Question

Let $\left|\begin{array}{ccc}x& 2& x\\ x^2& x& 6\\ x& x& 6\end{array}\right|=a{x}^4+b{x}^3+c{x}^2+dx+e$

Then, the value of $5a+4b+3c+2d+e$ is equal to
(a) 0
(b) − 16
(c) 16
(d) none of these

Answer 1

(d) none of these

$$\begin{gathered} ∆=\left|x2x \\ {x}^{2}x6 \\ xx6\right| \\ =x\left|x6 \\ x6\right|-x^2\left|2x \\ x6\right|+x\left|2x \\ x6\right|\left[\mathrm{Expanding}\mathrm{along}{C}_1\right] \\ =0-x^2\left(12-x^2\right)+x\left(12-x^2\right) \\ =x^4-12x^2+12x-x^3 \\ ∆=a{x}^4+b{x}^3+c{x}^2+dx+e\left[\mathrm{Given}\right] \\ \Rightarrow x^4-12x^2+12x-x^3=a{x}^4+b{x}^3+c{x}^2+dx+e \\ \Rightarrow a=1,b=-1,c=-12,d=12,e=0 \\ \mathrm{Thus}, \\ 5a+4b+3c+2d+e=5-4-36+24+0=-11 \end{gathered}$$