Question

If $f(x) = \begin{vmatrix} x - 3 & 2x^2 - 18 & 2x^3 - 81 \\ x - 5 & 2x^2 - 50 & 4x^2 - 500 \\ 1 & 2 & 3 \end{vmatrix}$, then $f(1) \cdot f(3) \cdot f(5) + f(5) \cdot f(1)$ is:

• A. 1
• B. 0
• C. 2
• D. None of these

Answer 1

The Correct Option is B

1. Understand the problem:

Given the determinant function \( f(x) \), we need to evaluate \( f(1) \cdot f(3) + f(3) \cdot f(5) + f(5) \cdot f(1) \).

2. Compute \( f(1) \), \( f(3) \), and \( f(5) \):

Substitute \( x = 1, 3, 5 \) into the determinant:

For \( x = 1 \):

\[ f(1) = \begin{vmatrix} -2 & -16 & -79 \\ -4 & -48 & -496 \\ 1 & 2 & 3 \end{vmatrix} \]

For \( x = 3 \):

\[ f(3) = \begin{vmatrix} 0 & 0 & -27 \\ -2 & -32 & -392 \\ 1 & 2 & 3 \end{vmatrix} \]

For \( x = 5 \):

\[ f(5) = \begin{vmatrix} 2 & 0 & -250 \\ 0 & 0 & 0 \\ 1 & 2 & 3 \end{vmatrix} \]

3. Evaluate the determinants:

Using properties of determinants:

- \( f(3) = 0 \) (first row is zero).

- \( f(5) = 0 \) (second row is zero).

- For \( f(1) \), expand along the third row:

\[ f(1) = 1 \cdot (-16 \cdot -496 - (-79) \cdot -48) - 2 \cdot (-2 \cdot -496 - (-79) \cdot -4) + 3 \cdot (-2 \cdot -48 - (-16) \cdot -4) = \text{Non-zero value} \]

4. Compute the required expression:

Since \( f(3) = f(5) = 0 \), the expression simplifies to:

\[ f(1) \cdot 0 + 0 \cdot 0 + 0 \cdot f(1) = 0 \]

Correct Answer: (B) 0

Answer 2

The determinant simplifies to $f(x)$. Compute $f(1)$, $f(3)$, and $f(5)$ explicitly. First, evaluate $f(1)$: \[ f(1) = \begin{vmatrix} 1 - 3 & 2(1)^2 - 18 & 2(1)^3 - 81 \\ 1 - 5 & 2(1)^2 - 50 & 4(1)^2 - 500 \\ 1 & 2 & 3 \end{vmatrix} = \begin{vmatrix} -2 & -16 & -79 \\ -4 & -48 & -496 \\ 1 & 2 & 3 \end{vmatrix} \] Expanding this determinant, we find $f(1) = 0$. Similarly, evaluating $f(3)$ and $f(5)$ will also result in $f(3) = 0$ and $f(5) = 0$. Substituting into the given expression: \[ f(1) \cdot f(3) \cdot f(5) + f(5) \cdot f(1) = 0 \cdot 0 \cdot 0 + 0 \cdot 0 = 0 \] Hence, the correct answer is 0.