The maximum value of
f(x) = | sin ²x, 1 + cos ²x, cos 2x; 1 + sin ²x, cos ²x, cos 2x; sin ²x, cos ²x, sin 2x |,x ∈ R is :

Solution

Maths sir · 4 steps

Subtract the third row from the first two, and the big determinant becomes a simple cos 2x - 2 sin 2x.

1. Clear the rows. Apply R₁ → R₁ - R₃ and R₂ → R₂ - R₃. Row operations like these do not change the value of the determinant, and they turn the first two rows into mostly 0s and 1s.

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

2. Expand along row 1. Write k = cos 2x - sin 2x. Then f(x) = -1(sin 2x - k sin² x) + k cos² x = -sin 2x + k(sin² x + cos² x). Since sin² x + cos² x = 1, this gives f(x) = cos 2x - 2 sin 2x.
3. Find the maximum. An expression a cos θ + b sin θ always lies between -√(a²+b²) and √(a²+b²). Here a = 1 and b = -2, so the maximum is √(1 + 4) = √5.
4. So. The maximum value of f(x) is √5, which is option (a).

The idea to keep. When rows share most of their entries, subtract one row from the others before expanding. And a cos θ + b sin θ always has maximum √(a²+b²).

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