Let f(x) = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mn>1</mn><mo>+</mo><msup><mi>sin</mi><mn>2</mn></msup><mi>x</mi></mtd><mtd><msup><mi>cos</mi><mn>2</mn></msup><mi>x</mi></mtd><mtd><mn>4</mn><mi>sin</mi><mn>2</mn><mi>x</mi></mtd></mtr><mtr><mtd><msup><mi>sin</mi><mn>2</mn></msup><mi>x</mi></mtd><mtd><mn>1</mn><mo>+</mo><msup><mi>cos</mi><mn>2</mn></msup><mi>x</mi></mtd><mtd><mn>4</mn><mi>sin</mi><mn>2</mn><mi>x</mi></mtd></mtr><mtr><mtd><msup><mi>sin</mi><mn>2</mn></msup><mi>x</mi></mtd><mtd><msup><mi>cos</mi><mn>2</mn></msup><mi>x</mi></mtd><mtd><mn>1</mn><mo>+</mo><mn>4</mn><mi>sin</mi><mn>2</mn><mi>x</mi></mtd></mtr></mtable></mfenced></math>, then the maximum value of f(x), is-
Step-by-Step Solution
Key Concept: Apply row operations R1 -> R1 - R3 and R2 -> R2 - R3 to simplify the determinant. The determinant reduces to 1 + 4sin(2x). The maximum value of sin(2x) is 1, so the maximum value is 1 + 4(1) = 5. Wait, checking the determinant expansion: R1-R3 gives (1, 0, -1) and R2-R3 gives (0, 1, -1). Expanding along R1 gives 1(1+4sin2x - 4sin2x) - 0 + (-1)(sin^2x - sin^2x(1+4sin2x)) = 1 + 4sin^2x. Let's re-evaluate. Actually, R1 -> R1-R3 and R2 -> R2-R3 gives |(1, 0, -1), (0, 1, -1), (sin^2x, cos^2x, 1+4sin2x)|. Expanding: 1(1+4sin2x + cos^2x) - 1(0 - sin^2x) = 1 + 4sin2x + cos^2x + sin^2x = 2 + 4sin2x. Max value is 2 + 4(1) = 6.
Applying R1 -> R1 - R3 and R2 -> R2 - R3, the determinant becomes |(1, 0, -1), (0, 1, -1), (sin^2x, cos^2x, 1+4sin2x)|. Expanding along R1: 1(1+4sin2x + cos^2x) - 1(0 - sin^2x) = 1 + 4sin2x + cos^2x + sin^2x = 2 + 4sin2x. Since the maximum value of sin2x is 1, the maximum value of f(x) is 2 + 4(1) = 6.
Correct Answer: 3