Question

Let $f\left(x\right)=\tan \sqrt{m}x$ where $m=\left[p\right]$ = greatest integer less than or equal to $p$ and Principal period of $f\left(x\right)$ is $\mathrm{\pi }$, then

• A. $2\le p\le 3$
• B. $1\le p\le 2$
• C. $1\le p<2$
• D. $3\le p<4$

Answer 1

The correct option is C

$1\le p<2$

Explanation for the correct answer:

Finding the value of $m=\left[p\right]$:

Principle period of $f\left(x\right)=\frac{\mathrm{\pi }}{\sqrt{m}}...\left(1\right)$

But the period of $f\left(x\right)=\mathrm{\pi }...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} \frac{\mathrm{\pi }}{\sqrt{m}}=\mathrm{\pi } \\ \Rightarrow \sqrt{\mathrm{m}}=1 \\ \Rightarrow \mathrm{m}=1 \\ \Rightarrow \left[\mathrm{p}\right]=1 \\ \Rightarrow 1\le \mathrm{p}<2 \end{gathered}$$

Therefore, the correct answer is option (C).