Question

If $f\left(x\right)={\tan }^2\frac{\pi x}{n^2-5n+8}+\cot (n+m)\pi x;(n\in N,m\in Q)$, is a periodic function with $2$ as its fundamental period, then $m$ can't belong to

• A. $(-\infty ,-2)\cup (-1,\infty )$
• B. $(-\infty ,-3)\cup (-2,\infty )$
• C. $(-2,-1)\cup (-3,-2)$
• D. $(-3,-\frac{5}{2})\cup (-\frac{5}{2},-2)$

Answer 1

Answer: (C)

$(-2,-1)\cup (-3,-2)$

Period of LCM of $n^2-5n+8$ and $\frac{1}{n+m}$.
$\Rightarrow n^2-5n+8=1$ or $n^2-5n+8=2$
$\Rightarrow n^2-5n+7=0$ or $n^2-5n+6=0$
Since, $nϵN,\therefore n=2,3$
and $\left(\frac{1}{n+m}\right)K_1=2,\left(K_1ϵI\right)\Rightarrow \frac{1}{n+m}\text{/}>1$
$\therefore \frac{1}{2+m}\text{/}>$ or $\frac{1}{3+m}\text{/}>1$
$\Rightarrow m\text{/}ϵ(-2,-1)\cup (-3,-2)$
Hence, (c) is the correct answer.