Question

If the period of the function $f\left(x\right)=\sin \left(\sqrt{\left[n\right]}x\right)$, where $\left[n\right]$ 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

The correct option is A $1\le n<2$

$\sin x$ is a periodic function with period $2\pi$,

therefore $\sin \left(\sqrt{\left[n\right]}x\right)$ is a periodic function with period $\frac{2\pi }{\sqrt{\left[n\right]}}$.
But the period of $f\left(x\right)$ is $2\pi$ (given).
$\therefore \frac{2\pi }{\sqrt{\left[n\right]}}=2\pi$

$\Rightarrow \sqrt{\left[n\right]}=1$

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

$\Rightarrow 1\le n<2$