Question

If f(x) = \(\begin{vmatrix} x^3-x & a+x & b+x \\ x-a & x^2-x & c+x \\ x-b & x-c & 0 \\ \end{vmatrix}\) then

• A. f(1) = 0
• B. f(2) = 0
• C. f(0) = 0
• D. f(-1) = 0

Answer 1

The Correct Option is C

We are given \(f(x) = \begin{vmatrix} x^3-x & a+x & b+x \\ x-a & x^2-x & c+x \\ x-b & x-c & 0 \end{vmatrix}\) and need to determine for which value of x, \(f(x) = 0\).

Let's try substituting x = 0:

\(f(0) = \begin{vmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{vmatrix}\)

Expanding the determinant along the first row:

\(f(0) = 0 \cdot \begin{vmatrix} 0 & c \\ -c & 0 \end{vmatrix} - a \cdot \begin{vmatrix} -a & c \\ -b & 0 \end{vmatrix} + b \cdot \begin{vmatrix} -a & 0 \\ -b & -c \end{vmatrix}\)

\(f(0) = 0 - a(0 - (-bc)) + b(ac - 0)\)

\(f(0) = -abc + abc = 0\)

Therefore, \(f(0) = 0\).

Thus, the correct option is (C) f(0) = 0.