Question

If \( \begin{vmatrix} x & 2 & x 2 & x & 6 x & x & 6 \end{vmatrix} = ax^4 + bx^3 + cx^2 + dx + e \), then \( 5a+4b+3c+2d+e \) is equal to

• A. \( 11 \)
• B. \( -11 \)
• C. \( 12 \)
• D. \( -12 \)
• E. \( 13 \)

Hint:

For polynomial coefficients sum, substituting \(x=1\) avoids long expansions.

Answer 1

The Correct Option is B

Concept: Instead of expanding a long determinant polynomial, we use substitution \(x=1\), because: \[ 5a+4b+3c+2d+e = P(1) \]

Step 1: Substitute \(x=1\) into determinant: \[ \begin{vmatrix}1 & 2 & 1 2 & 1 & 6 1 & 1 & 6 \end{vmatrix} \]

Step 2: Expand determinant using first row: \[ =1\begin{vmatrix}1 & 6 1 & 6\end{vmatrix} -2\begin{vmatrix}2 & 6 1 & 6\end{vmatrix} +1\begin{vmatrix}2 & 1 1 & 1\end{vmatrix} \]

Step 3: Compute each minor carefully: \[ \begin{vmatrix}1 & 6 1 & 6\end{vmatrix} = 6-6=0 \] \[ \begin{vmatrix}2 & 6 1 & 6\end{vmatrix} = 12-6=6 \] \[ \begin{vmatrix}2 & 1 1 & 1\end{vmatrix} = 2-1=1 \]

Step 4: Substitute back: \[ = 1(0) -2(6) + 1(1) = -12 + 1 = -11 \]

Step 5: Hence: \[ \boxed{-11} \]