Consider the function f (x)= x^2. Then - Vidyadip

MCQ

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

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

✓

Answer

Correct option: C.

f is not periodic

(C)
Let $f (x)$ is periodic with period T.
Then, $\cos (x+ T )^2=\cos x^2$ for all $x \in R$
$\Rightarrow \cos (x+ T )^2-\cos x^2=0$
$\Rightarrow-2 \sin \left(\frac{(x+ T )^2+x^2}{2}\right) \sin \left(\frac{(x+ T )^2-x^2}{2}\right)=0$
$\forall x \in R$
$\Rightarrow(x+ T )^2-x^2= n \pi$ or $(x+ T )^2+x^2= n \pi$
$\forall x \in R$
Here $T$ is denendent on the value of $r$
$\Rightarrow f (x)$ is not periodic.

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