Period of cos(x2) is

$\frac{\pi}{2}$ $2\pi$ $\pi$ none of these

none of these

Let $\cos x^2$ has periodic $T \cos (x+T)^2=\cos x^2 \Rightarrow(x+T)^2=2 n \pi \pm x^2$ $\Rightarrow(x+T)^2 \pm x^2=2 n \pi$ Since it is a quadratic in T and if solved, T will never be independent of x. Hence it is not periodic function. Hence (4) is the correct answer.