If f(x)=|x³, 2 x²+1, 1+3 x; 3 x²+2, 2 x, x³+6; x³-x, 4, x²-2| for all x ∈ ℝ , then 2 f(0)+f′(0) is equal to

Solution

Maths sir · 3 steps

Get f(0) by putting x = 0 directly, then get f'(0) by differentiating one row at a time and adding.

1. Find f(0). At x = 0 the rows are (0, 1, 1), (2, 0, 6) and (0, 4, -2). Expanding along the first row gives f(0) = 0 - 1(-4 - 0) + 1(8 - 0) = 12.
2. Rule for f'(x). The derivative of a determinant is the sum of three determinants. In each one, a single row is differentiated and the other two rows are left as they are. Put x = 0 in each of them.

   f'(0) = |0, 0, 3; 2, 0, 6; 0, 4, -2| + |0, 1, 1; 0, 2, 0; 0, 4, -2| + |0, 1, 1; 2, 0, 6; -1, 0, 0|

   Row differentiated	New row at x = 0	Value of determinant
   Row 1: (3x², 4x, 3)	(0, 0, 3)	3(2 × 4 - 0) = 24
   Row 2: (6x, 2, 3x²)	(0, 2, 0)	first column is all zero, so 0
   Row 3: (3x² - 1, 0, 2x)	(-1, 0, 0)	-1(1 × 6 - 1 × 0) = -6

3. So. f'(0) = 24 + 0 - 6 = 18, and 2f(0) + f'(0) = 24 + 18 = 42. The answer is (c).

The idea to keep. To differentiate a determinant, differentiate one row at a time, keep the other rows unchanged, and add the results. Put in the value of x before you expand, so the arithmetic stays small.

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