If f(x)= vmatrix x3 & 2x2+1 & 1+3x \ 3x2+2 & 2x & x3+6 \ x3-x & 4 & x2-2 vmatrix for all x R, then 2f(0)+f'(0) is equal to
- 24
- 18
- 42
- 48
Answer
(3) 42
Solution
f(0)= vmatrix 0 & 1 & 1 \ 2 & 0 & 6 \ 0 & 4 & -2 vmatrix=-1(-4-0)+1(8-0)=12
f'(x) = sum of three determinants, differentiating one row at a time; at x=0:
f'(0)= vmatrix 0 & 0 & 3 \ 2 & 0 & 6 \ 0 & 4 & -2 vmatrix+ vmatrix 0 & 1 & 1 \ 0 & 2 & 0 \ 0 & 4 & -2 vmatrix
+ vmatrix 0 & 1 & 1 \ 2 & 0 & 6 \ -1 & 0 & 0 vmatrix
=3(8)+0+(-6)=18 ( second determinant has a zero column)
2f(0)+f'(0)=24+18=42
Answer: (3)