The function f(x) satisfies the equation` f(x + 1) + f(x-1) = sqrt3 f(x) `. then the period of f(x) is

Options

A. 2
B. 6
C. 12
D. 4

Answer

C

Explanation

We have,
`f(x+1)+f(x-1)=sqrt(3)` f(x) for all ` x in R `
Replacing x by x+1 and x-1 respectively, we get
`f(x+2)+f(x)=sqrt(3)f(x+1)" "`….(i)
and ` , f(x)+f(x-2)=sqrt(3)f(x-1)" " ` ……..(ii)
Adding (i) and (ii), we get
`f(x+2)+f(x-2)+2f(x)=sqrt(3) [f(x+1)+f(x-1)]`
`implies f(x+2)+f(x-2)+2f(x)=sqrt(3){sqrt(3)f(x)} " " [ :' f(x+1)+f(x-1)=sqrt(3)f(x)]`
`implies f(x+2)+f(x-2)=f(x)" "` ......(iii)
Replacing x by x+2 , we get
f(x+4) +f(x)=f(x+2)`" " `....(iv)
Adding (iii) and (iv) ,we get
`f(x+2)+f(x-2)+f(x+4)+f(x)=f(x)+f(x+2)`
`implies f(x+4)+f(x-2)=0` for all ` x in R `
Replacing x be x+2 , we get
`f(x+6)+f(x)=0` for all ` x in R `
`implies f(x+6)=-f(x)` for all ` x in R " "`....(v)
Replacing x by x+6 , we get
`f(x+12)=-f(x+6)` for all ` x in R `
`implies f(x+12)=f(x) ` for all ` x in R " " `[Using (v)]
Hence , f(x) is a periodic function with period. 12.