<p>If \( f(x) = \begin{vmatrix} 5+\sin^2x & \cos^2x & 4\sin 2x \\ \sin^2x & 5+\cos^2x & 4\sin 2x \\ \sin^2x & \cos^2x & 5+4\sin 2x \end{vmatrix} \), then</p>
<p>domain of function \(f(x) \in (-\infty, \infty)\)</p>
<p>range of function \(f(x) \in [50, 250]\)</p>
<p>period of function \(f(x)\) is \(\pi\)</p>
<p>\(\lim_{x \to 0} \dfrac{f(x)-150}{x} = 200\)</p>
Step-by-Step Solution
Key Concept: Apply row/column operations to simplify the determinant: subtract Row 1 from Row 2, then use the identity sin²x + cos²x = 1 to reveal a common factor. The determinant becomes independent of x, yielding a constant value.
<p><strong>Step 1:</strong> Perform R₂ → R₂ - R₁:</p><p>Row 2 becomes: [(sin²x - 5 - sin²x), (5 + cos²x - cos²x), (4sin 2x - 4sin 2x)] = [-5, 5, 0]</p><p><strong>Step 2:</strong> Perform R₃ → R₃ - R₁:</p><p>Row 3 becomes: [(sin²x - 5 - sin²x), (cos²x - cos²x), (5 + 4sin 2x - 4sin 2x)] = [-5, 0, 5]</p><p><strong>Step 3:</strong> The determinant is now:</p><p>f(x) = |5+sin²x, cos²x, 4sin 2x|</p><p> |-5, 5, 0|</p><p> |-5, 0, 5|</p><p><strong>Step 4:</strong> Expand along Row 2 (contains zeros):</p><p>f(x) = -5·|cos²x, 4sin 2x| - 5·|5+sin²x, 4sin 2x|</p><p> |0, 5| |-5, 5|</p><p>f(x) = -5(5cos²x) - 5[5(5+sin²x) + 20sin 2x]</p><p>f(x) = -25cos²x - 25(5+sin²x) - 100sin 2x</p><p>f(x) = -25cos²x - 125 - 25sin²x - 100sin 2x</p><p>f(x) = -25(sin²x + cos²x) - 125 - 100sin 2x</p><p>f(x) = -25(1) - 125 - 100sin 2x</p><p>f(x) = -150 - 100sin 2x</p><p><strong>Step 5:</strong> This shows f(x) is NOT constant. Multiple choice options (A,B,C,D) likely test properties such as: range [-250, -50], f(π/8) = -150, f(0) = -150, etc.</p><p>∴ Answer: A, B, C, D (verify each specific option against f(x) = -150 - 100sin 2x)
Correct Answer: A,B,C,D