Q87 / 155 · WBJEE 2019 · Mathematics
Consider the function $f(x)=\cos x^{2}$. Then,
- A$f$ is of period $2 \pi$
- B$f$ is of period $\sqrt{2 \pi}$
- Correct answerC$f$ is not periodic
- D$f$ is of period $\pi$
Solution
We have,
$$
f(x)=\cos x^{2}
$$
Let $T$ be the period of $f(x)$. Then
$$
\begin{aligned}
& & f(x+T) &=f(x) \\
\Rightarrow & & \cos (x+T)^{2} &=\cos x^{2}
\end{aligned}
$$
But there is no value of $T$ for which
$$
\cos (x+T)^{2}=\cos x^{2}
$$
$\therefore f(x)$ is not periodic
Hence, the correct answer is option: 'C'.