Let f(x) = cos(α₁ + x) + (1/3)cos(α₂ + x) + (1/3²)cos(α₃ + x) + ... + (1/3^{n-1})cos(αₙ + x), where α₁, α₂, ..., αₙ ∈ ℝ. If f(x₁) = f(x₂) = 0, then |x₁ - x₂| can be:
A. π
B. 5π/2
C. 3π/2
D. 4π
Answer (2)
The function given has terms that oscillate with a periodicity related to cosine functions. The simplest valid interval between zero crossings, making sense from the periodic nature, is |x₁ - x₂| = π. Thus the chosen option is A. π.
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To solve this problem, the given function f ( x ) is a finite series
f ( x ) = cos ( α 1 + x ) + 3 1 cos ( α 2 + x ) + 3 2 1 cos ( α 3 + x ) + … + 3 n − 1 1 cos ( α n + x )
We are given that f ( x 1 ) = f ( x 2 ) = 0 . This means the function has the same value (zero) at two different points x 1 and x 2 . To find ∣ x 1 − x 2 ∣ , we need to determine the difference between these two points, which is equivalent to finding the period or half-period of the function.
Since the function consists of cosine terms, it is periodic with the period of the base cosine function being 2 π . The coefficients 3 k 1 in the series do not affect the period of cosine but scale the amplitude of each term.
Given that f ( x 1 ) = f ( x 2 ) = 0 , a possible value for ∣ x 1 − x 2 ∣ could be half of the common period of the cosine terms. For the cosine function cos ( θ ) , this value is typically π , since the cosine function achieves every value twice in a full period of 2 π .
Based on this reasoning, the interval between two consecutive zeros (or equal function values that are zero in this context) could logically be π , which corresponds with one of the multiple-choice options given.
Thus, the correct answer is π . We choose the option π , as this represents a possible period difference between points x 1 and x 2 where the function zeros occur.
