Find the period of `f(x)=sinx+tanx/2+sinx/(2^2)+t a n x/(2^3)+....+sinx/(2^(n-1))+tanx/(2^n)`

Options

A. `2pi `
B. `2^(n-1)pi `
C. `2^(n)pi`
D. `n pi `

Answer

C

Explanation

We have, `f(x)=sum_(r=1)^(n)(sin""(x)/(2^(r-1))+tan""(x)/(2^(r)))`
`implies f(x)=sum_(r=1)^(n)f_(r)(x), where" " f_(r)(x)= sin""(x)/(2^(r-1))+tan""(x)/(2^(r))`
since each of ` sin""(x)/(2^(r-1)) and (x)/(2^(r ))` are periodic functions with period `2^(r)pi`. Therefore, `f_(r)(x)` is periodic with period ` 2^(r) pi`.
But , `f(x)=sum_(r=1)^(n)f_(r)(x)` .
Therefore, , f(x) is periodic with period T given by
T=LCM of `(2pi, 2^(2)pi, ....,2^(n)pi)=2^(n)pi`