Question

If the period of the function $f\left(x\right)=\sin \left(\sqrt{\left[n\right]}\right)x$, where[n] denotes the greatest integer less than or equal to $n$, is $2\pi$, then

• A. $1\le n<2$
• B. $1<n<2$
• C. $1\le n\le 2$
• D. None of these

Answer 1

Answer: (A)

$1\le n<2$

We know period of $\sin ax$ is $\frac{2\pi }{a}$
Thus period of $f\left(x\right)=\sin \left(\sqrt{\left[n\right]}\right)x$ is $\frac{2\pi }{\sqrt{\left[n\right]}}=2\pi$ (given)
$\Rightarrow \sqrt{\left[n\right]}=1$

$\Rightarrow \left[n\right]=1$

$\Rightarrow n\in [1,2)$

$\Rightarrow 1\le n<2$