$f (x) = cos (x^{2})$
If $f (x)$ is periodic enter $1$ else enter $0$

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Solution

Let $T$ be the period of $f (x)$

$\Rightarrow f (x + T) = f (x) \Rightarrow cos (x + T)^{2} = cos x^{2}$

$\Rightarrow cos (x + T)^{2} - cos x^{2} = 0 \Rightarrow 2 sin \frac{(x + T)^{2} + x^{2}}{2} . sin \frac{(x + T)^{2} - x^{2}}{2} = 0$

$\Rightarrow x^{2} = (x + T)^{2} \Rightarrow T^{2} + 2 T x = 0 \Rightarrow T = 0$

But we know that period of any periodic function is a positive real number.
Hence given function is not periodic.

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