<p>If <em>θ</em> ∈ R, then maximum value of <br><br> \(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1 & 1 & 1+\cos\theta \end{vmatrix}\) is</p>
Step-by-Step Solution
Key Concept: Apply row/column operations to simplify the determinant, then recognize that the resulting expression involves a product of sine and cosine terms whose maximum can be found using calculus or AM-GM inequality.
<p><strong>Step 1:</strong> Apply row operations to simplify. Let R₂ → R₂ - R₁ and R₃ → R₃ - R₁:</p><p>Δ = |1 1 1|</p><p> |0 sin θ 0|</p><p> |0 0 cos θ|</p><p><strong>Step 2:</strong> This is an upper triangular matrix, so determinant equals product of diagonal elements:</p><p>Δ = 1 · sin θ · cos θ = sin θ cos θ</p><p><strong>Step 3:</strong> Use the double angle formula: sin θ cos θ = (1/2)sin(2θ)</p><p>Δ = (1/2)sin(2θ)</p><p><strong>Step 4:</strong> Since the maximum value of sin(2θ) is 1, the maximum value of Δ is:</p><p>Δ<sub>max</sub> = 1/2</p><p>∴ Answer: B</p>
Correct Answer: B