If $f(x)=\tan^2\left(\dfrac{\pi x}{n^2-5n+8}\right)+\cot(\left(n+m)\pi x\right),\;(n\in N, m\in Q)$ is a periodic function with its fundamental period as $2$. Then find interval in which $m$ cannot belong to
(a) $(-\infty,-2)\cup (-1, \infty)$
(b) $(-\infty,-3)\cup (-2, \infty)$
(c) $(-3,-2)\cup(-2,-1)$
(d) $\left(-3,-\dfrac{5}{2}\right)\cup\left(-\dfrac{5}{2},-2\right)$
My Approach:
Period of $\tan^2\left(\dfrac{\pi x}{n^2-5n+8}\right)$ is $n^2-5n+8$ and Period of $\cot(\left(n+m)\pi x\right)$ is $\dfrac{1}{(m+n)}$
Period of overall function will be LCM of $n^2-5n+8$ and $\dfrac{1}{(m+n)}$
Let $m=\frac{p}{q}$ where $\gcd(p,q)=1$ and $p,q \in I$
$\text{LCM}\left(n^2-5n+8,\dfrac{1}{\frac{p}{q}+n}\right)$
$\implies \text{LCM}\left(n^2-5n+8,\dfrac{q}{p+nq}\right)$ $\implies \dfrac{\text{LCM}(n^2-5n+8,q)}{\text{HCF}(1,np+q)}=\text{LCM}\left(n^2-5n+8,q\right)$
So we want $\text{LCM}\left(n^2-5n+8,q\right)=2$
Now I don 't Know how to proceed further and How to find interval of $m$
Note:$\text{LCM}\left(\dfrac{x}{y},\dfrac{r}{t}\right)=\dfrac{\text{LCM}(x,r)}{\text{HCF}(y,t)}$ and $\text{HCF}\left(\dfrac{x}{y},\dfrac{r}{t}\right)=\dfrac{\text{HCF}(x,r)}{\text{LCM}(y,t)}$