Let m and M be respectively the minimum and maximum values of
| cos ²x, 1 + sin ²x, sin 2x; 1 + cos ²x, sin ²x, sin 2x; cos ²x, sin ²x, 1 + sin 2x |
Then the ordered pair (m, M) is equal to :

Solution

Maths sir · 4 steps

Subtract the third row from the other two, and the determinant collapses to -(2 + sin 2x).

1. Clear the rows. Apply R₁ → R₁ - R₃ and R₂ → R₂ - R₃. The value does not change, and all the trigonometric terms in the first two rows cancel.

   Δ = |0, 1, -1; 1, 0, -1; cos² x, sin² x, 1 + sin 2x|

2. Expand along row one. Δ = -1[(1 + sin 2x) + cos² x] - 1[sin² x] = -(1 + cos² x + sin² x + sin 2x). Since cos² x + sin² x = 1, this gives Δ = -(2 + sin 2x).
3. Find the range. sin 2x runs from -1 to 1, so 2 + sin 2x runs from 1 to 3. The minus sign flips this, so Δ runs from -3 to -1.
4. So. m = -3 (when sin 2x = 1) and M = -1 (when sin 2x = -1). The ordered pair is (-3, -1), option (a).

The idea to keep. When every row shares the same messy entries, subtract one row from the others to kill them. Reduce the determinant to a single trig function first, and only then read off its range.

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