Question

Consider the function $f\left(x\right)=\cos x^2$. Then

• A. $f$ is of period $2\pi$
• B. $f$ is of period $\sqrt{2\pi }$
• C. $f$ is not periodic
• D. $f$ is of periodic $\pi$

Answer 1

The correct option is C $f$ is not periodic

Let $f\left(x\right)$ be periodic with period equal to $T$
Then, $f(x+T)=f\left(x\right)$
$\Rightarrow \cos (x+T{)}^2=\cos x^2\forall x\in R$

Using general solution
$(x+T{)}^2=2n\pi \pm x^2\forall x\in R$
$(x+T{)}^2=2n\pi +x^2$ or $(x+T{)}^2=2n\pi -x^2\forall x\in R$

There can be maximum $4$ values of $x$ satifying either of them.
$\Rightarrow f\left(x\right)$ is not periodic.