Question

Period of $f\left(x\right)=\sin \frac{\pi x}{(n-1)!}+\cos \frac{\pi x}{n!}$ is

• A. $n!$
• B. $2(n!)$
• C. $2(n-1)!$
• D. Does not exist

Answer 1

The correct option is B $2(n!)$

We have $g\left(x\right)=\sin \frac{\pi x}{(n-1)!}$ is a periodic function of $2(n-1)!$.

As $g\left\{2\right(n-1)!+x\}=\sin \pi \left(\frac{2(n-1)!+x}{(n-1)!}\right)=\sin (2\pi +\frac{\pi x}{(n-1)!})=\sin \frac{\pi x}{(n-1)!}=g\left(x\right)\forall x$.

Again $\cos \frac{\pi x}{n!}$ is periodic function of period $2n!$.

As, let $h\left(x\right)=\cos \frac{\pi x}{n!}$

$\Rightarrow h(2n!+x)=\cos (2\pi +\frac{\pi x}{2n!})=\cos \frac{\pi x}{2n!}=h\left(x\right)\forall x$

Now we have $f\left(x\right)=g\left(x\right)+h\left(x\right)$.

Then the period of the function $f\left(x\right)$ is LCM of $2(n-1)!$ and $2n!$ and it is $2n!$.

So option (B) is correct.