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Question: Period of \(f\left( x \right)=\sin \dfrac{\pi x}{\left( n-1 \right)!}+\cos \dfrac{\pi x}{n!}\) is ...

Period of f(x)=sin⁡πx(n−1)!+cos⁡πxn!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

Explanation

Solution

Hint: First represent f(x) as g(x) + h(x) where g(x) is sin⁡πx(n−1)!\sin \dfrac{\pi x}{\left( n-1 \right)!} and h(x) is cos⁡πxn!\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(x)=sin⁡πx(n−1)!+cos⁡πxn!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⁡πx(n−1)!\sin \dfrac{\pi x}{\left( n-1 \right)!} as function g(x) and cos⁡πxn!\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⁡πx(n−1)!\sin \dfrac{\pi x}{\left( n-1 \right)!} by using formula that if there is a function is in form of sin⁡(kx)\sin \left( kx \right) where K is a constant then the period of sin⁡(kx)\sin \left( kx \right) is 2πk\dfrac{2\pi }{k} So in the function g(x) the value of K is π(n−1)!\dfrac{\pi }{\left( n-1 \right)!}. So the period of g(x) or sin⁡(πx(n−1)!)\sin \left( \dfrac{\pi x}{\left( n-1 \right)!} \right)will be 2ππ(n−1)!\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{aligned} & 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{aligned}
On simplifying we get,
=sin⁡π(2+x(n−1)!)=\sin \pi \left( 2+\dfrac{x}{\left( n-1 \right)!} \right)
This is equal to sin⁡(2π+πx(n−1)!)\sin \left( 2\pi +\dfrac{\pi x}{\left( n-1 \right)!} \right)
So, here we will use fact that sin⁡(2π+θ)=sin⁡θ\sin \left( 2\pi +\theta \right)=\sin \theta
So, sin⁡(2π+πx(n−1)!)=sin⁡(πx(n−1)!)\sin \left( 2\pi +\dfrac{\pi x}{\left( n-1 \right)!} \right)=\sin \left( \dfrac{\pi x}{\left( n-1 \right)!} \right)
Here g(2(n−1)!+x)=g(x)g\left( 2\left( n-1 \right)!+x \right)=g\left( x \right).
Now let’s find periodicity of h(x) which is cos⁡πxn!\cos \dfrac{\pi x}{n!} which can be find out using formula if there is function cos⁡(kx)\cos \left( kx \right) exists then its period will be ′2πk′'\dfrac{2\pi }{k}' if K is a constant.
So for function h(x) which is equal to cos⁡πxn!\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(2(n)!+x)=cos⁡π(2(n)!+xn!)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⁡π(2+x(n)!)\cos \pi \left( 2+\dfrac{x}{\left( n \right)!} \right)
So, it can be written as cos⁡(2π+πxn!)\cos \left( 2\pi +\dfrac{\pi x}{n!} \right)
Hence, we can use the identity cos⁡(2π+θ)=cos⁡θ\cos \left( 2\pi +\theta \right)=\cos \theta
So, cos⁡(2π+πxn!)=cos⁡πxn!\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.