Question

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

• A. 11
• B. -11
• C. 12
• D. -12
• E. 13

Answer 1

Answer: (B)

Given that, $\begin{vmatrix}x & 2 & x \\ x^{2} & x & 6 \\ x & x & 6\end{vmatrix}=a x^{4}+b x^{3}+c x^{2}+d x +e$
$\Rightarrow x(6 x-6 x)-2\left(6 x^{2}-6 x\right)+x\left(x^{3}-x^{2}\right)$
$=a x^{4}+b x^{3}+c x^{2}+d x +e$
$\Rightarrow -12 x^{2}+12 x +x^{4}-x^{3}=a x^{4}+b x^{3}+c x^{2}+d x +e$
$\Rightarrow \, x^{4}-x^{3}-12 x^{2}+12 x=a x^{4}+b x^{3}+c x^{2}+d x +e$
On equating the coefficient of both sides, we get
$a=1, b=-1, c=-12, d=12, e=0$
$\therefore 5 a+4 b+3 c+2 d +e= 5 \times 1+4 \times(-1)+3(-12)$
$+2(12)+0$
$= 5-4-36+24$
$= 29-40=-11$