<p>Let \(f(x) = a_1 \cos(\alpha_1 + x) + a_2 \cos(\alpha_2 + x) + \ldots + a_n \cos(\alpha_n + x)\). If \(f(x)\) vanishes for \(x = 0\) and \(x = x_1\) (where \(x_1 \neq k\pi,\ k \in \mathbb{Z}\)), then</p>
<p>(a) \(a_1 \cos\alpha_1 + a_2 \cos\alpha_2 + \ldots + a_n \cos\alpha_n = 0\)</p>
<p>(b) \(a_1 \sin\alpha_1 + a_2 \sin\alpha_2 + \ldots + a_n \sin\alpha_n = 0\)</p>
<p>(c) \(f(x) = 0\) has only two solutions \(0,\ x_1\)</p>
<p>(d) \(f(x)\) is identically \(0\ \forall\ x\)</p>
Step-by-Step Solution
Key Concept: A linear combination of cosines that vanishes at two distinct non-trivial points must satisfy severe constraints on its coefficients. Use the condition f(0) = 0 and f(x₁) = 0 to derive relationships between the amplitudes and phases, recognizing that for non-special values of x₁, only degenerate cases allow this.
<p><strong>Step 1:</strong> Given f(x) = Σ aᵢ cos(αᵢ + x), write f(0) = 0:<br/>Σ aᵢ cos(αᵢ) = 0 ... (1)</p><p><strong>Step 2:</strong> Also f(x₁) = 0 gives:<br/>Σ aᵢ cos(αᵢ + x₁) = 0 ... (2)</p><p><strong>Step 3:</strong> Expand equation (2) using cos(αᵢ + x₁) = cos(αᵢ)cos(x₁) - sin(αᵢ)sin(x₁):<br/>cos(x₁)·Σ aᵢ cos(αᵢ) - sin(x₁)·Σ aᵢ sin(αᵢ) = 0<br/>Using (1): -sin(x₁)·Σ aᵢ sin(αᵢ) = 0</p><p><strong>Step 4:</strong> Since x₁ ≠ kπ, we have sin(x₁) ≠ 0, therefore:<br/>Σ aᵢ sin(αᵢ) = 0 ... (3)</p><p><strong>Step 5:</strong> Equations (1) and (3) mean the vector (a₁cos α₁, a₁sin α₁) + ... + (aₙcos αₙ, aₙsin αₙ) = (0,0). Combined with the constraint structure, this severely limits the form of f(x).</p><p><strong>Step 6:</strong> The key result is that f(x) must be identically zero OR the system forces specific relationships. This eliminates many general statements, making A, B, D the statements that hold universally for the given conditions.</p><p>∴ Answer: A,B,D</p>
Correct Answer: A,B,D