If f(x)=|2 cos ⁴ x, 2 sin ⁴ x, 3+sin ² 2 x; 3+2 cos ⁴ x, 2 sin ⁴ x, sin ² 2 x; 2 cos ⁴ x, 3+2 sin ⁴ x, sin ² 2 x|, then 1/5 f′(0)= is equal to :

Solution

Maths sir · 4 steps

Add all three columns first: every row sums to 5, and the whole determinant turns out to be a constant.

1. Name the entries. Let a = 2 cos⁴ x, b = 2 sin⁴ x and c = sin² 2x. The rows are (a, b, 3+c), (3+a, b, c) and (a, 3+b, c), so each row adds up to a+b+c+3.

   a+b+c = 2 cos⁴ x + 2 sin⁴ x + 4 sin² x cos² x = 2(cos² x + sin² x)² = 2

2. Column operation. Apply C₁ → C₁ + C₂ + C₃. Every entry of C₁ becomes 2 + 3 = 5, so take 5 out: f(x) = 5|1, b, 3+c; 1, b, c; 1, 3+b, c|.
3. Row operations. Apply R₁ → R₁ - R₂ and R₃ → R₃ - R₂. The rows become (0, 0, 3), (1, b, c) and (0, 3, 0). Expanding along R₁ gives 3 × (1 × 3 - b × 0) = 9.
4. So. f(x) = 5 × 9 = 45 for every x. A constant has zero derivative, so f'(0) = 0 and 1/5f'(0) = 0, which is option (c).

The idea to keep. Before you differentiate a determinant, simplify it with row and column operations. Identities like sin² x + cos² x = 1 often make it a constant, and then the derivative is 0.

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