Let f(x) = <table><tr><td>|</td><td>1+sin^2 x</td><td>cos^2 x</td><td>4sin 2x</td><td>|</td></tr><tr><td>|</td><td>sin^2 x</td><td>1+cos^2 x</td><td>4sin 2x</td><td>|</td></tr><tr><td>|</td><td>sin^2 x</td><td>cos^2 x</td><td>1+4sin 2x</td><td>|</td></tr></table>, then the maximum value of f(x), is-
Step-by-Step Solution
Key Concept: Use row operations to simplify the determinant. Perform R1 -> R1 - R3 and R2 -> R2 - R3 to reduce the determinant to a simpler form, then expand.
Applying R1 -> R1 - R3 and R2 -> R2 - R3, we get f(x) = |(1, 0, -1), (0, 1, -1), (sin^2 x, cos^2 x, 1+4sin 2x)|. Expanding along the first row: 1(1+4sin 2x + cos^2 x) - 0 + (-1)(0 - sin^2 x) = 1 + 4sin 2x + cos^2 x + sin^2 x = 1 + 4sin 2x + 1 = 2 + 4sin 2x. The maximum value of sin 2x is 1, so the maximum value of f(x) is 2 + 4(1) = 6.
Correct Answer: 3