Question

$f\left(x\right)=\cos \left(x^2\right)$
If $f\left(x\right)$ is periodic enter $1$ else enter $0$

Answer 1

Let $T$ be the period of $f\left(x\right)$

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

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

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

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