \cos x + \cos (\sqrt{2} x)$ attains its maximum is

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