Let `f(x)=cos(a_1+x)+1/2cos(a_2+x)+1/(2^2)cos(a_1+x)++1/(2^(n-1))cos(a_n+x)` where `a)1,a_2 a_n in Rdot` If `f(x_1)=f(x_2)=0,t h e n|x_2-x_1|` may be equal to `pi` (b) `2pi` (c) `3pi` (d) `pi/2`

Options

A. `pi`
B. `2pi`
C. `3pi`
D. `pi//2`

Answer

A, B, C

Explanation

`f(x)=(cos a_(1)+(cos a_(2))/2+...+(cos a_(n))/2^(n-1))cos x-((sin a_(1))1+(sin a_(2))/2+...+(sin a_(n))/2^(n-1)) sin x`
`rArr f(x)=A cos x-B sin x`
Now `f(x_(1))=f(x_(2))=0`
`rArr {:(A cos x_(1)- B sin x_(1)=0),(A cos x_(2)-B sin x_(2)=0):}}`
`rArr tan x_(1) = tan x_(2)`
`rArr x_(1)=n pi +x_(2) rArr x_(1)-x_(2)=n pi`