Let f(x) = | sin ²x, - 2 + cos ²x, cos 2x; 2 + sin ²x, cos ²x, cos 2x; sin ²x, cos ²x, 1 + cos 2x |,x ∈ [0,π ]. Then the maximum value of f(x) is equal to ______________.

Solution

Maths sir · 3 steps

Subtract the third row from the other two and the trig terms vanish, leaving zeros to expand along.

1. Clear the rows. Apply R₁ → R₁ - R₃ and R₂ → R₂ - R₃. Row operations like these do not change the value of a determinant. The first row becomes (0, -2, -1) and the second becomes (2, 0, -1), with no trig left in either.

   f(x) = |0, -2, -1; 2, 0, -1; sin² x, cos² x, 1 + cos 2x|

2. Expand along row 1. f(x) = 2[2(1 + cos 2x) + sin² x] - 2 cos² x = 4 + 4 cos 2x - 2(cos² x - sin² x). Since cos² x - sin² x = cos 2x, this gives f(x) = 4 + 2 cos 2x.
3. So. On [0, π], cos 2x reaches its largest value 1 at x = 0 and at x = π. The maximum is 4 + 2 × 1 = 6.

The idea to keep. When two rows share the same trig entries, subtract one from the other first. The variable parts cancel, you get constant rows with zeros, and expanding the determinant becomes short.

Answer from the NTA final answer key. Solution by PrepWiser.