<p>If \(y = \sin mx\), then the value of the determinant \[\begin{vmatrix} y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8 \end{vmatrix}\] where \(y_n = \dfrac{d^n y}{dx^n}\) is</p>
Step-by-Step Solution
Key Concept: Recognize that successive derivatives of y = sin(mx) follow a cyclic pattern repeating every 4 derivatives (with factor m^n), which makes all rows linearly dependent, resulting in a zero determinant.
<p><strong>Step 1:</strong> Find successive derivatives of y = sin(mx).</p><p>y₀ = sin(mx)</p><p>y₁ = m·cos(mx) = m·sin(mx + π/2)</p><p>y₂ = -m²·sin(mx) = m²·sin(mx + π)</p><p>y₃ = -m³·cos(mx) = m³·sin(mx + 3π/2)</p><p>y₄ = m⁴·sin(mx) = m⁴·sin(mx + 2π)</p><p><strong>Step 2:</strong> Observe the cyclic pattern. Each derivative equals m^n times sin(mx + nπ/2), which repeats with period 4.</p><p><strong>Step 3:</strong> Notice that y₄ = m⁴·y₀, y₅ = m⁴·y₁, y₆ = m⁴·y₂, y₇ = m⁴·y₃, y₈ = m⁴·y₄.</p><p><strong>Step 4:</strong> Row 3 = m⁴(Row 1). This means rows are linearly dependent.</p><p><strong>Step 5:</strong> When rows are linearly dependent, the determinant equals zero.</p><p>∴ Answer: D (which is 0)</p>
Correct Answer: D