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

The given function is:

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

We need to determine the value of $ f(x) $ for specific values of $ x $.

Step 1: Evaluate the determinant when $ x = 0 $.

Substitute $ x = 0 $ into the matrix:

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

Now, calculate the determinant of this 3x3 matrix. Using cofactor expansion along the first row:

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

The 2x2 determinants are calculated as follows:

$$ \begin{vmatrix} 0 & c \\ -c & 0 \end{vmatrix} = 0 \cdot 0 - (-c \cdot c) = c^2 $$ $$ \begin{vmatrix} -a & c \\ -b & 0 \end{vmatrix} = (-a)(0) - (c)(-b) = bc $$ $$ \begin{vmatrix} -a & 0 \\ -b & -c \end{vmatrix} = (-a)(-c) - (0)(-b) = ac $$

Substitute these values back into the determinant expression:

$$ f(0) = 0 - a(bc) + b(ac) $$ $$ f(0) = -abc + abc = 0 $$

Answer: (C) $ f(0) = 0 $.

Answer 2

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.