Let `f(x)=cos(a_1+x)+1/2cos(a_2+x)+1/(2^2)cos(a_3+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 (a) `pi` (b) `2pi` (c) `3pi` (d) `pi/2`

Options

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

Explanation

To solve the problem, we start with the function given:

\[ f(x) = \cos(a_1 + x) + \frac{1}{2} \cos(a_2 + x) + \frac{1}{2^2} \cos(a_3 + x) + \ldots + \frac{1}{2^{n-1}} \cos(a_n + x) \]

We need to find the absolute difference \( |x_2 - x_1| \) given that \( f(x_1) = f(x_2) = 0 \).

### Step-by-Step Solution:

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