Let f(x)=tanmx where m=p = greatest integer less than or equal to p and Principal period of f(x) is π, then
2≤p≤3
1≤p≤2
1≤p<2
3≤p<4
Explanation for the correct answer:
Finding the value of m=p:
Principle period of f(x)=πm...1
But the period of f(x)=π...2
From 1 and 2
πm=π⇒m=1⇒m=1⇒p=1⇒1≤p<2
Therefore, the correct answer is option (C).