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:

Updated On: Apr 8, 2025
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The Correct Option is B

Approach Solution - 1

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

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Approach Solution -2

Let the given determinant be $$ f(x) = \begin{vmatrix} x-3 & 2x^2 - 18 & 2x^3 - 81 \\ x-5 & 2x^2 - 50 & 4x^3 - 500 \\ 1 & 2 & 3 \end{vmatrix} $$ We observe that $2x^2 - 18 = 2(x^2 - 9) = 2(x-3)(x+3)$ and $2x^3 - 81 = 2(x^3 - 3^4) = 2(x-3)(x^2 + 3x + 9)$. Then $$ f(3) = \begin{vmatrix} 0 & 0 & 0 \\ x-5 & 2x^2 - 50 & 4x^3 - 500 \\ 1 & 2 & 3 \end{vmatrix} = 0 $$ Since $f(3) = 0$, the expression $f(1)f(3) + f(3)f(5) + f(5)f(1)$ simplifies to $f(5)f(1)$. Now let us compute $f(1)$ and $f(5)$: $$ f(1) = \begin{vmatrix} -2 & -16 & -80 \\ -4 & -48 & -496 \\ 1 & 2 & 3 \end{vmatrix} $$ $$ f(5) = \begin{vmatrix} 2 & 32 & 170 \\ 0 & 0 & 0 \\ 1 & 2 & 3 \end{vmatrix} = 0 $$ Since $f(5) = 0$, the expression becomes 0. 

Therefore, $f(1)f(3) + f(3)f(5) + f(5)f(1) = 0$.

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