Question

Period of $f\left( x \right)=\sin \dfrac{\pi x}{\left( n-1 \right)!}+\cos \dfrac{\pi x}{n!}$ is
a) n!
b) 2(n!)
c) 2(n-1)!
d) Does not exist

Answer 1

Hint: First represent f(x) as g(x) + h(x) where g(x) is $\sin \dfrac{\pi x}{\left( n-1 \right)!}$ and h(x) is $\cos \dfrac{\pi x}{n!}$ then find the period of functions separately then take the L.C.M of the two function to get the desired result.

Complete step-by-step answer:
We have been given function $f\left( x \right)=\sin \dfrac{\pi x}{\left( n-1 \right)!}+\cos \dfrac{\pi x}{n!}$ and we have to find the period of f(x).
Now let’s assume $\sin \dfrac{\pi x}{\left( n-1 \right)!}$ as function g(x) and $\cos \dfrac{\pi x}{n!}$ as function h(x)
So we can represent f(x) = g(x) + h(x)
Now we will find the periodicity of function g(x) which is equal to $\sin \dfrac{\pi x}{\left( n-1 \right)!}$ by using formula that if there is a function is in form of $\sin \left( kx \right)$ where K is a constant then the period of $\sin \left( kx \right)$ is $\dfrac{2\pi }{k}$ So in the function g(x) the value of K is $\dfrac{\pi }{\left( n-1 \right)!}$. So the period of g(x) or $\sin \left( \dfrac{\pi x}{\left( n-1 \right)!} \right)$will be $\dfrac{2\pi }{\dfrac{\pi }{\left( n-1 \right)!}}$which is equal to 2(n - 1)!
So now we will check periodicity by putting x as 2(n - 1)! + x in function g(x).
So,
$\begin{align}
  & g\left\{ 2\left( n-1 \right)!+x \right\} \\
 & =\sin \pi \left( \dfrac{2\left( n-1 \right)!+x}{\left( n-1 \right)!} \right) \\
\end{align}$
On simplifying we get,
$=\sin \pi \left( 2+\dfrac{x}{\left( n-1 \right)!} \right)$
This is equal to $\sin \left( 2\pi +\dfrac{\pi x}{\left( n-1 \right)!} \right)$
So, here we will use fact that $\sin \left( 2\pi +\theta \right)=\sin \theta $
So, $\sin \left( 2\pi +\dfrac{\pi x}{\left( n-1 \right)!} \right)=\sin \left( \dfrac{\pi x}{\left( n-1 \right)!} \right)$
Here $g\left( 2\left( n-1 \right)!+x \right)=g\left( x \right)$.
Now let’s find periodicity of h(x) which is $\cos \dfrac{\pi x}{n!}$ which can be find out using formula if there is function $\cos \left( kx \right)$ exists then its period will be $'\dfrac{2\pi }{k}'$ if K is a constant.
So for function h(x) which is equal to $\cos \dfrac{\pi x}{n!}$ the period will be 2(n)!
So we will check periodicity by replacing x by 2(n)! + x in function h(x),
So, $h\left( 2\left( n \right)!+x \right)=\cos \pi \left( \dfrac{2\left( n \right)!+x}{n!} \right)$
So, it can be simplified as,
$\cos \pi \left( 2+\dfrac{x}{\left( n \right)!} \right)$
So, it can be written as $\cos \left( 2\pi +\dfrac{\pi x}{n!} \right)$
Hence, we can use the identity $\cos \left( 2\pi +\theta \right)=\cos \theta $
So, $\cos \left( 2\pi +\dfrac{\pi x}{n!} \right)=\cos \dfrac{\pi x}{n!}$
Hence, h(2(n)! + x) = h(x)
Now as we know that f(x) = g(x) + h(x) then the period of f(x) is L.C.M of g(x) and h(x) period which is 2(n)! as 2(n-1)! Is a factor of 2(n)!.
So the correct option is ‘B’.

Note: Be careful while finding out periods of function g(x) and h(x) separately as they are places where students tend to make mistakes. Now while taking L.C.M analyzes both the numbers properly to avoid any calculation mistakes.