Question

Let $\Delta \left(x\right)=\left|\begin{array}{ccc}(x-2)& (x-1{)}^2& x^3\\ (x-1)& x^2& (x+1{)}^3\\ x& (x+1{)}^2& (x+2{)}^3\end{array}\right|$ Then the coefficient of x in $\Delta$(x) is $-k$ . Find k.

• A. -1
• B. 2
• C. -2
• D. 1

Answer 1

The correct option is C -2

Given $\Delta \left(x\right)=\left|\begin{array}{ccc}(x-2)& (x-1{)}^2& x^3\\ (x-1)& x^2& (x+1{)}^3\\ x& (x+1{)}^2& (x+2{)}^3\end{array}\right|$
Let $\Delta \left(x\right)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+.....$
$\left|\begin{array}{ccc}(x-2)& (x-1{)}^2& x^3\\ (x-1)& x^2& (x+1{)}^3\\ x& (x+1{)}^2& (x+2{)}^3\end{array}\right|=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+.....$ .....(1)
Put $x=0$ in (1)
$\Rightarrow \left|\begin{array}{ccc}-2& 1& 0\\ -1& 0& 1\\ 0& 1& 8\end{array}\right|=a_0$
$\Rightarrow a_0=-2(-1)-1(-8)=10$
So, $\Delta \left(x\right)=\left|\begin{array}{ccc}(x-2)& (x-1{)}^2& x^3\\ (x-1)& x^2& (x+1{)}^3\\ x& (x+1{)}^2& (x+2{)}^3\end{array}\right|=10+a_{1}x+a_2{x}^2+a_3{x}^3+.....$ ....(2)
Differentiating (2) w.r.t x, we get
$\left|\begin{array}{ccc}1& 2(x-1)& 3x^2\\ (x-1)& x^2& (x+1{)}^3\\ x& (x+1{)}^2& (x+2{)}^3\end{array}\right|+\left|\begin{array}{ccc}(x-2)& (x-1{)}^2& x^3\\ 1& 2x& 3(x+1{)}^2\\ x& (x+1{)}^2& (x+2{)}^3\end{array}\right|+\left|\begin{array}{ccc}(x-2)& (x-1{)}^2& x^3\\ (x-1)& x^2& (x+1{)}^3\\ 1& 2(x+1)& 3(x+2{)}^2\end{array}\right|=a_1+2a_{2}x+3a_3{x}^2+.....$
Put $x=0$ in above equation
$\Rightarrow \left|\begin{array}{ccc}1& -2& 0\\ -1& 0& 1\\ 0& 1& 8\end{array}\right|+\left|\begin{array}{ccc}-2& 1& 0\\ 1& 0& 3\\ 0& 1& 8\end{array}\right|+\left|\begin{array}{ccc}-2& 1& 0\\ -1& 0& 1\\ 1& 2& 12\end{array}\right|=a_1$
$\Rightarrow a_1=-17-2+17=-2$