Question

The period of $f\left(x\right)=\sin \frac{\pi x}{n!}-\cos \frac{\pi x}{(n+1)!}$ is $a(n+1)!$. Find $a$

Answer 1

Period of $\sin \frac{\pi x}{n!}$ is $\frac{n!.2\pi }{\pi }=2.(n!)=T_1$

and period of $\cos \frac{\pi x}{(n+1)!}$ is $\frac{(n+1)!.2\pi }{\pi }=2.(n+1)!==2(n+1)n!=T_2$

Hence, the period of $f\left(x\right)$ is L.C.M of $T_1$ and $T_2$, which is

$=2.(n+1)!$

by comparing it with $a(n+1)!$ we get

$\therefore a=2$